https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Scrabbler94&feedformat=atom AoPS Wiki - User contributions [en] 2021-07-31T18:34:45Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_8_Problems/Problem_3&diff=150547 2013 AMC 8 Problems/Problem 3 2021-03-29T18:47:29Z <p>Scrabbler94: more detailed solution</p> <hr /> <div>==Problem==<br /> What is the value of &lt;math&gt;4 \cdot (-1+2-3+4-5+6-7+\cdots+1000)&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ -10 \qquad \textbf{(B)}\ 0 \qquad \textbf{(C)}\ 1 \qquad \textbf{(D)}\ 500 \qquad \textbf{(E)}\ 2000&lt;/math&gt;<br /> <br /> ==Solution==<br /> We group the addends inside the parentheses two at a time: <br /> &lt;cmath&gt;<br /> \begin{align*}<br /> -1 + 2 - 3 + 4 - 5 + 6 - 7 + \ldots + 1000 &amp;= (-1 + 2) + (-3 + 4) + (-5 + 6) + \ldots + (-999 + 1000) \\<br /> &amp;= \underbrace{1+1+1+\ldots + 1}_{\text{500 1's}} \\<br /> &amp;= 500.<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> Then the desired answer is &lt;math&gt;4 \times 500 = \boxed{\textbf{(E)}\ 2000}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2013|num-b=2|num-a=4}}<br /> {{MAA Notice}}</div> Scrabbler94 https://artofproblemsolving.com/wiki/index.php?title=2021_AIME_I_Problems/Problem_14&diff=149127 2021 AIME I Problems/Problem 14 2021-03-12T01:23:17Z <p>Scrabbler94: </p> <hr /> <div>==Problem==<br /> For any positive integer &lt;math&gt;a, \sigma(a)&lt;/math&gt; denotes the sum of the positive integer divisors of &lt;math&gt;a&lt;/math&gt;. Let &lt;math&gt;n&lt;/math&gt; be the least positive integer such that &lt;math&gt;\sigma(a^n)-1&lt;/math&gt; is divisible by &lt;math&gt;2021&lt;/math&gt; for all positive integers &lt;math&gt;a&lt;/math&gt;. What is the sum of the prime factors in the prime factorization of &lt;math&gt;n&lt;/math&gt;?<br /> <br /> ==Solution==<br /> We first claim that &lt;math&gt;n&lt;/math&gt; must be divisible by 42. Since &lt;math&gt;\sigma(a^n)-1&lt;/math&gt; is divisible by 2021 for all positive integers &lt;math&gt;a&lt;/math&gt;, we can first consider the special case where &lt;math&gt;a \neq 0,1 \pmod{43}&lt;/math&gt;.<br /> <br /> Then &lt;math&gt;\sigma(a^n)-1 = \sum_{i=1}^n a^i = a\left(\frac{a^n - 1}{a-1}\right)&lt;/math&gt;. In order for this expression to be divisible by &lt;math&gt;2021=43\times 47&lt;/math&gt;, a necessary condition is &lt;math&gt;a^n - 1 \equiv 0 \pmod{43}&lt;/math&gt;. By [[Fermat's Little Theorem]], &lt;math&gt;a^{42} \equiv 1 \pmod{43}&lt;/math&gt;. Moreover, if &lt;math&gt;a&lt;/math&gt; is a primitive root modulo 43, then &lt;math&gt;\text{ord}_{43}(a) = 42&lt;/math&gt;, so &lt;math&gt;n&lt;/math&gt; must be divisible by 42.<br /> <br /> By similar reasoning, &lt;math&gt;n&lt;/math&gt; must be divisible by 46, by considering &lt;math&gt;a \neq 0,1 \pmod{47}&lt;/math&gt;.<br /> <br /> We next claim that &lt;math&gt;n&lt;/math&gt; must be divisible by 43 and 47. Consider the case &lt;math&gt;a=2022&lt;/math&gt;. Then &lt;math&gt;\sigma(a^n) \equiv n \pmod{2021}&lt;/math&gt;, so &lt;math&gt;\sigma(2022^n)-1&lt;/math&gt; is divisible by 2021 if and only if &lt;math&gt;n&lt;/math&gt; is divisible by 2021.<br /> <br /> Lastly, we claim that if &lt;math&gt;n = \text{lcm}(42, 46, 43, 47)&lt;/math&gt;, then &lt;math&gt;\sigma(a^n) - 1&lt;/math&gt; is divisible by 2021 for all positive integers &lt;math&gt;a&lt;/math&gt;. The claim is trivially true for &lt;math&gt;a=1&lt;/math&gt; so suppose &lt;math&gt;a&gt;1&lt;/math&gt;. Let &lt;math&gt;a = p_1^{e_1}\ldots p_k^{e_k}&lt;/math&gt; be the prime factorization of &lt;math&gt;a&lt;/math&gt;. Since &lt;math&gt;\sigma&lt;/math&gt; is [[multiplicative function|multiplicative]], we have<br /> &lt;cmath&gt;\sigma(a^n) - 1 = \prod_{i=1}^k \sigma(p_i^{e_in}) - 1.&lt;/cmath&gt;<br /> We can show that &lt;math&gt;\sigma(p_i^{e_in}) \equiv 1 \pmod{2021}&lt;/math&gt; for all primes &lt;math&gt;p_i&lt;/math&gt; and integers &lt;math&gt;e_i \ge 1&lt;/math&gt;, where<br /> &lt;cmath&gt;\sigma(p_i^{e_in}) = 1 + (p_i + p_i^2 + \ldots + p_i^n) + (p_i^{n+1} + \ldots + p_i^{2n}) + \ldots + (p_i^{n(e_i-1)+1} + \ldots + p_i^{e_in})&lt;/cmath&gt;<br /> where each expression in parentheses contains &lt;math&gt;n&lt;/math&gt; terms. It is easy to verify that if &lt;math&gt;p_i = 43&lt;/math&gt; or &lt;math&gt;p_i = 47&lt;/math&gt; then &lt;math&gt;\sigma(p_i^{e_in}) \equiv 1 \pmod{2021}&lt;/math&gt; for this choice of &lt;math&gt;n&lt;/math&gt;, so suppose &lt;math&gt;p_i \not\equiv 0 \pmod{43}&lt;/math&gt; and &lt;math&gt;p_i \not\equiv 0 \pmod{47}&lt;/math&gt;. Each expression in parentheses equals &lt;math&gt;\frac{p_i^n - 1}{p_i - 1}&lt;/math&gt; multiplied by some power of &lt;math&gt;p_i&lt;/math&gt;. If &lt;math&gt;p_i \neq 1 \pmod{43}&lt;/math&gt;, then FLT implies &lt;math&gt;p_i^n - 1 \equiv 0 \pmod{43}&lt;/math&gt;, and if &lt;math&gt;p_i \equiv 1 \pmod{43}&lt;/math&gt;, then &lt;math&gt;p_i + p_i^2 + \ldots + p_i^n \equiv 1 + 1 + \ldots + 1 \equiv 0 \pmod{43}&lt;/math&gt; (since &lt;math&gt;n&lt;/math&gt; is also a multiple of 43, by definition). Similarly, we can show &lt;math&gt;\sigma(p_i^{e_in}) \equiv 1 \pmod{47}&lt;/math&gt;, and a simple [[Chinese Remainder Theorem|CRT]] argument shows &lt;math&gt;\sigma(p_i^{e_in}) \equiv 1 \pmod{2021}&lt;/math&gt;. Then &lt;math&gt;\sigma(a^n) \equiv 1^k \equiv 1 \pmod{2021}&lt;/math&gt;.<br /> <br /> Then the prime factors of &lt;math&gt;n&lt;/math&gt; are &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;7&lt;/math&gt;, &lt;math&gt;23&lt;/math&gt;, &lt;math&gt;43&lt;/math&gt;, and &lt;math&gt;47&lt;/math&gt;, and the answer is &lt;math&gt;2+3+7+23+43+47 = \boxed{125}&lt;/math&gt;. ~scrabbler94<br /> ==See also==<br /> {{AIME box|year=2021|n=I|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Scrabbler94 https://artofproblemsolving.com/wiki/index.php?title=2021_AIME_I_Problems/Problem_14&diff=149126 2021 AIME I Problems/Problem 14 2021-03-12T01:22:26Z <p>Scrabbler94: add a solution</p> <hr /> <div>==Problem==<br /> For any positive integer &lt;math&gt;a, \sigma(a)&lt;/math&gt; denotes the sum of the positive integer divisors of &lt;math&gt;a&lt;/math&gt;. Let &lt;math&gt;n&lt;/math&gt; be the least positive integer such that &lt;math&gt;\sigma(a^n)-1&lt;/math&gt; is divisible by &lt;math&gt;2021&lt;/math&gt; for all positive integers &lt;math&gt;a&lt;/math&gt;. What is the sum of the prime factors in the prime factorization of &lt;math&gt;n&lt;/math&gt;?<br /> <br /> ==Solution==<br /> We first claim that &lt;math&gt;n&lt;/math&gt; must be divisible by 42. Since &lt;math&gt;\sigma(a^n)-1&lt;/math&gt; is divisible by 2021 for all positive integers &lt;math&gt;a&lt;/math&gt;, we can first consider the special case where &lt;math&gt;a \neq 0,1 \pmod{43}&lt;/math&gt;.<br /> <br /> Then &lt;math&gt;\sigma(a^n)-1 = \sum_{i=1}^n a^i = a\left(\frac{a^n - 1}{a-1}\right)&lt;/math&gt;. In order for this expression to be divisible by &lt;math&gt;2021=43\times 47&lt;/math&gt;, a necessary condition is &lt;math&gt;a^n - 1 \equiv 0 \pmod{43}&lt;/math&gt;. By [[Fermat's Little Theorem]], &lt;math&gt;a^{42} \equiv 1 \pmod{43}&lt;/math&gt;. Moreover, if &lt;math&gt;a&lt;/math&gt; is a primitive root modulo 43, then &lt;math&gt;\text{ord}_{43}(a) = 42&lt;/math&gt;, so &lt;math&gt;n&lt;/math&gt; must be divisible by 42.<br /> <br /> By similar reasoning, &lt;math&gt;n&lt;/math&gt; must be divisible by 46, by considering &lt;math&gt;a \neq 0,1 \pmod{47}&lt;/math&gt;.<br /> <br /> We next claim that &lt;math&gt;n&lt;/math&gt; must be divisible by 43 and 47. Consider the case &lt;math&gt;a=2022&lt;/math&gt;. Then &lt;math&gt;\sigma(a^n) \equiv n \pmod{2021}&lt;/math&gt;, so &lt;math&gt;\sigma(2022^n)-1&lt;/math&gt; is divisible by 2021 if and only if &lt;math&gt;n&lt;/math&gt; is divisible by 2021.<br /> <br /> Lastly, we claim that if &lt;math&gt;n = \text{lcm}(42, 46, 43, 47)&lt;/math&gt;, then &lt;math&gt;\sigma(a^n) - 1&lt;/math&gt; is divisible by 2021 for all positive integers &lt;math&gt;a&lt;/math&gt;. The claim is trivially true for &lt;math&gt;a=1&lt;/math&gt; so suppose &lt;math&gt;a&gt;1&lt;/math&gt;. Let &lt;math&gt;a = p_1^{e_1}\ldots p_k^{e_k}&lt;/math&gt; be the prime factorization of &lt;math&gt;a&lt;/math&gt;. Since &lt;math&gt;\sigma&lt;/math&gt; is multiplicative, we have<br /> &lt;cmath&gt;\sigma(a^n) - 1 = \prod_{i=1}^k \sigma(p_i^{e_in}) - 1.&lt;/cmath&gt;<br /> We can show that &lt;math&gt;\sigma(p_i^{e_in}) \equiv 1 \pmod{2021}&lt;/math&gt; for all primes &lt;math&gt;p_i&lt;/math&gt; and integers &lt;math&gt;e_i \ge 1&lt;/math&gt;, where<br /> &lt;cmath&gt;\sigma(p_i^{e_in}) = 1 + (p_i + p_i^2 + \ldots + p_i^n) + (p_i^{n+1} + \ldots + p_i^{2n}) + \ldots + (p_i^{n(e_i-1)+1} + \ldots + p_i^{e_in})&lt;/cmath&gt;<br /> where each expression in parentheses contains &lt;math&gt;n&lt;/math&gt; terms. It is easy to verify that if &lt;math&gt;p_i = 43&lt;/math&gt; or &lt;math&gt;p_i = 47&lt;/math&gt; then &lt;math&gt;\sigma(p_i^{e_in}) \equiv 1 \pmod{2021}&lt;/math&gt; for this choice of &lt;math&gt;n&lt;/math&gt;, so suppose &lt;math&gt;p_i \not\equiv 0 \pmod{43}&lt;/math&gt; and &lt;math&gt;p_i \not\equiv 0 \pmod{47}&lt;/math&gt;. Each expression in parentheses equals &lt;math&gt;\frac{p_i^n - 1}{p_i - 1}&lt;/math&gt; multiplied by some power of &lt;math&gt;p_i&lt;/math&gt;. If &lt;math&gt;p_i \neq 1 \pmod{43}&lt;/math&gt;, then FLT implies &lt;math&gt;p_i^n - 1 \equiv 0 \pmod{43}&lt;/math&gt;, and if &lt;math&gt;p_i \equiv 1 \pmod{43}&lt;/math&gt;, then &lt;math&gt;p_i + p_i^2 + \ldots + p_i^n \equiv 1 + 1 + \ldots + 1 \equiv 0 \pmod{43}&lt;/math&gt; (since &lt;math&gt;n&lt;/math&gt; is also a multiple of 43, by definition). Similarly, we can show &lt;math&gt;\sigma(p_i^{e_in}) \equiv 1 \pmod{47}&lt;/math&gt;, and a simple [[Chinese Remainder Theorem|CRT]] argument shows &lt;math&gt;\sigma(p_i^{e_in}) \equiv 1 \pmod{2021}&lt;/math&gt;. Then &lt;math&gt;\sigma(a^n) \equiv 1^k \equiv 1 \pmod{2021}&lt;/math&gt;.<br /> <br /> Then the prime factors of &lt;math&gt;n&lt;/math&gt; are &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;7&lt;/math&gt;, &lt;math&gt;23&lt;/math&gt;, &lt;math&gt;43&lt;/math&gt;, and &lt;math&gt;47&lt;/math&gt;, and the answer is &lt;math&gt;2+3+7+23+43+47 = \boxed{125}&lt;/math&gt;.<br /> ==See also==<br /> {{AIME box|year=2021|n=I|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Scrabbler94 https://artofproblemsolving.com/wiki/index.php?title=2021_AIME_I_Problems/Problem_3&diff=149110 2021 AIME I Problems/Problem 3 2021-03-11T23:46:32Z <p>Scrabbler94: more rigorous solution</p> <hr /> <div>==Problem==<br /> Find the number of positive integers less than &lt;math&gt;1000&lt;/math&gt; that can be expressed as the difference of two integral powers of &lt;math&gt;2.&lt;/math&gt;<br /> <br /> ==Solution==<br /> We want to find the number of positive integers &lt;math&gt;n&lt;1000&lt;/math&gt; which can be written in the form &lt;math&gt;n = 2^a - 2^b&lt;/math&gt; for some non-negative integers &lt;math&gt;a &gt; b \ge 0&lt;/math&gt; (note that if &lt;math&gt;a=b&lt;/math&gt;, then &lt;math&gt;2^a-2^b = 0&lt;/math&gt;). We first observe &lt;math&gt;a&lt;/math&gt; must be at most 10; if &lt;math&gt;a \ge 11&lt;/math&gt;, then &lt;math&gt;2^a - 2^b \ge 2^{10} &gt; 1000&lt;/math&gt;. As &lt;math&gt;2^{10} = 1024 \approx 1000&lt;/math&gt;, we can first choose two different numbers &lt;math&gt;a &gt; b&lt;/math&gt; from the set &lt;math&gt;\{0,1,2,\ldots,10\}&lt;/math&gt; in &lt;math&gt;\binom{10}{2}=55&lt;/math&gt; ways. This includes &lt;math&gt;(a,b) = (10,0)&lt;/math&gt;, &lt;math&gt;(10,1)&lt;/math&gt;, &lt;math&gt;(10,2)&lt;/math&gt;, &lt;math&gt;(10,3)&lt;/math&gt;, &lt;math&gt;(10,4)&lt;/math&gt; which are invalid as &lt;math&gt;2^a - 2^b &gt; 1000&lt;/math&gt; in this case. For all other choices &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;, the value of &lt;math&gt;2^a - 2^b&lt;/math&gt; is less than 1000.<br /> <br /> We claim that for all other choices of &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;, the values of &lt;math&gt;2^a - 2^b&lt;/math&gt; are pairwise distinct. More specifically, if &lt;math&gt;(a_1,b_1) \neq (a_2,b_2)&lt;/math&gt; where &lt;math&gt;10 \ge a_1 &gt; b_1 \ge 0&lt;/math&gt; and &lt;math&gt;10 \ge a_2 &gt; b_2 \ge 0&lt;/math&gt;, we must show that &lt;math&gt;2^{a_1}-2^{b_1} \neq 2^{a_2} - 2^{b_2}&lt;/math&gt;. Suppose otherwise for sake of contradiction; rearranging yields &lt;math&gt;2^{a_1}+2^{b_2} = 2^{a_2}+2^{b_1}&lt;/math&gt;. We use the fact that every positive integer has a unique binary representation:<br /> <br /> If &lt;math&gt;a_1 \neq b_2&lt;/math&gt; then &lt;math&gt;\{a_1,b_2\} = \{a_2,b_1\}&lt;/math&gt;; from here we can deduce either &lt;math&gt;a_1=a_2&lt;/math&gt; and &lt;math&gt;b_1=b_2&lt;/math&gt; (contradicting the assumption that &lt;math&gt;(a_1,b_1) \neq (a_2,b_2)&lt;/math&gt;, or &lt;math&gt;a_1=b_1&lt;/math&gt; and &lt;math&gt;a_2=b_2&lt;/math&gt; (contradicting the assumption &lt;math&gt;a_1&gt;b_1&lt;/math&gt; and &lt;math&gt;a_2&gt;b_2&lt;/math&gt;).<br /> <br /> If &lt;math&gt;a_1 = b_2&lt;/math&gt; then &lt;math&gt;2^{a_1}+2^{b_2} = 2 \times 2^{a_1}&lt;/math&gt;, and it follows that &lt;math&gt;a_1=a_2=b_1=b_2&lt;/math&gt;, also contradicting the assumption &lt;math&gt;(a_1,b_1) \neq (a_2,b_2)&lt;/math&gt;. Hence we obtain contradiction.<br /> <br /> Then there are &lt;math&gt;\binom{10}{2}-5&lt;/math&gt; choices for &lt;math&gt;(a,b)&lt;/math&gt; for which &lt;math&gt;2^a - 2^b&lt;/math&gt; is a positive integer less than 1000; by the above claim, each choice of &lt;math&gt;(a,b)&lt;/math&gt; results in a different positive integer &lt;math&gt;n&lt;/math&gt;. Then there are &lt;math&gt;55-5 = \boxed{050}&lt;/math&gt; integers which can be expressed as a difference of two powers of 2.<br /> ==See also==<br /> {{AIME box|year=2021|n=I|num-b=2|num-a=4}}<br /> <br /> [[Category:Introductory Number Theory Problems]]<br /> {{MAA Notice}}</div> Scrabbler94 https://artofproblemsolving.com/wiki/index.php?title=User:Scrabbler94&diff=149087 User:Scrabbler94 2021-03-11T22:48:30Z <p>Scrabbler94: </p> <hr /> <div>My username is Scrabbler94.<br /> <br /> &lt;b&gt;Bio:&lt;/b&gt;<br /> *MIT '16, major 18-C (Mathematics with computer science)<br /> *Ph.D. candidate in computer science. Research interests: graph theory, analysis of algorithms, complexity theory.<br /> *I play in NASPA-sanctioned Scrabble tournaments (as you can probably guess by my username). Rated ~1700-1800.<br /> *Amateur bowler (average ~175)<br /> <br /> &lt;b&gt;Past Contests:&lt;/b&gt;<br /> *2nd MATHCOUNTS State (2008), two-time MATHCOUNTS National qualifier (2007-2008)<br /> *AMC 10 perfect scorer (2009)<br /> *Six-time AIME qualifier (2007-2012)<br /> *USAMO qualifier, score: 13 (2010)<br /> *Putnam score: 20 (2014) - 1 point away from top 500 :(<br /> <br /> &lt;b&gt;AoPS Mock Contests:&lt;/b&gt;<br /> Contests held over AoPS.<br /> *[https://artofproblemsolving.com/community/c594864h1569848 Mock 2018 AMC 10]<br /> *[https://artofproblemsolving.com/community/c594864h1579782_released_mock_state_mathcounts_scrabbler94 Mock 2018 MATHCOUNTS State Competition]<br /> *[https://artofproblemsolving.com/community/c594864h1759381_2nd_annual_mock_mathcounts_state_scrabbler94 Mock 2019 MATHCOUNTS State Competition]<br /> *[https://artofproblemsolving.com/community/c594864h2159345_2020_mock_amc_1012_scrabbler94 Mock 2020 AMC 10/12]</div> Scrabbler94 https://artofproblemsolving.com/wiki/index.php?title=2007_AIME_I_Problems/Problem_4&diff=149086 2007 AIME I Problems/Problem 4 2021-03-11T22:46:44Z <p>Scrabbler94: Undo revision 109540 by Brudder (talk) Solution 2 (LCM/GCF) is wrong; this method would obtain 105 years even if the periods were 6000, 8400, 14000 for example.</p> <hr /> <div>== Problem ==<br /> Three planets orbit a star circularly in the same plane. Each moves in the same direction and moves at [[constant]] speed. Their periods are 60, 84, and 140 years. The three planets and the star are currently [[collinear]]. What is the fewest number of years from now that they will all be collinear again?<br /> <br /> == Solution ==<br /> <br /> Denote the planets &lt;math&gt;A, B, C &lt;/math&gt; respectively. Let &lt;math&gt;a(t), b(t), c(t) &lt;/math&gt; denote the angle which each of the respective planets makes with its initial position after &lt;math&gt;t &lt;/math&gt; years. These are given by &lt;math&gt; a(t) = \frac{t \pi}{30} &lt;/math&gt;, &lt;math&gt; b(t) = \frac{t \pi}{42} &lt;/math&gt;, &lt;math&gt;c(t) = \frac{t \pi}{70}&lt;/math&gt;.<br /> <br /> In order for the planets and the central star to be collinear, &lt;math&gt;a(t)&lt;/math&gt;, &lt;math&gt;b(t) &lt;/math&gt;, and &lt;math&gt;c(t) &lt;/math&gt; must differ by a multiple of &lt;math&gt;\pi &lt;/math&gt;. Note that &lt;math&gt; a(t) - b(t) = \frac{t \pi}{105}&lt;/math&gt; and &lt;math&gt; b(t) - c(t) = \frac{t \pi}{105}&lt;/math&gt;, so &lt;math&gt; a(t) - c(t) = \frac{ 2 t \pi}{105} &lt;/math&gt;. These are simultaneously multiples of &lt;math&gt;\pi &lt;/math&gt; exactly when &lt;math&gt;t &lt;/math&gt; is a multiple of &lt;math&gt;105&lt;/math&gt;, so the planets and the star will next be collinear in &lt;math&gt;\boxed{105}&lt;/math&gt; years.<br /> <br /> == See also ==<br /> {{AIME box|year=2007|n=I|num-b=3|num-a=5}}<br /> {{MAA Notice}}</div> Scrabbler94 https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_10B_Problems/Problem_2&diff=147735 2021 AMC 10B Problems/Problem 2 2021-02-22T18:18:36Z <p>Scrabbler94: move solution 2 before video solutions</p> <hr /> <div>==Problem==<br /> What is the value of &lt;cmath&gt;\sqrt{(3-2\sqrt{3})^2}+\sqrt{(3+2\sqrt{3})^2}?&lt;/cmath&gt;<br /> <br /> &lt;math&gt;\textbf{(A)} ~0 \qquad\textbf{(B)} ~4\sqrt{3}-6 \qquad\textbf{(C)} ~6 \qquad\textbf{(D)} ~4\sqrt{3} \qquad\textbf{(E)} ~4\sqrt{3}+6&lt;/math&gt;<br /> <br /> ==Solution==<br /> Note that the square root of any square is always the absolute value of the squared number because the square root function will only return a positive number. By squaring both &lt;math&gt;3&lt;/math&gt; and &lt;math&gt;2\sqrt{3}&lt;/math&gt;, we see that &lt;math&gt;2\sqrt{3}&gt;3&lt;/math&gt;, thus &lt;math&gt;3-2\sqrt{3}&lt;/math&gt; is negative, so we must take the absolute value of &lt;math&gt;3-2\sqrt{3}&lt;/math&gt;, which is just &lt;math&gt;2\sqrt{3}-3&lt;/math&gt;. Knowing this, the first term in the expression equals &lt;math&gt;2\sqrt{3}-3&lt;/math&gt; and the second term is &lt;math&gt;3+2\sqrt3&lt;/math&gt;, and summing the two gives &lt;math&gt;\boxed{\textbf{(D)} ~4\sqrt{3}}&lt;/math&gt;.<br /> <br /> ~bjc, abhinavg0627 and JackBocresion<br /> <br /> ==Solution 2==<br /> Let &lt;math&gt;x = \sqrt{(3-2\sqrt{3})^2}+\sqrt{(3+2\sqrt{3})^2}&lt;/math&gt;, then &lt;math&gt;x^2 = (3-2\sqrt{3})^2+2\sqrt{(-3)^2}+(3+2\sqrt3)^2&lt;/math&gt;. The &lt;math&gt;2\sqrt{(-3)^2}&lt;/math&gt; term is there due to difference of squares. Simplifying the expression gives us &lt;math&gt;x^2 = 48&lt;/math&gt;, so &lt;math&gt;x=\boxed{\textbf{(D)} ~4\sqrt{3}}&lt;/math&gt; ~ shrungpatel<br /> <br /> ==Video Solution==<br /> https://youtu.be/HHVdPTLQsLc<br /> ~Math Python<br /> <br /> == Video Solution by OmegaLearn ==<br /> https://youtu.be/Df3AIGD78xM<br /> <br /> ==Video Solution 3==<br /> https://youtu.be/v71C6cFbErQ<br /> <br /> ~savannahsolver<br /> <br /> ==Video Solution by TheBeautyofMath==<br /> https://youtu.be/gLahuINjRzU?t=154<br /> <br /> ~IceMatrix<br /> {{AMC10 box|year=2021|ab=B|num-b=1|num-a=3}}</div> Scrabbler94 https://artofproblemsolving.com/wiki/index.php?title=2020_AIME_II_Problems/Problem_12&diff=147357 2020 AIME II Problems/Problem 12 2021-02-18T06:19:36Z <p>Scrabbler94: /* Solution 2 (Official MAA) */</p> <hr /> <div>==Problem==<br /> Let &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; be odd integers greater than &lt;math&gt;1.&lt;/math&gt; An &lt;math&gt;m\times n&lt;/math&gt; rectangle is made up of unit squares where the squares in the top row are numbered left to right with the integers &lt;math&gt;1&lt;/math&gt; through &lt;math&gt;n&lt;/math&gt;, those in the second row are numbered left to right with the integers &lt;math&gt;n + 1&lt;/math&gt; through &lt;math&gt;2n&lt;/math&gt;, and so on. Square &lt;math&gt;200&lt;/math&gt; is in the top row, and square &lt;math&gt;2000&lt;/math&gt; is in the bottom row. Find the number of ordered pairs &lt;math&gt;(m,n)&lt;/math&gt; of odd integers greater than &lt;math&gt;1&lt;/math&gt; with the property that, in the &lt;math&gt;m\times n&lt;/math&gt; rectangle, the line through the centers of squares &lt;math&gt;200&lt;/math&gt; and &lt;math&gt;2000&lt;/math&gt; intersects the interior of square &lt;math&gt;1099&lt;/math&gt;.<br /> <br /> ==Solution==<br /> Let us take some cases. Since &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are odds, and &lt;math&gt;200&lt;/math&gt; is in the top row and &lt;math&gt;2000&lt;/math&gt; in the bottom, &lt;math&gt;m&lt;/math&gt; has to be &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;5&lt;/math&gt;, &lt;math&gt;7&lt;/math&gt;, or &lt;math&gt;9&lt;/math&gt;. Also, taking a look at the diagram, the slope of the line connecting those centers has to have an absolute value of &lt;math&gt; &lt; 1&lt;/math&gt;. Therefore, &lt;math&gt;m &lt; 1800 \mod n &lt; 1800-m&lt;/math&gt;. <br /> <br /> If &lt;math&gt;m=3&lt;/math&gt;, &lt;math&gt;n&lt;/math&gt; can range from &lt;math&gt;667&lt;/math&gt; to &lt;math&gt;999&lt;/math&gt;. However, &lt;math&gt;900&lt;/math&gt; divides &lt;math&gt;1800&lt;/math&gt;, so looking at mods, we can easily eliminate &lt;math&gt;899&lt;/math&gt; and &lt;math&gt;901&lt;/math&gt;. Now, counting these odd integers, we get &lt;math&gt;167 - 2 = 165&lt;/math&gt;.<br /> <br /> Similarly, let &lt;math&gt;m=5&lt;/math&gt;. Then &lt;math&gt;n&lt;/math&gt; can range from &lt;math&gt;401&lt;/math&gt; to &lt;math&gt;499&lt;/math&gt;. However, &lt;math&gt;450|1800&lt;/math&gt;, so one can remove &lt;math&gt;449&lt;/math&gt; and &lt;math&gt;451&lt;/math&gt;. Counting odd integers, we get &lt;math&gt;50 - 2 = 48&lt;/math&gt;.<br /> <br /> Take &lt;math&gt;m=7&lt;/math&gt;. Then, &lt;math&gt;n&lt;/math&gt; can range from &lt;math&gt;287&lt;/math&gt; to &lt;math&gt;333&lt;/math&gt;. However, &lt;math&gt;300|1800&lt;/math&gt;, so one can verify and eliminate &lt;math&gt;299&lt;/math&gt; and &lt;math&gt;301&lt;/math&gt;. Counting odd integers, we get &lt;math&gt;24 - 2 = 22&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;m = 9&lt;/math&gt;. Then &lt;math&gt;n&lt;/math&gt; can vary from &lt;math&gt;223&lt;/math&gt; to &lt;math&gt;249&lt;/math&gt;. However, &lt;math&gt;225|1800&lt;/math&gt;. Checking that value and the values around it, we can eliminate &lt;math&gt;225&lt;/math&gt;. Counting odd integers, we get &lt;math&gt;14 - 1 = 13&lt;/math&gt;.<br /> <br /> Add all of our cases to get &lt;cmath&gt; 165+48+22+13 = \boxed{248} &lt;/cmath&gt;<br /> <br /> -Solution by thanosaops<br /> <br /> ==Solution 2 (Official MAA)==<br /> Because square &lt;math&gt;2000&lt;/math&gt; is in the bottom row, it follows that &lt;math&gt;\frac{2000}m \le n &lt; \frac{2000}{m-1}&lt;/math&gt;. Moreover, because square &lt;math&gt;200&lt;/math&gt; is in the top row, and square &lt;math&gt;2000&lt;/math&gt; is not in the top row, &lt;math&gt;1 &lt; m \le 10&lt;/math&gt;. In particular, because the number of rows in the rectangle must be odd, &lt;math&gt;m&lt;/math&gt; must be one of &lt;math&gt;3, 5, 7,&lt;/math&gt; or &lt;math&gt;9.&lt;/math&gt;<br /> <br /> For each possible choice of &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt;, let &lt;math&gt;\ell_{m,n}&lt;/math&gt; denote the line through the centers of squares &lt;math&gt;200&lt;/math&gt; and &lt;math&gt;2000.&lt;/math&gt; Note that for odd values of &lt;math&gt;m&lt;/math&gt;, the line &lt;math&gt;\ell_{m,n}&lt;/math&gt; passes through the center of square &lt;math&gt;1100.&lt;/math&gt; Thus &lt;math&gt;\ell_{m,n}&lt;/math&gt; intersects the interior of cell &lt;math&gt;1099&lt;/math&gt; exactly when its slope is strictly between &lt;math&gt;-1&lt;/math&gt; and &lt;math&gt;1&lt;/math&gt;. The line &lt;math&gt;\ell_{m,n}&lt;/math&gt; is vertical whenever square &lt;math&gt;2000&lt;/math&gt; is the &lt;math&gt;200&lt;/math&gt;th square in the bottom row of the rectangle. This would happen for &lt;math&gt;m = 3, 5, 7, 9&lt;/math&gt; when &lt;math&gt;n = 900, 450, 300, 225&lt;/math&gt;, respectively. When &lt;math&gt;n&lt;/math&gt; is 1 greater than or 1 less than these numbers, the slope of &lt;math&gt;\ell_{m,n}&lt;/math&gt; is &lt;math&gt;1&lt;/math&gt; or &lt;math&gt;-1&lt;/math&gt;, respectively. In all other cases the slope is strictly between &lt;math&gt;-1&lt;/math&gt; and &lt;math&gt;1.&lt;/math&gt; The admissible values for &lt;math&gt;n&lt;/math&gt; for each possible value of &lt;math&gt;m&lt;/math&gt; are given in the following table.<br /> &lt;cmath&gt;\begin{tabular}{|c|c|c|c|c|}\hline<br /> m &amp; minimum n &amp; maximum n &amp; avoided n &amp; number of odd n\\\hline<br /> 3&amp;667&amp;999&amp;899, 900, 901&amp;165\\\hline<br /> 5&amp;400&amp;499&amp;449, 450, 451&amp;48\\\hline<br /> 7&amp;286&amp;333&amp;299, 300, 301&amp;22\\\hline<br /> 9&amp;223&amp;249&amp;224, 225, 226&amp;13\\\hline<br /> \end{tabular}&lt;/cmath&gt;<br /> This accounts for &lt;math&gt;165 + 48 + 22 + 13 = 248&lt;/math&gt; rectangles.<br /> <br /> ==Video Solution 1==<br /> https://www.youtube.com/watch?v=MrtKoO16XLQ ~ MathEx<br /> <br /> ==Video Solution 2==<br /> https://youtu.be/v58SLOoAKTw<br /> <br /> ==See Also==<br /> {{AIME box|year=2020|n=II|num-b=11|num-a=13}}<br /> {{MAA Notice}}</div> Scrabbler94 https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12A_Problems/Problem_25&diff=146558 2021 AMC 12A Problems/Problem 25 2021-02-14T03:52:00Z <p>Scrabbler94: /* Solution 2 (Fast) */</p> <hr /> <div>==Problem==<br /> Let &lt;math&gt;d(n)&lt;/math&gt; denote the number of positive integers that divide &lt;math&gt;n&lt;/math&gt;, including &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt;. For example, &lt;math&gt;d(1)=1,d(2)=2,&lt;/math&gt; and &lt;math&gt;d(12)=6&lt;/math&gt;. (This function is known as the divisor function.) Let&lt;cmath&gt;f(n)=\frac{d(n)}{\sqrt n}.&lt;/cmath&gt;There is a unique positive integer &lt;math&gt;N&lt;/math&gt; such that &lt;math&gt;f(N)&gt;f(n)&lt;/math&gt; for all positive integers &lt;math&gt;n\ne N&lt;/math&gt;. What is the sum of the digits of &lt;math&gt;N?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }5 \qquad \textbf{(B) }6 \qquad \textbf{(C) }7 \qquad \textbf{(D) }8\qquad \textbf{(E) }9&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Consider the prime factorization &lt;cmath&gt;n={p_1}^{e_1}{p_2}^{e_2}{p_3}^{e_3}\cdots{p_k}^{e_k}.&lt;/cmath&gt; By the Multiplication Principle, &lt;cmath&gt;d(n)=(e_1+1)(e_2+1)(e_3+1)\cdots(e_k+1).&lt;/cmath&gt; Now, we rewrite &lt;math&gt;f(n)&lt;/math&gt; as &lt;cmath&gt;f(n)=\frac{d(n)}{\sqrt n}=\frac{(e_1+1)(e_2+1)(e_3+1)\cdots(e_k+1)}{{p_1}^{{e_1}/3}{p_2}^{{e_2}/3}{p_3}^{{e_3}/3}\cdots{p_k}^{{e_k}/3}}=\left(\frac{e_1+1}{{p_1}^{{e_1}/3}}\right)\left(\frac{e_2+1}{{p_2}^{{e_2}/3}}\right)\left(\frac{e_3+1}{{p_3}^{{e_3}/3}}\right)\cdots\left(\frac{e_k+1}{{p_k}^{{e_k}/3}}\right).&lt;/cmath&gt; As &lt;math&gt;f(n)&gt;0&lt;/math&gt; for all positive integers &lt;math&gt;n,&lt;/math&gt; it follows that for all positive integers &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;, &lt;math&gt;f(a)&gt;f(b)&lt;/math&gt; if and only if &lt;math&gt;f(a)^3&gt;f(b)^3.&lt;/math&gt; So, &lt;math&gt;f(n)&lt;/math&gt; is maximized if and only if &lt;cmath&gt;f(n)^3=\left(\frac{(e_1+1)^3}{{p_1}^{e_1}}\right)\left(\frac{(e_2+1)^3}{{p_2}^{e_2}}\right)\left(\frac{(e_3+1)^3}{{p_3}^{e_3}}\right)\cdots\left(\frac{(e_k+1)^3}{{p_k}^{e_k}}\right)&lt;/cmath&gt; is maximized.<br /> <br /> For every factor &lt;math&gt;\frac{(e_i+1)^3}{{p_i}^{e_i}}&lt;/math&gt; with a fixed &lt;math&gt;p_i&lt;/math&gt; where &lt;math&gt;1\leq i\leq k,&lt;/math&gt; the denominator grows faster than the numerator, as exponential functions grow faster than polynomial functions. For each prime &lt;math&gt;p_i=2,3,5,7,\cdots,&lt;/math&gt; we look for the &lt;math&gt;e_i&lt;/math&gt; for which &lt;math&gt;\frac{(e_i+1)^3}{{p_i}^{e_i}}&lt;/math&gt; is a relative maximum:<br /> &lt;cmath&gt;\begin{array}{ c c c c }<br /> p_i &amp; e_i &amp; (e_i+1)^3/\left({p_i}^{e_i}\right) &amp; \text{max?} \\ <br /> \hline\hline<br /> 2 &amp; 0 &amp; 1 &amp; \\ <br /> 2 &amp; 1 &amp; 4 &amp; \\<br /> 2 &amp; 2 &amp; 27/4 &amp;\\<br /> 2 &amp; 3 &amp; 8 &amp; \text{yes}\\<br /> 2 &amp; 4 &amp; 125/16 &amp; \\<br /> \hline<br /> 3 &amp; 0 &amp; 1 &amp;\\<br /> 3 &amp; 1 &amp; 8/3 &amp; \\<br /> 3 &amp; 2 &amp; 3 &amp; \text{yes}\\<br /> 3 &amp; 3 &amp; 64/27 &amp; \\<br /> \hline<br /> 5 &amp; 0 &amp; 1 &amp; \\<br /> 5 &amp; 1 &amp; 8/5 &amp; \text{yes}\\<br /> 5 &amp; 2 &amp; 27/25 &amp; \\<br /> \hline<br /> 7 &amp; 0 &amp; 1 &amp; \\<br /> 7 &amp; 1 &amp; 8/7 &amp; \text{yes}\\<br /> 7 &amp; 2 &amp; 27/49 &amp; \\<br /> \hline<br /> 11 &amp; 0 &amp; 1 &amp; \text{yes} \\<br /> 11 &amp; 1 &amp; 8/11 &amp; \\<br /> \hline<br /> \cdots &amp; \cdots &amp; \cdots &amp; <br /> \end{array}&lt;/cmath&gt;<br /> <br /> Finally, the number we seek is &lt;math&gt;N=2^3 3^2 5^1 7^1 = 2520.&lt;/math&gt; The sum of its digits is &lt;math&gt;2+5+2+0=\boxed{\textbf{(E) }9}.&lt;/math&gt;<br /> <br /> Actually, once we get that &lt;math&gt;3^2&lt;/math&gt; is a factor of &lt;math&gt;N,&lt;/math&gt; we know that the sum of the digits of &lt;math&gt;N&lt;/math&gt; must be a multiple of &lt;math&gt;9.&lt;/math&gt; Only choice &lt;math&gt;\textbf{(E)}&lt;/math&gt; is possible.<br /> <br /> ~MRENTHUSIASM<br /> <br /> ==Solution 2 (Fast)==<br /> Using the answer choices to our advantage, we can show that &lt;math&gt;N&lt;/math&gt; must be divisible by 9 without explicitly computing &lt;math&gt;N&lt;/math&gt;, by exploiting the following fact:<br /> <br /> '''Claim''': If &lt;math&gt;n&lt;/math&gt; is not divisible by 3, then &lt;math&gt;f(9n) &gt; f(3n) &gt; f(n)&lt;/math&gt;.<br /> <br /> '''Proof''': Since &lt;math&gt;d(\cdot)&lt;/math&gt; is a [[multiplicative function]], we have &lt;math&gt;d(3n) = d(3)d(n) = 2d(n)&lt;/math&gt; and &lt;math&gt;d(9n) = 3d(n)&lt;/math&gt;. Then<br /> <br /> &lt;cmath&gt;\begin{align*}<br /> f(3n) &amp;= \frac{2d(n)}{\sqrt{3n}} \approx 1.38 f(n)\\<br /> f(9n) &amp;= \frac{3d(n)}{\sqrt{9n}} \approx 1.44 f(n)<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> Note that the values &lt;math&gt;\frac{2}{\sqrt{3}}&lt;/math&gt; and &lt;math&gt;\frac{3}{\sqrt{9}}&lt;/math&gt; do not have to be explicitly computed; we only need the fact that &lt;math&gt;\frac{3}{\sqrt{9}} &gt; \frac{2}{\sqrt{3}} &gt; 1&lt;/math&gt; which is easy to show by hand.<br /> <br /> The above claim automatically implies &lt;math&gt;N&lt;/math&gt; is a multiple of 9: if &lt;math&gt;N&lt;/math&gt; was not divisible by 9, then &lt;math&gt;f(9N) &gt; f(N)&lt;/math&gt; which is a contradiction, and if &lt;math&gt;N&lt;/math&gt; was divisible by 3 and not 9, then &lt;math&gt;f(3N) &gt; f(N) &gt; f\left(\frac{N}{3}\right)&lt;/math&gt;, also a contradiction. Then the sum of digits of &lt;math&gt;N&lt;/math&gt; must be a multiple of 9, so only choice &lt;math&gt;\boxed{\textbf{(E) } 9}&lt;/math&gt; works.<br /> <br /> -scrabbler94<br /> <br /> == Video Solution by OmegaLearn (Multiplicative function properties + Meta-solving ) ==<br /> https://youtu.be/6P-0ZHAaC_A<br /> <br /> ~ pi_is_3.14<br /> <br /> ==See also==<br /> {{AMC12 box|year=2021|ab=A|num-b=24|after=Last problem}}<br /> <br /> [[Category:Intermediate Number Theory Problems]]<br /> {{MAA Notice}}</div> Scrabbler94 https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12A_Problems/Problem_25&diff=146557 2021 AMC 12A Problems/Problem 25 2021-02-14T03:51:03Z <p>Scrabbler94: /* Solution 2 (Fast) */</p> <hr /> <div>==Problem==<br /> Let &lt;math&gt;d(n)&lt;/math&gt; denote the number of positive integers that divide &lt;math&gt;n&lt;/math&gt;, including &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt;. For example, &lt;math&gt;d(1)=1,d(2)=2,&lt;/math&gt; and &lt;math&gt;d(12)=6&lt;/math&gt;. (This function is known as the divisor function.) Let&lt;cmath&gt;f(n)=\frac{d(n)}{\sqrt n}.&lt;/cmath&gt;There is a unique positive integer &lt;math&gt;N&lt;/math&gt; such that &lt;math&gt;f(N)&gt;f(n)&lt;/math&gt; for all positive integers &lt;math&gt;n\ne N&lt;/math&gt;. What is the sum of the digits of &lt;math&gt;N?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }5 \qquad \textbf{(B) }6 \qquad \textbf{(C) }7 \qquad \textbf{(D) }8\qquad \textbf{(E) }9&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Consider the prime factorization &lt;cmath&gt;n={p_1}^{e_1}{p_2}^{e_2}{p_3}^{e_3}\cdots{p_k}^{e_k}.&lt;/cmath&gt; By the Multiplication Principle, &lt;cmath&gt;d(n)=(e_1+1)(e_2+1)(e_3+1)\cdots(e_k+1).&lt;/cmath&gt; Now, we rewrite &lt;math&gt;f(n)&lt;/math&gt; as &lt;cmath&gt;f(n)=\frac{d(n)}{\sqrt n}=\frac{(e_1+1)(e_2+1)(e_3+1)\cdots(e_k+1)}{{p_1}^{{e_1}/3}{p_2}^{{e_2}/3}{p_3}^{{e_3}/3}\cdots{p_k}^{{e_k}/3}}=\left(\frac{e_1+1}{{p_1}^{{e_1}/3}}\right)\left(\frac{e_2+1}{{p_2}^{{e_2}/3}}\right)\left(\frac{e_3+1}{{p_3}^{{e_3}/3}}\right)\cdots\left(\frac{e_k+1}{{p_k}^{{e_k}/3}}\right).&lt;/cmath&gt; As &lt;math&gt;f(n)&gt;0&lt;/math&gt; for all positive integers &lt;math&gt;n,&lt;/math&gt; it follows that for all positive integers &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;, &lt;math&gt;f(a)&gt;f(b)&lt;/math&gt; if and only if &lt;math&gt;f(a)^3&gt;f(b)^3.&lt;/math&gt; So, &lt;math&gt;f(n)&lt;/math&gt; is maximized if and only if &lt;cmath&gt;f(n)^3=\left(\frac{(e_1+1)^3}{{p_1}^{e_1}}\right)\left(\frac{(e_2+1)^3}{{p_2}^{e_2}}\right)\left(\frac{(e_3+1)^3}{{p_3}^{e_3}}\right)\cdots\left(\frac{(e_k+1)^3}{{p_k}^{e_k}}\right)&lt;/cmath&gt; is maximized.<br /> <br /> For every factor &lt;math&gt;\frac{(e_i+1)^3}{{p_i}^{e_i}}&lt;/math&gt; with a fixed &lt;math&gt;p_i&lt;/math&gt; where &lt;math&gt;1\leq i\leq k,&lt;/math&gt; the denominator grows faster than the numerator, as exponential functions grow faster than polynomial functions. For each prime &lt;math&gt;p_i=2,3,5,7,\cdots,&lt;/math&gt; we look for the &lt;math&gt;e_i&lt;/math&gt; for which &lt;math&gt;\frac{(e_i+1)^3}{{p_i}^{e_i}}&lt;/math&gt; is a relative maximum:<br /> &lt;cmath&gt;\begin{array}{ c c c c }<br /> p_i &amp; e_i &amp; (e_i+1)^3/\left({p_i}^{e_i}\right) &amp; \text{max?} \\ <br /> \hline\hline<br /> 2 &amp; 0 &amp; 1 &amp; \\ <br /> 2 &amp; 1 &amp; 4 &amp; \\<br /> 2 &amp; 2 &amp; 27/4 &amp;\\<br /> 2 &amp; 3 &amp; 8 &amp; \text{yes}\\<br /> 2 &amp; 4 &amp; 125/16 &amp; \\<br /> \hline<br /> 3 &amp; 0 &amp; 1 &amp;\\<br /> 3 &amp; 1 &amp; 8/3 &amp; \\<br /> 3 &amp; 2 &amp; 3 &amp; \text{yes}\\<br /> 3 &amp; 3 &amp; 64/27 &amp; \\<br /> \hline<br /> 5 &amp; 0 &amp; 1 &amp; \\<br /> 5 &amp; 1 &amp; 8/5 &amp; \text{yes}\\<br /> 5 &amp; 2 &amp; 27/25 &amp; \\<br /> \hline<br /> 7 &amp; 0 &amp; 1 &amp; \\<br /> 7 &amp; 1 &amp; 8/7 &amp; \text{yes}\\<br /> 7 &amp; 2 &amp; 27/49 &amp; \\<br /> \hline<br /> 11 &amp; 0 &amp; 1 &amp; \text{yes} \\<br /> 11 &amp; 1 &amp; 8/11 &amp; \\<br /> \hline<br /> \cdots &amp; \cdots &amp; \cdots &amp; <br /> \end{array}&lt;/cmath&gt;<br /> <br /> Finally, the number we seek is &lt;math&gt;N=2^3 3^2 5^1 7^1 = 2520.&lt;/math&gt; The sum of its digits is &lt;math&gt;2+5+2+0=\boxed{\textbf{(E) }9}.&lt;/math&gt;<br /> <br /> Actually, once we get that &lt;math&gt;3^2&lt;/math&gt; is a factor of &lt;math&gt;N,&lt;/math&gt; we know that the sum of the digits of &lt;math&gt;N&lt;/math&gt; must be a multiple of &lt;math&gt;9.&lt;/math&gt; Only choice &lt;math&gt;\textbf{(E)}&lt;/math&gt; is possible.<br /> <br /> ~MRENTHUSIASM<br /> <br /> ==Solution 2 (Fast)==<br /> Using the answer choices to our advantage, we can show that &lt;math&gt;N&lt;/math&gt; must be divisible by 9 without explicitly computing &lt;math&gt;N&lt;/math&gt;, by exploiting the following fact:<br /> <br /> '''Claim''': If &lt;math&gt;n&lt;/math&gt; is not divisible by 3, then &lt;math&gt;f(9n) &gt; f(3n) &gt; f(n)&lt;/math&gt;.<br /> <br /> '''Proof''': Since &lt;math&gt;d(\cdot)&lt;/math&gt; is a [[multiplicative function]], we have &lt;math&gt;d(3n) = d(3)d(n) = 2d(n)&lt;/math&gt; and &lt;math&gt;d(9n) = 3d(n)&lt;/math&gt;. Then<br /> <br /> &lt;cmath&gt;\begin{align*}<br /> f(3n) &amp;= \frac{2d(n)}{\sqrt{3n}} \approx 1.38 f(n)\\<br /> f(9n) &amp;= \frac{3d(n)}{\sqrt{9n}} \approx 1.44 f(n)<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> Note that the values &lt;math&gt;\frac{2}{\sqrt{3}}&lt;/math&gt; and &lt;math&gt;\frac{3}{\sqrt{9}}&lt;/math&gt; do not have to be explicitly computed; we only need the fact that &lt;math&gt;\frac{3}{\sqrt{9}} &gt; \frac{2}{\sqrt{3}} &gt; 1&lt;/math&gt; which is easy to show by hand.<br /> <br /> The above claim automatically implies &lt;math&gt;N&lt;/math&gt; is a multiple of 9: if &lt;math&gt;N&lt;/math&gt; was not divisible by 9, then &lt;math&gt;f(9N) &gt; f(N)&lt;/math&gt; which is a contradiction, and if &lt;math&gt;N&lt;/math&gt; was divisible by 3 and not 9, then &lt;math&gt;f(3N) &gt; f(N) &gt; f\left(\frac{N}{3}\right)&lt;/math&gt;, also a contradiction. Then the sum of digits of &lt;math&gt;N&lt;/math&gt; must be a multiple of 9, so only choice &lt;math&gt;\boxed{\textbf{(E) } 9}&lt;/math&gt; works.<br /> <br /> == Video Solution by OmegaLearn (Multiplicative function properties + Meta-solving ) ==<br /> https://youtu.be/6P-0ZHAaC_A<br /> <br /> ~ pi_is_3.14<br /> <br /> ==See also==<br /> {{AMC12 box|year=2021|ab=A|num-b=24|after=Last problem}}<br /> <br /> [[Category:Intermediate Number Theory Problems]]<br /> {{MAA Notice}}</div> Scrabbler94 https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12A_Problems/Problem_25&diff=146556 2021 AMC 12A Problems/Problem 25 2021-02-14T03:48:33Z <p>Scrabbler94: combine solutions 2 and 3 with more rigorous justification why we can say N is a multiple of 9</p> <hr /> <div>==Problem==<br /> Let &lt;math&gt;d(n)&lt;/math&gt; denote the number of positive integers that divide &lt;math&gt;n&lt;/math&gt;, including &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt;. For example, &lt;math&gt;d(1)=1,d(2)=2,&lt;/math&gt; and &lt;math&gt;d(12)=6&lt;/math&gt;. (This function is known as the divisor function.) Let&lt;cmath&gt;f(n)=\frac{d(n)}{\sqrt n}.&lt;/cmath&gt;There is a unique positive integer &lt;math&gt;N&lt;/math&gt; such that &lt;math&gt;f(N)&gt;f(n)&lt;/math&gt; for all positive integers &lt;math&gt;n\ne N&lt;/math&gt;. What is the sum of the digits of &lt;math&gt;N?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }5 \qquad \textbf{(B) }6 \qquad \textbf{(C) }7 \qquad \textbf{(D) }8\qquad \textbf{(E) }9&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Consider the prime factorization &lt;cmath&gt;n={p_1}^{e_1}{p_2}^{e_2}{p_3}^{e_3}\cdots{p_k}^{e_k}.&lt;/cmath&gt; By the Multiplication Principle, &lt;cmath&gt;d(n)=(e_1+1)(e_2+1)(e_3+1)\cdots(e_k+1).&lt;/cmath&gt; Now, we rewrite &lt;math&gt;f(n)&lt;/math&gt; as &lt;cmath&gt;f(n)=\frac{d(n)}{\sqrt n}=\frac{(e_1+1)(e_2+1)(e_3+1)\cdots(e_k+1)}{{p_1}^{{e_1}/3}{p_2}^{{e_2}/3}{p_3}^{{e_3}/3}\cdots{p_k}^{{e_k}/3}}=\left(\frac{e_1+1}{{p_1}^{{e_1}/3}}\right)\left(\frac{e_2+1}{{p_2}^{{e_2}/3}}\right)\left(\frac{e_3+1}{{p_3}^{{e_3}/3}}\right)\cdots\left(\frac{e_k+1}{{p_k}^{{e_k}/3}}\right).&lt;/cmath&gt; As &lt;math&gt;f(n)&gt;0&lt;/math&gt; for all positive integers &lt;math&gt;n,&lt;/math&gt; it follows that for all positive integers &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;, &lt;math&gt;f(a)&gt;f(b)&lt;/math&gt; if and only if &lt;math&gt;f(a)^3&gt;f(b)^3.&lt;/math&gt; So, &lt;math&gt;f(n)&lt;/math&gt; is maximized if and only if &lt;cmath&gt;f(n)^3=\left(\frac{(e_1+1)^3}{{p_1}^{e_1}}\right)\left(\frac{(e_2+1)^3}{{p_2}^{e_2}}\right)\left(\frac{(e_3+1)^3}{{p_3}^{e_3}}\right)\cdots\left(\frac{(e_k+1)^3}{{p_k}^{e_k}}\right)&lt;/cmath&gt; is maximized.<br /> <br /> For every factor &lt;math&gt;\frac{(e_i+1)^3}{{p_i}^{e_i}}&lt;/math&gt; with a fixed &lt;math&gt;p_i&lt;/math&gt; where &lt;math&gt;1\leq i\leq k,&lt;/math&gt; the denominator grows faster than the numerator, as exponential functions grow faster than polynomial functions. For each prime &lt;math&gt;p_i=2,3,5,7,\cdots,&lt;/math&gt; we look for the &lt;math&gt;e_i&lt;/math&gt; for which &lt;math&gt;\frac{(e_i+1)^3}{{p_i}^{e_i}}&lt;/math&gt; is a relative maximum:<br /> &lt;cmath&gt;\begin{array}{ c c c c }<br /> p_i &amp; e_i &amp; (e_i+1)^3/\left({p_i}^{e_i}\right) &amp; \text{max?} \\ <br /> \hline\hline<br /> 2 &amp; 0 &amp; 1 &amp; \\ <br /> 2 &amp; 1 &amp; 4 &amp; \\<br /> 2 &amp; 2 &amp; 27/4 &amp;\\<br /> 2 &amp; 3 &amp; 8 &amp; \text{yes}\\<br /> 2 &amp; 4 &amp; 125/16 &amp; \\<br /> \hline<br /> 3 &amp; 0 &amp; 1 &amp;\\<br /> 3 &amp; 1 &amp; 8/3 &amp; \\<br /> 3 &amp; 2 &amp; 3 &amp; \text{yes}\\<br /> 3 &amp; 3 &amp; 64/27 &amp; \\<br /> \hline<br /> 5 &amp; 0 &amp; 1 &amp; \\<br /> 5 &amp; 1 &amp; 8/5 &amp; \text{yes}\\<br /> 5 &amp; 2 &amp; 27/25 &amp; \\<br /> \hline<br /> 7 &amp; 0 &amp; 1 &amp; \\<br /> 7 &amp; 1 &amp; 8/7 &amp; \text{yes}\\<br /> 7 &amp; 2 &amp; 27/49 &amp; \\<br /> \hline<br /> 11 &amp; 0 &amp; 1 &amp; \text{yes} \\<br /> 11 &amp; 1 &amp; 8/11 &amp; \\<br /> \hline<br /> \cdots &amp; \cdots &amp; \cdots &amp; <br /> \end{array}&lt;/cmath&gt;<br /> <br /> Finally, the number we seek is &lt;math&gt;N=2^3 3^2 5^1 7^1 = 2520.&lt;/math&gt; The sum of its digits is &lt;math&gt;2+5+2+0=\boxed{\textbf{(E) }9}.&lt;/math&gt;<br /> <br /> Actually, once we get that &lt;math&gt;3^2&lt;/math&gt; is a factor of &lt;math&gt;N,&lt;/math&gt; we know that the sum of the digits of &lt;math&gt;N&lt;/math&gt; must be a multiple of &lt;math&gt;9.&lt;/math&gt; Only choice &lt;math&gt;\textbf{(E)}&lt;/math&gt; is possible.<br /> <br /> ~MRENTHUSIASM<br /> <br /> ==Solution 2 (Fast)==<br /> Using the answer choices to our advantage, we can show that &lt;math&gt;N&lt;/math&gt; must be divisible by 9 without explicitly computing &lt;math&gt;N&lt;/math&gt;, by exploiting the following fact:<br /> <br /> '''Claim''': If &lt;math&gt;n&lt;/math&gt; is not divisible by 3, then &lt;math&gt;f(9n) &gt; f(3n) &gt; f(n)&lt;/math&gt;.<br /> <br /> '''Proof''': Since &lt;math&gt;d(\cdot)&lt;/math&gt; is a [[multiplicative function]], we have &lt;math&gt;d(3n) = d(3)d(n) = 2d(n)&lt;/math&gt; and &lt;math&gt;d(9n) = 3d(n)&lt;/math&gt;. Then<br /> <br /> &lt;cmath&gt;\begin{align*}<br /> f(3n) &amp;= \frac{2d(n)}{\sqrt{3n}} \approx 1.38 f(n)\\<br /> f(9n) &amp;= \frac{3d(n)}{\sqrt{9n}} \approx 1.44 f(n)<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> Note that the values &lt;math&gt;\frac{2}{\sqrt{3}}&lt;/math&gt; and &lt;math&gt;\frac{3}{\sqrt{9}}&lt;/math&gt; do not have to be explicitly computed; we only need the fact that &lt;math&gt;\frac{3}{\sqrt{9}} &gt; \frac{2}{\sqrt{3}} &gt; 1&lt;/math&gt; which is easy to show by hand.<br /> <br /> The above claim automatically implies &lt;math&gt;N&lt;/math&gt; is a multiple of 9: if &lt;math&gt;N&lt;/math&gt; was not divisible by 9, then &lt;math&gt;f(9N) &gt; f(N)&lt;/math&gt; which is a contradiction, and if &lt;math&gt;N&lt;/math&gt; was divisible by 3 and not 9, then &lt;math&gt;f(3N) &gt; f(N) &gt; f\left(\frac{N}{3}\right)&lt;/math&gt;, also a contradiction. Then the sum of digits of &lt;math&gt;N&lt;/math&gt; must be a multiple of 9, so only choice &lt;math&gt;\textbf{(E) } 9&lt;/math&gt; works.<br /> <br /> == Video Solution by OmegaLearn (Multiplicative function properties + Meta-solving ) ==<br /> https://youtu.be/6P-0ZHAaC_A<br /> <br /> ~ pi_is_3.14<br /> <br /> ==See also==<br /> {{AMC12 box|year=2021|ab=A|num-b=24|after=Last problem}}<br /> <br /> [[Category:Intermediate Number Theory Problems]]<br /> {{MAA Notice}}</div> Scrabbler94 https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12A_Problems&diff=146553 2021 AMC 12A Problems 2021-02-14T03:23:24Z <p>Scrabbler94: /* Problem 21 */ fix typo</p> <hr /> <div>{{AMC12 Problems|year=2021|ab=A}}<br /> ==Problem 1==<br /> What is the value of&lt;cmath&gt;2^{1+2+3}-(2^1+2^2+2^3)?&lt;/cmath&gt;<br /> &lt;math&gt;\textbf{(A) }0 \qquad \textbf{(B) }50 \qquad \textbf{(C) }52 \qquad \textbf{(D) }54 \qquad \textbf{(E) }57&lt;/math&gt;<br /> <br /> <br /> [[2021 AMC 12A Problems/Problem 1|Solution]]<br /> <br /> ==Problem 2==<br /> Under what conditions does &lt;math&gt;\sqrt{a^2+b^2}=a+b&lt;/math&gt; hold, where &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are real numbers?<br /> <br /> &lt;math&gt;\textbf{(A) }&lt;/math&gt; It is never true.<br /> <br /> &lt;math&gt;\textbf{(B) }&lt;/math&gt; It is true if and only if &lt;math&gt;ab=0&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{(C) }&lt;/math&gt; It is true if and only if &lt;math&gt;a+b\ge 0&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{(D) }&lt;/math&gt; It is true if and only if &lt;math&gt;ab=0&lt;/math&gt; and &lt;math&gt;a+b\ge 0&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{(E) }&lt;/math&gt; It is always true.<br /> <br /> [[2021 AMC 12A Problems/Problem 2|Solution]]<br /> <br /> ==Problem 3==<br /> The sum of two natural numbers is &lt;math&gt;17,402&lt;/math&gt;. One of the two numbers is divisible by &lt;math&gt;10&lt;/math&gt;. If the units digit of that number is erased, the other number is obtained. What is the difference of these two numbers?<br /> <br /> &lt;math&gt;\textbf{(A) }10,272 \qquad \textbf{(B) }11,700 \qquad \textbf{(C) }13,362 \qquad \textbf{(D) }14,238 \qquad \textbf{(E) }15,462&lt;/math&gt;<br /> <br /> [[2021 AMC 12A Problems/Problem 3|Solution]]<br /> <br /> ==Problem 4==<br /> Tom has a collection of &lt;math&gt;13&lt;/math&gt; snakes, &lt;math&gt;4&lt;/math&gt; of which are purple and &lt;math&gt;5&lt;/math&gt; of which are happy. He observes that<br /> <br /> &lt;math&gt;\quad\bullet&lt;/math&gt; all of his happy snakes can add,<br /> <br /> &lt;math&gt;\quad\bullet&lt;/math&gt; none of his purple snakes can subtract, and<br /> <br /> &lt;math&gt;\quad\bullet&lt;/math&gt; all of his snakes that can't subtract also can't add.<br /> <br /> Which of these conclusions can be drawn about Tom's snakes?<br /> <br /> <br /> &lt;math&gt;\textbf{(A) }&lt;/math&gt; Purple snakes can add.<br /> <br /> &lt;math&gt;\textbf{(B) }&lt;/math&gt; Purple snakes are happy.<br /> <br /> &lt;math&gt;\textbf{(C) }&lt;/math&gt; Snakes that can add are purple.<br /> <br /> &lt;math&gt;\textbf{(D) }&lt;/math&gt; Happy snakes are not purple.<br /> <br /> &lt;math&gt;\textbf{(E) }&lt;/math&gt; Happy snakes can't subtract.<br /> <br /> [[2021 AMC 12A Problems/Problem 4|Solution]]<br /> <br /> ==Problem 5==<br /> When a student multiplied the number &lt;math&gt;66&lt;/math&gt; by the repeating decimal&lt;cmath&gt;\underline{1}.\underline{a}\underline{b}\underline{a}\underline{b}...=\underline{1}.\overline{\underline{a}\underline{b}},&lt;/cmath&gt;where &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are digits, he did not notice the notation and just multiplied &lt;math&gt;66&lt;/math&gt; times &lt;math&gt;\underline{1}.\underline{a}\underline{b}&lt;/math&gt;. Later he found that his answer is &lt;math&gt;0.5&lt;/math&gt; less than the correct answer. What is the &lt;math&gt;2&lt;/math&gt;-digit number &lt;math&gt;\underline{a}\underline{b}?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }15 \qquad \textbf{(B) }30 \qquad \textbf{(C) }45 \qquad \textbf{(D) }60 \qquad \textbf{(E) }75&lt;/math&gt;<br /> <br /> <br /> [[2021 AMC 12A Problems/Problem 5|Solution]]<br /> <br /> ==Problem 6==<br /> A deck of cards has only red cards and black cards. The probability of a randomly chosen card being red is &lt;math&gt;\frac13&lt;/math&gt;. When &lt;math&gt;4&lt;/math&gt; black cards are added to the deck, the probability of choosing red becomes &lt;math&gt;\frac14&lt;/math&gt;. How many cards were in the deck originally?<br /> <br /> &lt;math&gt;\textbf{(A) }6 \qquad \textbf{(B) }9 \qquad \textbf{(C) }12 \qquad \textbf{(D) }15 \qquad \textbf{(E) }18&lt;/math&gt;<br /> <br /> [[2021 AMC 12A Problems/Problem 6|Solution]]<br /> <br /> ==Problem 7==<br /> What is the least possible value of &lt;math&gt;(xy-1)^2+(x+y)^2&lt;/math&gt; for all real numbers &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }0 \qquad \textbf{(B) }\frac14 \qquad \textbf{(C) }\frac12 \qquad \textbf{(D) }1 \qquad \textbf{(E) }2&lt;/math&gt;<br /> <br /> [[2021 AMC 12A Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8==<br /> A sequence of numbers is defined by &lt;math&gt;D_0=0,D_1=0,D_2=1&lt;/math&gt; and &lt;math&gt;D_n=D_{n-1}+D_{n-3}&lt;/math&gt; for &lt;math&gt;n\ge 3&lt;/math&gt;. What are the parities (evenness or oddness) of the triple of numbers &lt;math&gt;(D_{2021},D_{2022},D_{2023})&lt;/math&gt;, where &lt;math&gt;E&lt;/math&gt; denotes even and &lt;math&gt;O&lt;/math&gt; denotes odd?<br /> <br /> &lt;math&gt;\textbf{(A) }(O,E,O) \qquad \textbf{(B) }(E,E,O) \qquad \textbf{(C) }(E,O,E) \qquad \textbf{(D) }(O,O,E) \qquad \textbf{(E) }(O,O,O)&lt;/math&gt;<br /> <br /> <br /> [[2021 AMC 12A Problems/Problem 8|Solution]]<br /> <br /> ==Problem 9==<br /> Which of the following is equilvalent to&lt;cmath&gt;(2+3)(2^2+3^2)(2^4+3^4)(2^8+3^8)(2^{16}+3^{16})(2^{32}+3^{32})(2^{64}+3^{64})?&lt;/cmath&gt;<br /> &lt;math&gt;\textbf{(A) }3^{127}+2^{127} \qquad \textbf{(B) }3^{127}+2^{127}+2\cdot 3^{63}+3\cdot 2^{63} \qquad \textbf{(C) }3^{128}-2^{128} \qquad \textbf{(D) }3^{128}+2^{128} \qquad \textbf{(E) }5^{127}&lt;/math&gt;<br /> <br /> [[2021 AMC 12A Problems/Problem 9|Solution]]<br /> <br /> ==Problem 10==<br /> Two right circular cones with vertices facing down as shown in the figure below contains the same amount of liquid. The radii of the tops of the liquid surfaces are &lt;math&gt;3&lt;/math&gt; cm and &lt;math&gt;6&lt;/math&gt; cm. Into each cone is dropped a spherical marble of radius &lt;math&gt;1&lt;/math&gt; cm, which sinks to the bottom and is completely submerged without spilling any liquid. What is the ratio of the rise of the liquid level in the narrow cone to the rise of the liquid level in the wide cone?<br /> <br /> &lt;asy&gt;<br /> size(350);<br /> defaultpen(linewidth(0.8));<br /> real h1 = 10, r = 3.1, s=0.75;<br /> pair P = (r,h1), Q = (-r,h1), Pp = s * P, Qp = s * Q;<br /> path e = ellipse((0,h1),r,0.9), ep = ellipse((0,h1*s),r*s,0.9);<br /> draw(ellipse(origin,r*(s-0.1),0.8));<br /> fill(ep,gray(0.8));<br /> fill(origin--Pp--Qp--cycle,gray(0.8));<br /> draw((-r,h1)--(0,0)--(r,h1)^^e);<br /> draw(subpath(ep,0,reltime(ep,0.5)),linetype(&quot;4 4&quot;));<br /> draw(subpath(ep,reltime(ep,0.5),reltime(ep,1)));<br /> draw(Qp--(0,Qp.y),Arrows(size=8));<br /> draw(origin--(0,12),linetype(&quot;4 4&quot;));<br /> draw(origin--(r*(s-0.1),0));<br /> label(&quot;$3$&quot;,(-0.9,h1*s),N,fontsize(10));<br /> <br /> real h2 = 7.5, r = 6, s=0.6, d = 14;<br /> pair P = (d+r-0.05,h2-0.15), Q = (d-r+0.05,h2-0.15), Pp = s * P + (1-s)*(d,0), Qp = s * Q + (1-s)*(d,0);<br /> path e = ellipse((d,h2),r,1), ep = ellipse((d,h2*s+0.09),r*s,1);<br /> draw(ellipse((d,0),r*(s-0.1),0.8));<br /> fill(ep,gray(0.8));<br /> fill((d,0)--Pp--Qp--cycle,gray(0.8));<br /> draw(P--(d,0)--Q^^e);<br /> draw(subpath(ep,0,reltime(ep,0.5)),linetype(&quot;4 4&quot;));<br /> draw(subpath(ep,reltime(ep,0.5),reltime(ep,1)));<br /> draw(Qp--(d,Qp.y),Arrows(size=8));<br /> draw((d,0)--(d,10),linetype(&quot;4 4&quot;));<br /> draw((d,0)--(d+r*(s-0.1),0));<br /> label(&quot;$6$&quot;,(d-r/4,h2*s-0.06),N,fontsize(10));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }1:1 \qquad \textbf{(B) }47:43 \qquad \textbf{(C) }2:1 \qquad \textbf{(D) }40:13 \qquad \textbf{(E) }4:1&lt;/math&gt;<br /> <br /> [[2021 AMC 12A Problems/Problem 10|Solution]]<br /> <br /> ==Problem 11==<br /> A laser is placed at the point &lt;math&gt;(3,5)&lt;/math&gt;. The laser bean travels in a straight line. Larry wants the beam to hit and bounce off the &lt;math&gt;y&lt;/math&gt;-axis, then hit and bounce off the &lt;math&gt;x&lt;/math&gt;-axis, then hit the point &lt;math&gt;(7,5)&lt;/math&gt;. What is the total distance the beam will travel along this path?<br /> <br /> &lt;math&gt;\textbf{(A) }2\sqrt{10} \qquad \textbf{(B) }5\sqrt2 \qquad \textbf{(C) }10\sqrt2 \qquad \textbf{(D) }15\sqrt2 \qquad \textbf{(E) }10\sqrt5&lt;/math&gt;<br /> <br /> [[2021 AMC 12A Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> All the roots of the polynomial &lt;math&gt;z^6-10z^5+Az^4+Bz^3+Cz^2+Dz+16&lt;/math&gt; are positive integers, possibly repeated. What is the value of &lt;math&gt;B&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }-88 \qquad \textbf{(B) }-80 \qquad \textbf{(C) }-64 \qquad \textbf{(D) }-41\qquad \textbf{(E) }-40&lt;/math&gt;<br /> <br /> [[2021 AMC 12A Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> Of the following complex numbers &lt;math&gt;z&lt;/math&gt;, which one has the property that &lt;math&gt;z^5&lt;/math&gt; has the greatest real part?<br /> <br /> &lt;math&gt;\textbf{(A) }-2 \qquad \textbf{(B) }-\sqrt3+i \qquad \textbf{(C) }-\sqrt2+\sqrt2 i \qquad \textbf{(D) }-1+\sqrt3 i\qquad \textbf{(E) }2i&lt;/math&gt;<br /> <br /> [[2021 AMC 12A Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> What is the value of&lt;cmath&gt;\left(\sum_{k=1}^{20} \log_{5^k} 3^{k^2}\right)\cdot\left(\sum_{k=1}^{100} \log_{9^k} 25^k\right)?&lt;/cmath&gt;<br /> &lt;math&gt;\textbf{(A) }21 \qquad \textbf{(B) }100\log_5 3 \qquad \textbf{(C) }200\log_3 5 \qquad \textbf{(D) }2,200\qquad \textbf{(E) }21,000&lt;/math&gt;<br /> <br /> [[2021 AMC 12A Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> A choir direction must select a group of singers from among his &lt;math&gt;6&lt;/math&gt; tenors and &lt;math&gt;8&lt;/math&gt; basses. The only<br /> requirements are that the difference between the numbers of tenors and basses must be a multiple<br /> of &lt;math&gt;4&lt;/math&gt;, and the group must have at least one singer. Let &lt;math&gt;N&lt;/math&gt; be the number of different groups that could be<br /> selected. What is the remainder when &lt;math&gt;N&lt;/math&gt; is divided by &lt;math&gt;100&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 47\qquad\textbf{(B) } 48\qquad\textbf{(C) } 83\qquad\textbf{(D) } 95\qquad\textbf{(E) } 96\qquad&lt;/math&gt;<br /> <br /> [[2021 AMC 12A Problems/Problem 15|Solution]]<br /> <br /> ==Problem 16==<br /> In the following list of numbers, the integer &lt;math&gt;n&lt;/math&gt; appears &lt;math&gt;n&lt;/math&gt; times in the list for &lt;math&gt;1\le n \le 200&lt;/math&gt;.&lt;cmath&gt;1,2,2,3,3,3,4,4,4,...,200,200,...,200&lt;/cmath&gt;What is the median of the numbers in this list?<br /> <br /> &lt;math&gt;\textbf{(A) }100.5 \qquad \textbf{(B) }134 \qquad \textbf{(C) }142 \qquad \textbf{(D) }150.5\qquad \textbf{(E) }167&lt;/math&gt;<br /> <br /> [[2021 AMC 12A Problems/Problem 16|Solution]]<br /> <br /> ==Problem 17==<br /> Trapezoid &lt;math&gt;ABCD&lt;/math&gt; has &lt;math&gt;\overline{AB}\parallel\overline{CD},BC=CD=43&lt;/math&gt;, and &lt;math&gt;\overline{AD}\perp\overline{BD}&lt;/math&gt;. Let &lt;math&gt;O&lt;/math&gt; be the intersection of the diagonals &lt;math&gt;\overline{AC}&lt;/math&gt; and &lt;math&gt;\overline{BD}&lt;/math&gt;, and let &lt;math&gt;P&lt;/math&gt; be the midpoint of &lt;math&gt;\overline{BD}&lt;/math&gt;. Given that &lt;math&gt;OP=11&lt;/math&gt;, the length of &lt;math&gt;AD&lt;/math&gt; can be written in the form &lt;math&gt;m\sqrt{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are positive integers and &lt;math&gt;n&lt;/math&gt; is not divisible by the quare of any prime. What is &lt;math&gt;m+n&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }65 \qquad \textbf{(B) }132 \qquad \textbf{(C) }157 \qquad \textbf{(D) }194\qquad \textbf{(E) }215&lt;/math&gt;<br /> <br /> [[2021 AMC 12A Problems/Problem 17|Solution]]<br /> <br /> ==Problem 18==<br /> Let &lt;math&gt;f&lt;/math&gt; be a function defined on the set of positive rational numbers with the property that &lt;math&gt;f(a\cdot b)=f(a)+f(b)&lt;/math&gt; for all positive rational numbers &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;. Furthermore, suppose that &lt;math&gt;f&lt;/math&gt; also has the property that &lt;math&gt;f(p)=p&lt;/math&gt; for every prime number &lt;math&gt;p&lt;/math&gt;. For which of the following numbers &lt;math&gt;x&lt;/math&gt; is &lt;math&gt;f(x)&lt;0&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{17}{32} \qquad \textbf{(B) }\frac{11}{16} \qquad \textbf{(C) }\frac79 \qquad \textbf{(D) }\frac76\qquad \textbf{(E) }\frac{25}{11}&lt;/math&gt;<br /> <br /> [[2021 AMC 12A Problems/Problem 18|Solution]]<br /> <br /> ==Problem 19==<br /> How many solutions does the equation &lt;math&gt;\sin \left( \frac{\pi}2 \cos x\right)=\cos \left( \frac{\pi}2 \sin x\right)&lt;/math&gt; have in the closed interval &lt;math&gt;[0,\pi]&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }0 \qquad \textbf{(B) }1 \qquad \textbf{(C) }2 \qquad \textbf{(D) }3\qquad \textbf{(E) }4&lt;/math&gt;<br /> <br /> [[2021 AMC 12A Problems/Problem 19|Solution]]<br /> <br /> ==Problem 20==<br /> Suppose that on a parabola with vertex &lt;math&gt;V&lt;/math&gt; and a focus &lt;math&gt;F&lt;/math&gt; there exists a point &lt;math&gt;A&lt;/math&gt; such that &lt;math&gt;AF=20&lt;/math&gt; and &lt;math&gt;AV=21&lt;/math&gt;. What is the sum of all possible values of the length &lt;math&gt;FV?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }13 \qquad \textbf{(B) }\frac{40}3 \qquad \textbf{(C) }\frac{41}3 \qquad \textbf{(D) }14\qquad \textbf{(E) }\frac{43}3&lt;/math&gt;<br /> <br /> [[2021 AMC 12A Problems/Problem 20|Solution]]<br /> <br /> ==Problem 21==<br /> The five solutions to the equation&lt;cmath&gt;(z-1)(z^2+2z+4)(z^2+4z+6)=0&lt;/cmath&gt;may be written in the form &lt;math&gt;x_k+y_ki&lt;/math&gt; for &lt;math&gt;1\le k\le 5,&lt;/math&gt; where &lt;math&gt;x_k&lt;/math&gt; and &lt;math&gt;y_k&lt;/math&gt; are real. Let &lt;math&gt;\mathcal E&lt;/math&gt; be the unique ellipse that passes through the points &lt;math&gt;(x_1,y_1),(x_2,y_2),(x_3,y_3),(x_4,y_4),&lt;/math&gt; and &lt;math&gt;(x_5,y_5)&lt;/math&gt;. The eccentricity of &lt;math&gt;\mathcal E&lt;/math&gt; can be written in the form &lt;math&gt;\sqrt{\frac mn}&lt;/math&gt; where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. What is &lt;math&gt;m+n&lt;/math&gt;? (Recall that the eccentricity of an ellipse &lt;math&gt;\mathcal E&lt;/math&gt; is the ratio &lt;math&gt;\frac ca&lt;/math&gt;, where &lt;math&gt;2a&lt;/math&gt; is the length of the major axis of &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;2c&lt;/math&gt; is the is the distance between its two foci.)<br /> <br /> &lt;math&gt;\textbf{(A) }7 \qquad \textbf{(B) }9 \qquad \textbf{(C) }11 \qquad \textbf{(D) }13\qquad \textbf{(E) }15&lt;/math&gt;<br /> <br /> [[2021 AMC 12A Problems/Problem 21|Solution]]<br /> <br /> ==Problem 22==<br /> Suppose that the roots of the polynomial &lt;math&gt;P(x)=x^3+ax^2+bx+c&lt;/math&gt; are &lt;math&gt;\cos \frac{2\pi}7,\cos \frac{4\pi}7,&lt;/math&gt; and &lt;math&gt;\cos \frac{6\pi}7&lt;/math&gt;, where angles are in radians. What is &lt;math&gt;abc&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }-\frac{3}{49} \qquad \textbf{(B) }-\frac{1}{28} \qquad \textbf{(C) }\frac{\sqrt7}{64} \qquad \textbf{(D) }\frac{1}{32}\qquad \textbf{(E) }\frac{1}{28}&lt;/math&gt;<br /> <br /> [[2021 AMC 12A Problems/Problem 22|Solution]]<br /> <br /> ==Problem 23==<br /> Frieda the frog begins a sequence of hops on a &lt;math&gt;3\times3&lt;/math&gt; grid of squares, moving one square on each hop and choosing at random the direction of each hop up, down, left, or right. She does not hop diagonally. When the direction of a hop would take Frieda off the grid, she &quot;wraps around&quot; and jumps to the opposite edge. For example if Frieda begins in the center square and makes two hops &quot;up&quot;, the first hop would place her in the top row middle square, and the second hop would cause Frieda to jump to the opposite edge, landing in the bottom row middle square. Suppose Frieda starts from the center square, makes at most four hops at random, and stops hopping if she lands on a corner square. What is the probability that she reaches a corner square on one of the four hops?<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{9}{16} \qquad \textbf{(B) }\frac{5}{8} \qquad \textbf{(C) }\frac34 \qquad \textbf{(D) }\frac{25}{32}\qquad \textbf{(E) }\frac{13}{16}&lt;/math&gt;<br /> <br /> [[2021 AMC 12A Problems/Problem 23|Solution]]<br /> <br /> ==Problem 24==<br /> Semicircle &lt;math&gt;\Gamma&lt;/math&gt; has diameter &lt;math&gt;\overline{AB}&lt;/math&gt; of length &lt;math&gt;14&lt;/math&gt;. Circle &lt;math&gt;\Omega&lt;/math&gt; lies tangent to &lt;math&gt;\overline{AB}&lt;/math&gt; at a point &lt;math&gt;P&lt;/math&gt; and intersects &lt;math&gt;\Gamma&lt;/math&gt; at points &lt;math&gt;Q&lt;/math&gt; and &lt;math&gt;R&lt;/math&gt;. If &lt;math&gt;QR=3\sqrt3&lt;/math&gt; and &lt;math&gt;\angle QPR=60^\circ&lt;/math&gt;, then the area of &lt;math&gt;\triangle PQR&lt;/math&gt; equals &lt;math&gt;\frac{a\sqrt{b}}{c}&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; are relatively prime positive integers, and &lt;math&gt;b&lt;/math&gt; is a positive integer not divisible by the square of any prime. What is &lt;math&gt;a+b+c&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }110 \qquad \textbf{(B) }114 \qquad \textbf{(C) }118 \qquad \textbf{(D) }122\qquad \textbf{(E) }126&lt;/math&gt;<br /> <br /> [[2021 AMC 12A Problems/Problem 24|Solution]]<br /> <br /> ==Problem 25==<br /> Let &lt;math&gt;d(n)&lt;/math&gt; denote the number of positive integers that divide &lt;math&gt;n&lt;/math&gt;, including &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt;. For example, &lt;math&gt;d(1)=1,d(2)=2,&lt;/math&gt; and &lt;math&gt;d(12)=6&lt;/math&gt;. (This function is known as the divisor function.) Let&lt;cmath&gt;f(n)=\frac{d(n)}{\sqrt n}.&lt;/cmath&gt;There is a unique positive integer &lt;math&gt;N&lt;/math&gt; such that &lt;math&gt;f(N)&gt;f(n)&lt;/math&gt; for all positive integers &lt;math&gt;n\ne N&lt;/math&gt;. What is the sum of the digits of &lt;math&gt;N?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }5 \qquad \textbf{(B) }6 \qquad \textbf{(C) }7 \qquad \textbf{(D) }8\qquad \textbf{(E) }9&lt;/math&gt;<br /> <br /> [[2021 AMC 12A Problems/Problem 25|Solution]]<br /> <br /> ==See also==<br /> {{AMC12 box|year=2021|ab=A|before=[[2020 AMC 12B Problems]]|after=[[2021 AMC 12B Problems]]}}<br /> <br /> [[Category:AMC 12 Problems]]<br /> {{MAA Notice}}</div> Scrabbler94 https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_10B_Problems/Problem_11&diff=146547 2021 AMC 10B Problems/Problem 11 2021-02-14T03:13:40Z <p>Scrabbler94: solution 2 is exactly the same as solution 1 except x,y are used instead of m,n. Added more detail to solution 1.</p> <hr /> <div>==Problem==<br /> Grandma has just finished baking a large rectangular pan of brownies. She is planning to make rectangular pieces of equal size and shape, with straight cuts parallel to the sides of the pan. Each cut must be made entirely across the pan. Grandma wants to make the same number of interior pieces as pieces along the perimeter of the pan. What is the greatest possible number of brownies she can produce?<br /> <br /> &lt;math&gt;\textbf{(A)} ~24 \qquad\textbf{(B)} ~30 \qquad\textbf{(C)} ~48 \qquad\textbf{(D)} ~60 \qquad\textbf{(E)} ~64&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Let the side lengths of the rectangular pan be &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt;. It follows that &lt;math&gt;(m-2)(n-2) = \frac{mn}{2}&lt;/math&gt;, since half of the brownie pieces are in the interior. This gives &lt;math&gt;2(m-2)(n-2) = mn \iff mn - 2m - 2n - 4 = 0&lt;/math&gt;. Adding 8 to both sides and applying [[Simon's Favorite Factoring Trick]], we obtain &lt;math&gt;(m-2)(n-2) = 8&lt;/math&gt;. Since &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are both positive, we obtain &lt;math&gt;(m, n) = (5, 12), (6, 8)&lt;/math&gt; (up to ordering). By inspection, &lt;math&gt;5 \cdot 12 = \boxed{\textbf{(D) }60}&lt;/math&gt; maximizes the number of brownies.<br /> <br /> ~ ike.chen<br /> <br /> ==Solution 2==<br /> Obviously, no side of the rectangular pan can have less than &lt;math&gt;5&lt;/math&gt; brownies beside it. We let one side of the pan have &lt;math&gt;5&lt;/math&gt; brownies, and let the number of brownies on its adjacent side be &lt;math&gt;x&lt;/math&gt;. Therefore, &lt;math&gt;5x=2\cdot3(x-2)&lt;/math&gt;, and solving yields &lt;math&gt;x=12&lt;/math&gt; and there are &lt;math&gt;5\cdot12=60&lt;/math&gt; brownies in the pan. &lt;math&gt;64&lt;/math&gt; is the only choice larger than &lt;math&gt;60&lt;/math&gt;, but it cannot be the answer since the only way to fit &lt;math&gt;64&lt;/math&gt; brownies in a pan without letting a side of it have less than &lt;math&gt;5&lt;/math&gt; brownies beside it is by forming a square of &lt;math&gt;8&lt;/math&gt; brownies on each side, which does not meet the requirement. Thus the answer is &lt;math&gt;\boxed{\textbf{(D) }60}&lt;/math&gt;.<br /> <br /> -SmileKat32<br /> <br /> == Video Solution by OmegaLearn (Simon's Favorite Factoring Trick) ==<br /> https://youtu.be/vWlRQiyt0c8<br /> <br /> ~ pi_is_3.14<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2021|ab=B|num-b=10|num-a=12}}<br /> {{MAA Notice}}</div> Scrabbler94 https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12A_Problems/Problem_10&diff=146105 2021 AMC 12A Problems/Problem 10 2021-02-12T18:14:03Z <p>Scrabbler94: add a quick solution</p> <hr /> <div>==Problem==<br /> Two right circular cones with vertices facing down as shown in the figure below contains the same amount of liquid. The radii of the tops of the liquid surfaces are &lt;math&gt;3&lt;/math&gt; cm and &lt;math&gt;6&lt;/math&gt; cm. Into each cone is dropped a spherical marble of radius &lt;math&gt;1&lt;/math&gt; cm, which sinks to the bottom and is completely submerged without spilling any liquid. What is the ratio of the rise of the liquid level in the narrow cone to the rise of the liquid level in the wide cone?<br /> <br /> &lt;asy&gt;<br /> size(350);<br /> defaultpen(linewidth(0.8));<br /> real h1 = 10, r = 3.1, s=0.75;<br /> pair P = (r,h1), Q = (-r,h1), Pp = s * P, Qp = s * Q;<br /> path e = ellipse((0,h1),r,0.9), ep = ellipse((0,h1*s),r*s,0.9);<br /> draw(ellipse(origin,r*(s-0.1),0.8));<br /> fill(ep,gray(0.8));<br /> fill(origin--Pp--Qp--cycle,gray(0.8));<br /> draw((-r,h1)--(0,0)--(r,h1)^^e);<br /> draw(subpath(ep,0,reltime(ep,0.5)),linetype(&quot;4 4&quot;));<br /> draw(subpath(ep,reltime(ep,0.5),reltime(ep,1)));<br /> draw(Qp--(0,Qp.y),Arrows(size=8));<br /> draw(origin--(0,12),linetype(&quot;4 4&quot;));<br /> draw(origin--(r*(s-0.1),0));<br /> label(&quot;$3$&quot;,(-0.9,h1*s),N,fontsize(10));<br /> <br /> real h2 = 7.5, r = 6, s=0.6, d = 14;<br /> pair P = (d+r-0.05,h2-0.15), Q = (d-r+0.05,h2-0.15), Pp = s * P + (1-s)*(d,0), Qp = s * Q + (1-s)*(d,0);<br /> path e = ellipse((d,h2),r,1), ep = ellipse((d,h2*s+0.09),r*s,1);<br /> draw(ellipse((d,0),r*(s-0.1),0.8));<br /> fill(ep,gray(0.8));<br /> fill((d,0)--Pp--Qp--cycle,gray(0.8));<br /> draw(P--(d,0)--Q^^e);<br /> draw(subpath(ep,0,reltime(ep,0.5)),linetype(&quot;4 4&quot;));<br /> draw(subpath(ep,reltime(ep,0.5),reltime(ep,1)));<br /> draw(Qp--(d,Qp.y),Arrows(size=8));<br /> draw((d,0)--(d,10),linetype(&quot;4 4&quot;));<br /> draw((d,0)--(d+r*(s-0.1),0));<br /> label(&quot;$6$&quot;,(d-r/4,h2*s-0.06),N,fontsize(10));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }1 \qquad \textbf{(B) }\frac{47}{43} \qquad \textbf{(C) }2 \qquad \textbf{(D) }\frac{40}{13} \qquad \textbf{(E) }4&lt;/math&gt;<br /> <br /> ==Solution (Fraction Trick)==<br /> &lt;b&gt;Initially:&lt;/b&gt;<br /> <br /> For the narrow cone liquid, the base radius is &lt;math&gt;3.&lt;/math&gt; Let its height be &lt;math&gt;h_1.&lt;/math&gt; By similar triangles, the ratio of base radius to height is &lt;math&gt;\frac{3}{h_1}.&lt;/math&gt; The volume is &lt;math&gt;\frac13\pi(3)^2h_1=3\pi h_1.&lt;/math&gt;<br /> <br /> For the wide cone liquid, the base radius is &lt;math&gt;6.&lt;/math&gt; Let its height be &lt;math&gt;h_2.&lt;/math&gt; By similar triangles, the ratio of base radius to height is &lt;math&gt;\frac{6}{h_2}.&lt;/math&gt; The volume is &lt;math&gt;\frac13\pi(6)^2h_2=12\pi h_2.&lt;/math&gt;<br /> <br /> Equating initial volumes gives &lt;math&gt;3\pi h_1=12\pi h_2,&lt;/math&gt; from which &lt;math&gt;\frac{h_1}{h_2}=4.&lt;/math&gt;<br /> <br /> &lt;b&gt;Finally:&lt;/b&gt;<br /> <br /> For the narrow cone liquid, the base radius is &lt;math&gt;3x,&lt;/math&gt; where &lt;math&gt;x&gt;1.&lt;/math&gt; By similar triangles, it follows that its height is &lt;math&gt;h_1x&lt;/math&gt; and its volume is &lt;math&gt;3\pi h_1 x^3.&lt;/math&gt;<br /> <br /> For the wide cone liquid, the base radius is &lt;math&gt;6y,&lt;/math&gt; where &lt;math&gt;y&gt;1.&lt;/math&gt; By similar triangles, it follows that its height is &lt;math&gt;h_2y&lt;/math&gt; and its volume is &lt;math&gt;12\pi h_2 y^3.&lt;/math&gt;<br /> <br /> Equating final volumes simplifies to &lt;math&gt;x^3=y^3,&lt;/math&gt; or &lt;math&gt;x=y.&lt;/math&gt;<br /> <br /> Lastly, the fraction we seek is &lt;cmath&gt;\frac{h_1 x - h_1}{h_2 y - h_2}=\frac{h_1 (x-1)}{h_2 (y-1)}=\frac{h_1}{h_2}=\boxed{\textbf{(E) }4}.&lt;/cmath&gt;<br /> <br /> &lt;b&gt;PS:&lt;/b&gt;<br /> <br /> 1. This problem uses the following fraction trick:<br /> <br /> For unequal positive numbers &lt;math&gt;a,b,c&lt;/math&gt; and &lt;math&gt;d,&lt;/math&gt; if &lt;math&gt;\frac ab = \frac cd = k,&lt;/math&gt; then &lt;math&gt;\frac{a\pm c}{b\pm d}=k.&lt;/math&gt;<br /> <br /> &lt;u&gt;Quick Proof&lt;/u&gt;<br /> <br /> From &lt;math&gt;\frac ab = \frac cd = k,&lt;/math&gt; we know that &lt;math&gt;a=bk&lt;/math&gt; and &lt;math&gt;c=dk&lt;/math&gt;. Therefore, &lt;cmath&gt;\frac{a\pm c}{b\pm d}=\frac{bk\pm dk}{b\pm d}=\frac{(b\pm d)k}{b\pm d}=k.&lt;/cmath&gt;<br /> <br /> 2. Most of the steps can be done through mental math. Also, drawing a table showing the initial and final measurements can effectively organize the work.<br /> <br /> ~MRENTHUSIASM<br /> <br /> ==Solution 2 (Quick and dirty)==<br /> The heights of the cones are not given, so suppose the heights are very large (i.e. tending towards infinity) in order to approximate the cones as cylinders with base radii 3 and 6 and infinitely large height. Then the base area of the wide cylinder is 4 times that of the narrow cylinder. Since we are dropping a ball of the same volume into each cylinder, the water level in the narrow cone/cylinder should rise &lt;math&gt;\boxed{\textbf{(E) } 4}&lt;/math&gt; times as much.<br /> <br /> -scrabbler94<br /> <br /> <br /> ==Video Solution by Hawk Math==<br /> https://www.youtube.com/watch?v=AjQARBvdZ20<br /> <br /> == Video Solution by OmegaLearn (Similar Triangles, 3D Geometry - Cones) ==<br /> https://youtu.be/4Iuo7cvGJr8<br /> <br /> ~ pi_is_3.14<br /> <br /> ==See also==<br /> {{AMC10 box|year=2021|ab=A|num-b=11|num-a=13}}<br /> {{AMC12 box|year=2021|ab=A|num-b=9|num-a=11}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> {{MAA Notice}}</div> Scrabbler94 https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_4&diff=144800 2017 AMC 10A Problems/Problem 4 2021-02-03T17:13:00Z <p>Scrabbler94: </p> <hr /> <div>==Problem==<br /> Mia is &quot;helping&quot; her mom pick up &lt;math&gt;30&lt;/math&gt; toys that are strewn on the floor. Mia’s mom manages to put &lt;math&gt;3&lt;/math&gt; toys into the toy box every &lt;math&gt;30&lt;/math&gt; seconds, but each time immediately after those &lt;math&gt;30&lt;/math&gt; seconds have elapsed, Mia takes &lt;math&gt;2&lt;/math&gt; toys out of the box. How much time, in minutes, will it take Mia and her mom to put all &lt;math&gt;30&lt;/math&gt; toys into the box for the first time?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 13.5\qquad\textbf{(B)}\ 14\qquad\textbf{(C)}\ 14.5\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 15.5&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Every &lt;math&gt;30&lt;/math&gt; seconds, &lt;math&gt;3&lt;/math&gt; toys are put in the box and &lt;math&gt;2&lt;/math&gt; toys are taken out, so the number of toys increases by &lt;math&gt;3-2=1&lt;/math&gt; every &lt;math&gt;30&lt;/math&gt; seconds. Then after &lt;math&gt;27 \times 30 = 810&lt;/math&gt; seconds (or &lt;math&gt;13 \frac{1}{2}&lt;/math&gt; minutes), there are &lt;math&gt;27&lt;/math&gt; toys in the box. Mia's mom will then put the remaining &lt;math&gt;3&lt;/math&gt; toys into the box after &lt;math&gt;30&lt;/math&gt; more seconds, so the total time taken is &lt;math&gt;27\times 30+30=840&lt;/math&gt; seconds, or &lt;math&gt;\boxed{(\textbf{B})\ 14}&lt;/math&gt; minutes.<br /> <br /> ==Video Solution==<br /> https://youtu.be/str7kmcRMY8<br /> <br /> https://youtu.be/1F0IB0y6578<br /> <br /> ~savannahsolver<br /> <br /> ==See also==<br /> <br /> {{AMC10 box|year=2017|ab=A|num-b=3|num-a=5}}<br /> {{MAA Notice}}</div> Scrabbler94 https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_4&diff=144799 2017 AMC 10A Problems/Problem 4 2021-02-03T17:12:23Z <p>Scrabbler94: /* Solution */</p> <hr /> <div>==Problem==<br /> Mia is &quot;helping&quot; her mom pick up &lt;math&gt;30&lt;/math&gt; toys that are strewn on the floor. Mia’s mom manages to put &lt;math&gt;3&lt;/math&gt; toys into the toy box every &lt;math&gt;30&lt;/math&gt; seconds, but each time immediately after those &lt;math&gt;30&lt;/math&gt; seconds have elapsed, Mia takes &lt;math&gt;2&lt;/math&gt; toys out of the box. How much time, in minutes, will it take Mia and her mom to put all &lt;math&gt;30&lt;/math&gt; toys into the box for the first time?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 13.5\qquad\textbf{(B)}\ 14\qquad\textbf{(C)}\ 14.5\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 15.5&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Every &lt;math&gt;30&lt;/math&gt; seconds, &lt;math&gt;3&lt;/math&gt; toys are put in the box and &lt;math&gt;2&lt;/math&gt; toys are taken out, so the number of toys increases by &lt;math&gt;3-2=1&lt;/math&gt; every &lt;math&gt;30&lt;/math&gt; seconds. Then after &lt;math&gt;27 \times 30 = 810&lt;/math&gt; seconds (or &lt;math&gt;13 \frac{1}{2}&lt;/math&gt; minutes), there are &lt;math&gt;27&lt;/math&gt; toys in the box. Mia's mom will then put &lt;math&gt;3&lt;/math&gt; toys into the box after &lt;math&gt;30&lt;/math&gt; more seconds, so the total time taken is &lt;math&gt;27\times 30+30=840&lt;/math&gt; seconds, or &lt;math&gt;\boxed{(\textbf{B})\ 14}&lt;/math&gt; minutes.<br /> <br /> ==Video Solution==<br /> https://youtu.be/str7kmcRMY8<br /> <br /> https://youtu.be/1F0IB0y6578<br /> <br /> ~savannahsolver<br /> <br /> ==See also==<br /> <br /> {{AMC10 box|year=2017|ab=A|num-b=3|num-a=5}}<br /> {{MAA Notice}}</div> Scrabbler94 https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_4&diff=144798 2017 AMC 10A Problems/Problem 4 2021-02-03T17:11:47Z <p>Scrabbler94: combine solutions 1 and 2 as they are essentially the same. also clarify that the number of toys increases by 1; not necessarily &quot;1 toy is placed in the box&quot; since different toys can be put in and taken out.</p> <hr /> <div>==Problem==<br /> Mia is &quot;helping&quot; her mom pick up &lt;math&gt;30&lt;/math&gt; toys that are strewn on the floor. Mia’s mom manages to put &lt;math&gt;3&lt;/math&gt; toys into the toy box every &lt;math&gt;30&lt;/math&gt; seconds, but each time immediately after those &lt;math&gt;30&lt;/math&gt; seconds have elapsed, Mia takes &lt;math&gt;2&lt;/math&gt; toys out of the box. How much time, in minutes, will it take Mia and her mom to put all &lt;math&gt;30&lt;/math&gt; toys into the box for the first time?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 13.5\qquad\textbf{(B)}\ 14\qquad\textbf{(C)}\ 14.5\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 15.5&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Every &lt;math&gt;30&lt;/math&gt; seconds, &lt;math&gt;3&lt;/math&gt; toys are put in the box and &lt;math&gt;2&lt;/math&gt; toys are taken out, so the number of toys increases by &lt;math&gt;3-2=1&lt;/math&gt; every 30 seconds. Then after &lt;math&gt;27 \times 30 = 810&lt;/math&gt; seconds (or &lt;math&gt;13 \frac{1}{2}&lt;/math&gt; minutes), there are 27 toys in the box. Mia's mom will then put &lt;math&gt;3&lt;/math&gt; toys into the box after 30 more seconds, so the total time taken is &lt;math&gt;27\cdot30+30=840&lt;/math&gt; seconds, or &lt;math&gt;\boxed{(\textbf{B})\ 14}&lt;/math&gt; minutes.<br /> <br /> ==Video Solution==<br /> https://youtu.be/str7kmcRMY8<br /> <br /> https://youtu.be/1F0IB0y6578<br /> <br /> ~savannahsolver<br /> <br /> ==See also==<br /> <br /> {{AMC10 box|year=2017|ab=A|num-b=3|num-a=5}}<br /> {{MAA Notice}}</div> Scrabbler94 https://artofproblemsolving.com/wiki/index.php?title=User:Pi-r2&diff=143753 User:Pi-r2 2021-01-29T16:44:07Z <p>Scrabbler94: Undo revision 143523 by Awesome3.14 (talk)</p> <hr /> <div>Hoi I am Pi</div> Scrabbler94 https://artofproblemsolving.com/wiki/index.php?title=2000_AMC_12_Answer_Key&diff=143750 2000 AMC 12 Answer Key 2021-01-29T16:41:17Z <p>Scrabbler94: Undo revision 143614 by Awesome3.14 (talk) please do not vandalize the AoPS wiki pages.</p> <hr /> <div>#E<br /> #A<br /> #B<br /> #C<br /> #C<br /> #C<br /> #E<br /> #C<br /> #C<br /> #E<br /> #E<br /> #E<br /> #C<br /> #E<br /> #B<br /> #D<br /> #D<br /> #A<br /> #C<br /> #B<br /> #D<br /> #C<br /> #B<br /> #D<br /> #E</div> Scrabbler94 https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10B_Problems/Problem_23&diff=138946 2011 AMC 10B Problems/Problem 23 2020-12-03T01:07:11Z <p>Scrabbler94: need gcd(a,1000) = 1</p> <hr /> <div>==Problem==<br /> <br /> What is the hundreds digit of &lt;math&gt;2011^{2011}?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 1 \qquad \textbf{(B) } 4 \qquad \textbf{(C) }5 \qquad \textbf{(D) } 6 \qquad \textbf{(E) } 9&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> Since &lt;math&gt;2011 \equiv 11 \pmod{1000},&lt;/math&gt; we know that &lt;math&gt;2011^{2011} \equiv 11^{2011} \pmod{1000}.&lt;/math&gt;<br /> <br /> To compute this, we use a clever application of the [[binomial theorem]].<br /> <br /> &lt;math&gt;\begin{aligned} 11^{2011} &amp;= (1+10)^{2011} \\ &amp;= 1 + \dbinom{2011}{1} \cdot 10 + \dbinom{2011}{2} \cdot 10^2 + \cdots \end{aligned}&lt;/math&gt;<br /> <br /> In all of the other terms, the power of &lt;math&gt;10&lt;/math&gt; is greater than &lt;math&gt;3&lt;/math&gt; and so is equivalent to &lt;math&gt;0&lt;/math&gt; modulo &lt;math&gt;1000,&lt;/math&gt; which means we can ignore it. We have:<br /> <br /> &lt;math&gt;\begin{aligned}11^{2011} &amp;\equiv 1 + 2011\cdot 10 + \dfrac{2011 \cdot 2010}{2} \cdot 100 \\ &amp;\equiv 1+20110 + \dfrac{11\cdot 10}{2} \cdot 100\\ &amp;= 1 + 20110 + 5500\\ &amp;\equiv 1 + 110 + 500\\&amp;=611 \pmod{1000} \end{aligned}&lt;/math&gt;<br /> <br /> Therefore, the hundreds digit is &lt;math&gt;\boxed{\textbf{(D) } 6}.&lt;/math&gt;<br /> <br /> Side note: By [[Euler's Totient Theorem]], &lt;math&gt;a^{\phi (1000)} \equiv 1 \pmod{1000}&lt;/math&gt; for any &lt;math&gt;a&lt;/math&gt; relatively prime with 1000, so &lt;math&gt;a^{400} \equiv 1 \pmod{1000}&lt;/math&gt; and &lt;math&gt;11^{2011} \equiv 11^{11} \pmod{1000}&lt;/math&gt;. We can then proceed using the clever application of the Binomial Theorem.<br /> <br /> == Solution 2 ==<br /> <br /> We need to compute &lt;math&gt;2011^{2011} \pmod{1000}.&lt;/math&gt; By the [[Chinese Remainder Theorem]], it suffices to compute &lt;math&gt;2011^{2011} \pmod{8}&lt;/math&gt; and &lt;math&gt;2011^{2011} \pmod{125}.&lt;/math&gt;<br /> <br /> In modulo &lt;math&gt;8,&lt;/math&gt; we have &lt;math&gt;2011^4 \equiv 1 \pmod{8}&lt;/math&gt; by Euler's Theorem, and also &lt;math&gt;2011 \equiv 3 \pmod{8},&lt;/math&gt; so we have &lt;cmath&gt;2011^{2011} = (2011^4)^{502} \cdot 2011^3 \equiv 1^{502} \cdot 3^3 \equiv 3 \pmod{8}.&lt;/cmath&gt;<br /> <br /> In modulo &lt;math&gt;125,&lt;/math&gt; we have &lt;math&gt;2011^{100} \equiv 1 \pmod{125}&lt;/math&gt; by Euler's Theorem, and also &lt;math&gt;2011 \equiv 11 \pmod{125}.&lt;/math&gt; Therefore, we have &lt;math&gt;\begin{aligned} 2011^{2011} &amp;= (2011^{100})^{20} \cdot 2011^{11} \\ &amp;\equiv 1^{20} \cdot 11^{11} \\ &amp;= 121^5 \cdot 11 \\ &amp;= (-4)^5 \cdot 11 = -1024 \cdot 11 \\ &amp;\equiv -24 \cdot 11 = -264 \\ &amp;\equiv 111 \pmod{125}. \end{aligned} &lt;/math&gt;<br /> <br /> After finding the solution &lt;math&gt;2011^{2011} \equiv 611 \pmod{1000},&lt;/math&gt; we conclude it is the only one by the Chinese Remainder Theorem. Thus, the hundreds digit is &lt;math&gt;\boxed{\textbf{(D) } 6}.&lt;/math&gt;<br /> <br /> == Solution 3 ==<br /> <br /> Notice that the hundreds digit of &lt;math&gt;2011^{2011}&lt;/math&gt; won't be affected by &lt;math&gt;2000&lt;/math&gt;. Essentially we could solve the problem by finding the hundreds digit of &lt;math&gt;11^{2011}&lt;/math&gt;. Powers of &lt;math&gt;11&lt;/math&gt; are special because they can be represented by the Pascal's Triangle. Drawing the triangle, there is a theorem that states the powers of &lt;math&gt;11&lt;/math&gt; can be found by reading rows of the triangle and adding extra numbers up. [add source] For example, the sixth row of the triangle is &lt;math&gt;1, 5, 10, 10, 5,&lt;/math&gt; and &lt;math&gt;1&lt;/math&gt;. Adding all numbers from right to left, we get &lt;math&gt;161051&lt;/math&gt;, which is also &lt;math&gt;11^5&lt;/math&gt;. In other words, each number is &lt;math&gt;10^n&lt;/math&gt; steps from the right side of the row. The hundreds digit is &lt;math&gt;0&lt;/math&gt;. We can do the same for &lt;math&gt;11^{2011}&lt;/math&gt;, but we only need to find the &lt;math&gt;3&lt;/math&gt; digits from the right. Observing, every &lt;math&gt;3&lt;/math&gt; number from the right is &lt;math&gt;1 + 2 + 3... + n&lt;/math&gt;. So to find the third number from the right on the row of &lt;math&gt;11^{2011}&lt;/math&gt;, &lt;cmath&gt;f(11^n) = 1 + 2 + 3... + (n-1),&lt;/cmath&gt; or &lt;math&gt;\frac{(2010 \cdot 2011)}{2}&lt;/math&gt;, or &lt;math&gt;2021055&lt;/math&gt;. The last digit is five, but we must remember to add the number on the right of it, which, by observing other rows is obviously &lt;math&gt;2011&lt;/math&gt;. We must carry the &lt;math&gt;1&lt;/math&gt; in &lt;math&gt;2011&lt;/math&gt;'s tens digit to the &lt;math&gt;5&lt;/math&gt; in &lt;math&gt;2021055&lt;/math&gt;'s unit digit to get &lt;math&gt;\boxed{\textbf{(D) } 6}&lt;/math&gt;. The one at the very end of the row doesn't affect anything, so we can leave it alone.<br /> <br /> <br /> -jackshi2006<br /> <br /> ==Solution 4 (naive solution, EXTREMELY bashy) ==<br /> <br /> Since we are only looking at the last 3 digits and &lt;math&gt;2011^2&lt;/math&gt; has the same last 3 digits as &lt;math&gt;11^2&lt;/math&gt;, we can find &lt;math&gt;11^{2011}&lt;/math&gt; instead.<br /> After this, we can repeatedly multiply the last 3 digits by 11 and take the last 3 digits of that product. We discover that &lt;math&gt;11^{51}&lt;/math&gt;'s last 2 digits are -11, the same as &lt;math&gt;11^1&lt;/math&gt;. <br /> <br /> From this information, we can figure out &lt;math&gt;11^{11}&lt;/math&gt; and &lt;math&gt;11^{61}&lt;/math&gt; end in 611. Adding various multiples of 50 to the exponent gives us the fact that &lt;math&gt;11^{2011}&lt;/math&gt;'s last digits are 611. We get &lt;math&gt;\boxed{\textbf{(D) } 6}&lt;/math&gt;.<br /> -ThisUsernameIsTaken<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2011|ab=B|num-a=24|num-b=22}}<br /> {{MAA Notice}}</div> Scrabbler94 https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_8_Problems/Problem_24&diff=137866 2020 AMC 8 Problems/Problem 24 2020-11-19T21:34:54Z <p>Scrabbler94: add quick solution</p> <hr /> <div>A large square region is paved with &lt;math&gt;n^2&lt;/math&gt; gray square tiles, each measuring &lt;math&gt;s&lt;/math&gt; inches on a side. A border &lt;math&gt;d&lt;/math&gt; inches wide surrounds each tile. The figure below shows the case for &lt;math&gt;n=3&lt;/math&gt;. When &lt;math&gt;n=24&lt;/math&gt;, the &lt;math&gt;576&lt;/math&gt; gray tiles cover &lt;math&gt;64\%&lt;/math&gt; of the area of the large square region. What is the ratio &lt;math&gt;\frac{d}{s}&lt;/math&gt; for this larger value of &lt;math&gt;n?&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(13,0)--(13,13)--(0,13)--cycle);<br /> filldraw((1,1)--(4,1)--(4,4)--(1,4)--cycle, mediumgray);<br /> filldraw((1,5)--(4,5)--(4,8)--(1,8)--cycle, mediumgray);<br /> filldraw((1,9)--(4,9)--(4,12)--(1,12)--cycle, mediumgray);<br /> filldraw((5,1)--(8,1)--(8,4)--(5,4)--cycle, mediumgray);<br /> filldraw((5,5)--(8,5)--(8,8)--(5,8)--cycle, mediumgray);<br /> filldraw((5,9)--(8,9)--(8,12)--(5,12)--cycle, mediumgray);<br /> filldraw((9,1)--(12,1)--(12,4)--(9,4)--cycle, mediumgray);<br /> filldraw((9,5)--(12,5)--(12,8)--(9,8)--cycle, mediumgray);<br /> filldraw((9,9)--(12,9)--(12,12)--(9,12)--cycle, mediumgray);<br /> &lt;/asy&gt; <br /> <br /> &lt;math&gt;\textbf{(A) }\frac{6}{25} \qquad \textbf{(B) }\frac{1}{4} \qquad \textbf{(C) }\frac{9}{25} \qquad \textbf{(D) }\frac{7}{16} \qquad \textbf{(E) }\frac{9}{16}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> WLOG, let &lt;math&gt;s=1&lt;/math&gt;. Then, the total area of the squares of side &lt;math&gt;s&lt;/math&gt; is &lt;math&gt;576&lt;/math&gt;, &lt;math&gt;64\%&lt;/math&gt; of the area of the large square, which would be &lt;math&gt;900&lt;/math&gt;, making the side of the large square &lt;math&gt;30&lt;/math&gt;. Then, &lt;math&gt;25&lt;/math&gt; borders have a total length of &lt;math&gt;30-24=6&lt;/math&gt;. Since &lt;math&gt;\frac{d}{s}=d&lt;/math&gt; if &lt;math&gt;s=1&lt;/math&gt; is the value we're asked to find, the answer is &lt;math&gt;\boxed{\textbf{(A) }\frac{6}{25}}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> When &lt;math&gt;n=3&lt;/math&gt;, we see that the total height of the large square is &lt;math&gt;3s+4d&lt;/math&gt;. Similarly, when &lt;math&gt;n=24&lt;/math&gt;, the total height of the large square is &lt;math&gt;24s+25d&lt;/math&gt;. The total area of the &lt;math&gt;576&lt;/math&gt; gray tiles is &lt;math&gt;576s^2&lt;/math&gt; and the area of the large white square is &lt;math&gt;(24s+25d)^2&lt;/math&gt;. We are given that the ratio of the gray area to the area of the large square is &lt;math&gt;\frac{64}{100}=\frac{16}{25}&lt;/math&gt;. Thus, our equation becomes &lt;math&gt;\frac{576s^2}{(24s+25d)^2}=\frac{16}{25}&lt;/math&gt;. Square rooting both sides, we get &lt;math&gt;\frac{24s}{24s+25d}=\frac{4}{5}&lt;/math&gt;. Cross multiplying, we get &lt;math&gt;120s=96s+100d&lt;/math&gt;. Combining like terms, we get &lt;math&gt;24s=100d&lt;/math&gt;, which implies that &lt;math&gt;\frac{d}{s}=\frac{24}{100}=\frac{6}{25}\implies\boxed{\textbf{(A) }\frac{6}{25}}&lt;/math&gt;.&lt;br&gt;<br /> ~[http://artofproblemsolving.com/community/user/jmansuri junaidmansuri]<br /> <br /> ==Solution 3==<br /> <br /> The area of the shaded region is &lt;math&gt;(24s)^2&lt;/math&gt;. The total area of the square is &lt;math&gt;(24s+25d)^2&lt;/math&gt; because for each side of the square there one extra row/column of the border. Our equation is &lt;math&gt;\frac{(24s)^2}{(24x+25d)^2}=\frac{64}{100}&lt;/math&gt;. Taking the square root of both sides gives &lt;math&gt;\frac{24x}{24x+25d}=\frac 45&lt;/math&gt;. Cross multiplying and rearranging gives &lt;math&gt;\frac ds=\textbf{(A)} \frac{6}{25}&lt;/math&gt;.<br /> <br /> -franzliszt<br /> <br /> ==Solution 4==<br /> WLOG &lt;math&gt;s = 1&lt;/math&gt;. For large enough &lt;math&gt;n&lt;/math&gt;, we can approximate the percentage of the area covered by the gray tiles by subdividing most of the region into congruent squares, as shown:<br /> &lt;asy&gt;<br /> draw((0,0)--(13,0)--(13,13)--(0,13)--cycle);<br /> filldraw((1,1)--(4,1)--(4,4)--(1,4)--cycle, mediumgray);<br /> filldraw((1,5)--(4,5)--(4,8)--(1,8)--cycle, mediumgray);<br /> filldraw((1,9)--(4,9)--(4,12)--(1,12)--cycle, mediumgray);<br /> filldraw((5,1)--(8,1)--(8,4)--(5,4)--cycle, mediumgray);<br /> filldraw((5,5)--(8,5)--(8,8)--(5,8)--cycle, mediumgray);<br /> filldraw((5,9)--(8,9)--(8,12)--(5,12)--cycle, mediumgray);<br /> filldraw((9,1)--(12,1)--(12,4)--(9,4)--cycle, mediumgray);<br /> filldraw((9,5)--(12,5)--(12,8)--(9,8)--cycle, mediumgray);<br /> filldraw((9,9)--(12,9)--(12,12)--(9,12)--cycle, mediumgray);<br /> <br /> for(int i = 1; i &lt;= 13; i += 4){<br /> draw((1,i)--(13,i), red);<br /> draw((i,1)--(i,13), red);<br /> }<br /> &lt;/asy&gt; <br /> Each red square has side length &lt;math&gt;1+d&lt;/math&gt;, so by solving &lt;math&gt;\frac{1^2}{(1+d)^2} = \frac{64}{100} \iff \frac{1}{1+d} = \frac{4}{5}&lt;/math&gt;, we obtain &lt;math&gt;d = \frac{1}{4}&lt;/math&gt;. The actual fraction of area covered by the gray tiles is slightly less than &lt;math&gt;\frac{1}{(1+d)^2}&lt;/math&gt;, which implies &lt;math&gt;\frac{1}{(1+d)^2} &gt; \frac{64}{100} \iff \frac{1}{1+d} &gt; \frac{4}{5} \iff d &lt; \frac{1}{4}&lt;/math&gt;. Hence &lt;math&gt;d&lt;/math&gt; (and &lt;math&gt;\frac{d}{s}&lt;/math&gt;) is less than &lt;math&gt;\frac{1}{4}&lt;/math&gt;, and the only choice that satisfies this is &lt;math&gt;\boxed{\textbf{(A) }\frac{6}{25}}&lt;/math&gt;. ~scrabbler94<br /> <br /> ==See also==<br /> {{AMC8 box|year=2020|num-b=23|num-a=25}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> {{MAA Notice}}</div> Scrabbler94 https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_8_Problems/Problem_21&diff=137675 2020 AMC 8 Problems/Problem 21 2020-11-18T23:27:15Z <p>Scrabbler94: clarify solution 1</p> <hr /> <div>==Problem 21==<br /> A game board consists of &lt;math&gt;64&lt;/math&gt; squares that alternate in color between black and white. The figure below shows square &lt;math&gt;P&lt;/math&gt; in the bottom row and square &lt;math&gt;Q&lt;/math&gt; in the top row. A marker is placed at &lt;math&gt;P.&lt;/math&gt; A step consists of moving the marker onto one of the adjoining white squares in the row above. How many &lt;math&gt;7&lt;/math&gt;-step paths are there from &lt;math&gt;P&lt;/math&gt; to &lt;math&gt;Q?&lt;/math&gt; (The figure shows a sample path.)<br /> <br /> &lt;asy&gt;//diagram by SirCalcsALot<br /> size(200); int[] x = {6, 5, 4, 5, 6, 5, 6}; int[] y = {1, 2, 3, 4, 5, 6, 7}; int N = 7; for (int i = 0; i &lt; 8; ++i) { for (int j = 0; j &lt; 8; ++j) { draw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)); if ((i+j) % 2 == 0) { filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)--cycle,black); } } } for (int i = 0; i &lt; N; ++i) { draw(circle((x[i],y[i])+(0.5,0.5),0.35)); } label(&quot;$P$&quot;, (5.5, 0.5)); label(&quot;$Q$&quot;, (6.5, 7.5)); &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }28 \qquad \textbf{(B) }30 \qquad \textbf{(C) }32 \qquad \textbf{(D) }33 \qquad \textbf{(E) }35&lt;/math&gt;<br /> <br /> <br /> ==Solution 1==<br /> Noticing that we can only move along white squares, to get to a white square we can only go from the one or two white squares immediately beneath it. In the following diagram, each number represents the number of ways to move from &lt;math&gt;P&lt;/math&gt; to that square.<br /> &lt;asy&gt;<br /> int N = 7;<br /> for (int i = 0; i &lt; 8; ++i) {<br /> for (int j = 0; j &lt; 8; ++j) {<br /> draw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j));<br /> if ((i+j) % 2 == 0) {<br /> filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)--cycle,black);<br /> }<br /> }<br /> }<br /> label(&quot;$1$&quot;, (5.5, .5));<br /> label(&quot;$1$&quot;, (4.5, 1.5));<br /> label(&quot;$1$&quot;, (6.5, 1.5));<br /> label(&quot;$1$&quot;, (3.5, 2.5));<br /> label(&quot;$1$&quot;, (7.5, 2.5));<br /> label(&quot;$2$&quot;, (5.5, 2.5));<br /> label(&quot;$1$&quot;, (2.5, 3.5));<br /> label(&quot;$3$&quot;, (6.5, 3.5));<br /> label(&quot;$3$&quot;, (4.5, 3.5));<br /> label(&quot;$4$&quot;, (3.5, 4.5));<br /> label(&quot;$3$&quot;, (7.5, 4.5));<br /> label(&quot;$6$&quot;, (5.5, 4.5));<br /> label(&quot;$10$&quot;, (4.5, 5.5));<br /> label(&quot;$9$&quot;, (6.5, 5.5));<br /> label(&quot;$19$&quot;, (5.5, 6.5));<br /> label(&quot;$9$&quot;, (7.5, 6.5));<br /> label(&quot;$28$&quot;, (6.5, 7.5));<br /> &lt;/asy&gt;<br /> So the answer is &lt;math&gt;\boxed{\textbf{(A)}28}&lt;/math&gt;<br /> ~yofro (Diagram credits to franzliszt)<br /> <br /> ==Solution 2==<br /> Suppose we &quot;extend&quot; the chessboard indefinitely to the right:<br /> <br /> &lt;asy&gt;<br /> int N = 7;<br /> for (int i = 0; i &lt; 10; ++i) {<br /> for (int j = 0; j &lt; 8; ++j) {<br /> draw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j));<br /> if ((i+j) % 2 == 0) {<br /> filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)--cycle,black);<br /> }<br /> }<br /> }<br /> draw((8,0) -- (8,8),red);<br /> label(&quot;$P$&quot;, (5.5,.5));<br /> label(&quot;$Q$&quot;, (6.5,7.5));<br /> label(&quot;$X$&quot;, (8.5,3.5));<br /> label(&quot;$Y$&quot;, (8.5,5.5));<br /> &lt;/asy&gt;<br /> The total number of paths from &lt;math&gt;P&lt;/math&gt; to &lt;math&gt;Q&lt;/math&gt; (including invalid paths which cross over the red line) is &lt;math&gt;\binom{7}{3} = 35&lt;/math&gt;. We subtract the number of invalid paths that pass through &lt;math&gt;X&lt;/math&gt; or &lt;math&gt;Y&lt;/math&gt;. The number of paths that pass through &lt;math&gt;X&lt;/math&gt; is &lt;math&gt;\binom{3}{0}\binom{4}{3} = 4&lt;/math&gt; and the number of paths that pass through &lt;math&gt;Y&lt;/math&gt; is &lt;math&gt;\binom{5}{1}\binom{2}{2} = 5&lt;/math&gt;. However, we overcounted the invalid paths which pass through both &lt;math&gt;X&lt;/math&gt; and &lt;math&gt;Y&lt;/math&gt;, of which there are 2 paths. Hence, the number of invalid paths is &lt;math&gt;4+5-2=7&lt;/math&gt; and the number of valid paths from &lt;math&gt;P&lt;/math&gt; to &lt;math&gt;Q&lt;/math&gt; is &lt;math&gt;35-7 = \boxed{\textbf{(A)} 28}&lt;/math&gt;. -scrabbler94<br /> <br /> ==See also== <br /> {{AMC8 box|year=2020|num-b=20|num-a=22}}<br /> {{MAA Notice}}</div> Scrabbler94 https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_8_Problems/Problem_21&diff=137673 2020 AMC 8 Problems/Problem 21 2020-11-18T23:24:19Z <p>Scrabbler94: Undo revision 137613 by Franzliszt (talk) this is the exact same as solution 1.</p> <hr /> <div>==Problem 21==<br /> A game board consists of &lt;math&gt;64&lt;/math&gt; squares that alternate in color between black and white. The figure below shows square &lt;math&gt;P&lt;/math&gt; in the bottom row and square &lt;math&gt;Q&lt;/math&gt; in the top row. A marker is placed at &lt;math&gt;P.&lt;/math&gt; A step consists of moving the marker onto one of the adjoining white squares in the row above. How many &lt;math&gt;7&lt;/math&gt;-step paths are there from &lt;math&gt;P&lt;/math&gt; to &lt;math&gt;Q?&lt;/math&gt; (The figure shows a sample path.)<br /> <br /> &lt;asy&gt;//diagram by SirCalcsALot<br /> size(200); int[] x = {6, 5, 4, 5, 6, 5, 6}; int[] y = {1, 2, 3, 4, 5, 6, 7}; int N = 7; for (int i = 0; i &lt; 8; ++i) { for (int j = 0; j &lt; 8; ++j) { draw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)); if ((i+j) % 2 == 0) { filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)--cycle,black); } } } for (int i = 0; i &lt; N; ++i) { draw(circle((x[i],y[i])+(0.5,0.5),0.35)); } label(&quot;$P$&quot;, (5.5, 0.5)); label(&quot;$Q$&quot;, (6.5, 7.5)); &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }28 \qquad \textbf{(B) }30 \qquad \textbf{(C) }32 \qquad \textbf{(D) }33 \qquad \textbf{(E) }35&lt;/math&gt;<br /> <br /> <br /> ==Solution==<br /> We count paths. Noticing that we can only go along white squares, to get to a white square we can only go from the two whites beneath it. Here is a diagram:<br /> &lt;asy&gt;<br /> int N = 7;<br /> for (int i = 0; i &lt; 8; ++i) {<br /> for (int j = 0; j &lt; 8; ++j) {<br /> draw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j));<br /> if ((i+j) % 2 == 0) {<br /> filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)--cycle,black);<br /> }<br /> }<br /> }<br /> label(&quot;$1$&quot;, (5.5, .5));<br /> label(&quot;$1$&quot;, (4.5, 1.5));<br /> label(&quot;$1$&quot;, (6.5, 1.5));<br /> label(&quot;$1$&quot;, (3.5, 2.5));<br /> label(&quot;$1$&quot;, (7.5, 2.5));<br /> label(&quot;$2$&quot;, (5.5, 2.5));<br /> label(&quot;$1$&quot;, (2.5, 3.5));<br /> label(&quot;$3$&quot;, (6.5, 3.5));<br /> label(&quot;$3$&quot;, (4.5, 3.5));<br /> label(&quot;$4$&quot;, (3.5, 4.5));<br /> label(&quot;$3$&quot;, (7.5, 4.5));<br /> label(&quot;$6$&quot;, (5.5, 4.5));<br /> label(&quot;$10$&quot;, (4.5, 5.5));<br /> label(&quot;$9$&quot;, (6.5, 5.5));<br /> label(&quot;$19$&quot;, (5.5, 6.5));<br /> label(&quot;$9$&quot;, (7.5, 6.5));<br /> label(&quot;$28$&quot;, (6.5, 7.5));<br /> &lt;/asy&gt;<br /> So the answer is &lt;math&gt;\boxed{\textbf{(A)}28}&lt;/math&gt;<br /> ~yofro (Diagram credits to franzliszt)<br /> <br /> ==Solution 2==<br /> Suppose we &quot;extend&quot; the chessboard indefinitely to the right:<br /> <br /> &lt;asy&gt;<br /> int N = 7;<br /> for (int i = 0; i &lt; 10; ++i) {<br /> for (int j = 0; j &lt; 8; ++j) {<br /> draw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j));<br /> if ((i+j) % 2 == 0) {<br /> filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)--cycle,black);<br /> }<br /> }<br /> }<br /> draw((8,0) -- (8,8),red);<br /> label(&quot;$P$&quot;, (5.5,.5));<br /> label(&quot;$Q$&quot;, (6.5,7.5));<br /> label(&quot;$X$&quot;, (8.5,3.5));<br /> label(&quot;$Y$&quot;, (8.5,5.5));<br /> &lt;/asy&gt;<br /> The total number of paths from &lt;math&gt;P&lt;/math&gt; to &lt;math&gt;Q&lt;/math&gt; (including invalid paths which cross over the red line) is &lt;math&gt;\binom{7}{3} = 35&lt;/math&gt;. We subtract the number of invalid paths that pass through &lt;math&gt;X&lt;/math&gt; or &lt;math&gt;Y&lt;/math&gt;. The number of paths that pass through &lt;math&gt;X&lt;/math&gt; is &lt;math&gt;\binom{3}{0}\binom{4}{3} = 4&lt;/math&gt; and the number of paths that pass through &lt;math&gt;Y&lt;/math&gt; is &lt;math&gt;\binom{5}{1}\binom{2}{2} = 5&lt;/math&gt;. However, we overcounted the invalid paths which pass through both &lt;math&gt;X&lt;/math&gt; and &lt;math&gt;Y&lt;/math&gt;, of which there are 2 paths. Hence, the number of invalid paths is &lt;math&gt;4+5-2=7&lt;/math&gt; and the number of valid paths from &lt;math&gt;P&lt;/math&gt; to &lt;math&gt;Q&lt;/math&gt; is &lt;math&gt;35-7 = \boxed{\textbf{(A)} 28}&lt;/math&gt;. -scrabbler94<br /> <br /> ==See also== <br /> {{AMC8 box|year=2020|num-b=20|num-a=22}}<br /> {{MAA Notice}}</div> Scrabbler94 https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_8_Problems/Problem_21&diff=137578 2020 AMC 8 Problems/Problem 21 2020-11-18T16:27:24Z <p>Scrabbler94: /* Solution */ add alternate solution without using recursion</p> <hr /> <div>==Problem 21==<br /> A game board consists of &lt;math&gt;64&lt;/math&gt; squares that alternate in color between black and white. The figure below shows square &lt;math&gt;P&lt;/math&gt; in the bottom row and square &lt;math&gt;Q&lt;/math&gt; in the top row. A marker is placed at &lt;math&gt;P.&lt;/math&gt; A step consists of moving the marker onto one of the adjoining white squares in the row above. How many &lt;math&gt;7&lt;/math&gt;-step paths are there from &lt;math&gt;P&lt;/math&gt; to &lt;math&gt;Q?&lt;/math&gt; (The figure shows a sample path.)<br /> <br /> &lt;asy&gt;//diagram by SirCalcsALot<br /> size(200); int[] x = {6, 5, 4, 5, 6, 5, 6}; int[] y = {1, 2, 3, 4, 5, 6, 7}; int N = 7; for (int i = 0; i &lt; 8; ++i) { for (int j = 0; j &lt; 8; ++j) { draw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)); if ((i+j) % 2 == 0) { filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)--cycle,black); } } } for (int i = 0; i &lt; N; ++i) { draw(circle((x[i],y[i])+(0.5,0.5),0.35)); } label(&quot;$P$&quot;, (5.5, 0.5)); label(&quot;$Q$&quot;, (6.5, 7.5)); &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }28 \qquad \textbf{(B) }30 \qquad \textbf{(C) }32 \qquad \textbf{(D) }33 \qquad \textbf{(E) }35&lt;/math&gt;<br /> <br /> <br /> ==Solution==<br /> We count paths. Noticing that we can only go along white squares, to get to a white square we can only go from the two whites beneath it. Here is a diagram:<br /> &lt;asy&gt;<br /> int N = 7;<br /> for (int i = 0; i &lt; 8; ++i) {<br /> for (int j = 0; j &lt; 8; ++j) {<br /> draw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j));<br /> if ((i+j) % 2 == 0) {<br /> filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)--cycle,black);<br /> }<br /> }<br /> }<br /> label(&quot;$1$&quot;, (5.5, .5));<br /> label(&quot;$1$&quot;, (4.5, 1.5));<br /> label(&quot;$1$&quot;, (6.5, 1.5));<br /> label(&quot;$1$&quot;, (3.5, 2.5));<br /> label(&quot;$1$&quot;, (7.5, 2.5));<br /> label(&quot;$2$&quot;, (5.5, 2.5));<br /> label(&quot;$1$&quot;, (2.5, 3.5));<br /> label(&quot;$3$&quot;, (6.5, 3.5));<br /> label(&quot;$3$&quot;, (4.5, 3.5));<br /> label(&quot;$4$&quot;, (3.5, 4.5));<br /> label(&quot;$3$&quot;, (7.5, 4.5));<br /> label(&quot;$6$&quot;, (5.5, 4.5));<br /> label(&quot;$10$&quot;, (4.5, 5.5));<br /> label(&quot;$9$&quot;, (6.5, 5.5));<br /> label(&quot;$19$&quot;, (5.5, 6.5));<br /> label(&quot;$9$&quot;, (7.5, 6.5));<br /> label(&quot;$28$&quot;, (6.5, 7.5));<br /> &lt;/asy&gt;<br /> So the answer is &lt;math&gt;\boxed{\textbf{(A)}28}&lt;/math&gt;<br /> ~yofro (Diagram credits to franzliszt)<br /> <br /> ==Solution 2==<br /> Suppose we &quot;extend&quot; the chessboard indefinitely to the right:<br /> <br /> &lt;asy&gt;<br /> int N = 7;<br /> for (int i = 0; i &lt; 10; ++i) {<br /> for (int j = 0; j &lt; 8; ++j) {<br /> draw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j));<br /> if ((i+j) % 2 == 0) {<br /> filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)--cycle,black);<br /> }<br /> }<br /> }<br /> draw((8,0) -- (8,8),red);<br /> label(&quot;$P$&quot;, (5.5,.5));<br /> label(&quot;$Q$&quot;, (6.5,7.5));<br /> label(&quot;$X$&quot;, (8.5,3.5));<br /> label(&quot;$Y$&quot;, (8.5,5.5));<br /> &lt;/asy&gt;<br /> The total number of paths from &lt;math&gt;P&lt;/math&gt; to &lt;math&gt;Q&lt;/math&gt; (including invalid paths which cross over the red line) is &lt;math&gt;\binom{7}{3} = 35&lt;/math&gt;. We subtract the number of invalid paths that pass through &lt;math&gt;X&lt;/math&gt; or &lt;math&gt;Y&lt;/math&gt;. The number of paths that pass through &lt;math&gt;X&lt;/math&gt; is &lt;math&gt;\binom{3}{0}\binom{4}{3} = 4&lt;/math&gt; and the number of paths that pass through &lt;math&gt;Y&lt;/math&gt; is &lt;math&gt;\binom{5}{1}\binom{2}{2} = 5&lt;/math&gt;. However, we overcounted the invalid paths which pass through both &lt;math&gt;X&lt;/math&gt; and &lt;math&gt;Y&lt;/math&gt;, of which there are 2 paths. Hence, the number of invalid paths is &lt;math&gt;4+5-2=7&lt;/math&gt; and the number of valid paths from &lt;math&gt;P&lt;/math&gt; to &lt;math&gt;Q&lt;/math&gt; is &lt;math&gt;35-7 = \boxed{\textbf{(A)} 28}&lt;/math&gt;. -scrabbler94<br /> <br /> ==See also== <br /> {{AMC8 box|year=2020|num-b=20|num-a=22}}<br /> {{MAA Notice}}</div> Scrabbler94 https://artofproblemsolving.com/wiki/index.php?title=Tree_(graph_theory)&diff=136434 Tree (graph theory) 2020-11-02T20:51:26Z <p>Scrabbler94: </p> <hr /> <div>A '''tree''' is an undirected [[Graph (graph theory)|graph]] which is both connected and acyclic. Equivalently, a tree is a graph &lt;math&gt;G=(V,E)&lt;/math&gt; such that for any two vertices &lt;math&gt;u, v \in V&lt;/math&gt; with &lt;math&gt;u \neq v&lt;/math&gt;, there is exactly one path connecting &lt;math&gt;u&lt;/math&gt; and &lt;math&gt;v&lt;/math&gt; in &lt;math&gt;G&lt;/math&gt;. Every tree on &lt;math&gt;|V|=n&lt;/math&gt; vertices has exactly &lt;math&gt;n-1&lt;/math&gt; edges.</div> Scrabbler94 https://artofproblemsolving.com/wiki/index.php?title=Tree_(graph_theory)&diff=136432 Tree (graph theory) 2020-11-02T20:50:40Z <p>Scrabbler94: more complete description of a tree</p> <hr /> <div>A '''tree''' is an undirected graph which is both connected and acyclic. Equivalently, a tree is a graph &lt;math&gt;G=(V,E)&lt;/math&gt; such that for any two vertices &lt;math&gt;u, v \in V&lt;/math&gt; with &lt;math&gt;u \neq v&lt;/math&gt;, there is exactly one path connecting &lt;math&gt;u&lt;/math&gt; and &lt;math&gt;v&lt;/math&gt; in &lt;math&gt;G&lt;/math&gt;. Every tree on &lt;math&gt;|V|=n&lt;/math&gt; vertices has exactly &lt;math&gt;n-1&lt;/math&gt; edges.</div> Scrabbler94 https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_6&diff=136428 2020 AMC 10A Problems/Problem 6 2020-11-02T20:42:02Z <p>Scrabbler94: /* Solution 1 */</p> <hr /> <div>{{duplicate|[[2020 AMC 12A Problems|2020 AMC 12A #4]] and [[2020 AMC 10A Problems|2020 AMC 10A #6]]}}<br /> <br /> ==Problem==<br /> <br /> How many &lt;math&gt;4&lt;/math&gt;-digit positive integers (that is, integers between &lt;math&gt;1000&lt;/math&gt; and &lt;math&gt;9999&lt;/math&gt;, inclusive) having only even digits are divisible by &lt;math&gt;5?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 80 \qquad \textbf{(B) } 100 \qquad \textbf{(C) } 125 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 500&lt;/math&gt;<br /> <br /> == Solution==<br /> The units digit, for all numbers divisible by 5, must be either &lt;math&gt;0&lt;/math&gt; or &lt;math&gt;5&lt;/math&gt;. However, since all digits are even, the units digit must be &lt;math&gt;0&lt;/math&gt;. The middle two digits can be 0, 2, 4, 6, or 8, giving 5 choices for each. The first (thousands) digit can be 2, 4, 6, or 8, giving 4 choices. Therefore, using the multiplication rule, we get &lt;math&gt;4\times 5 \times 5 \times 1 = \boxed{\textbf{(B) } 100}&lt;/math&gt; possible 4-digit integers.<br /> <br /> ==Video Solution==<br /> https://youtu.be/JEjib74EmiY<br /> <br /> ~IceMatrix<br /> <br /> https://youtu.be/Ep6XF3VUO3E<br /> <br /> ~savannahsolver<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2020|ab=A|num-b=5|num-a=7}}<br /> {{AMC12 box|year=2020|ab=A|num-b=3|num-a=5}}<br /> {{MAA Notice}}</div> Scrabbler94 https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_6&diff=136427 2020 AMC 10A Problems/Problem 6 2020-11-02T20:41:45Z <p>Scrabbler94: improve solution 1; solutions 2 and 3 are identical to solution 1 and don't contribute anything</p> <hr /> <div>{{duplicate|[[2020 AMC 12A Problems|2020 AMC 12A #4]] and [[2020 AMC 10A Problems|2020 AMC 10A #6]]}}<br /> <br /> ==Problem==<br /> <br /> How many &lt;math&gt;4&lt;/math&gt;-digit positive integers (that is, integers between &lt;math&gt;1000&lt;/math&gt; and &lt;math&gt;9999&lt;/math&gt;, inclusive) having only even digits are divisible by &lt;math&gt;5?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 80 \qquad \textbf{(B) } 100 \qquad \textbf{(C) } 125 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 500&lt;/math&gt;<br /> <br /> == Solution 1==<br /> The units digit, for all numbers divisible by 5, must be either &lt;math&gt;0&lt;/math&gt; or &lt;math&gt;5&lt;/math&gt;. However, since all digits are even, the units digit must be &lt;math&gt;0&lt;/math&gt;. The middle two digits can be 0, 2, 4, 6, or 8, giving 5 choices for each. The first (thousands) digit can be 2, 4, 6, or 8, giving 4 choices. Therefore, using the multiplication rule, we get &lt;math&gt;4\times 5 \times 5 \times 1 = \boxed{\textbf{(B) } 100}&lt;/math&gt; possible 4-digit integers.<br /> <br /> ==Video Solution==<br /> https://youtu.be/JEjib74EmiY<br /> <br /> ~IceMatrix<br /> <br /> https://youtu.be/Ep6XF3VUO3E<br /> <br /> ~savannahsolver<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2020|ab=A|num-b=5|num-a=7}}<br /> {{AMC12 box|year=2020|ab=A|num-b=3|num-a=5}}<br /> {{MAA Notice}}</div> Scrabbler94 https://artofproblemsolving.com/wiki/index.php?title=Mock_AMC&diff=128381 Mock AMC 2020-07-16T04:38:40Z <p>Scrabbler94: add mock AMC 10/12</p> <hr /> <div>A '''Mock AMC''' is a contest intended to mimic an actual [[AMC]] (American Mathematics Competitions 8, 10, or 12) exam. A number of '''Mock AMC''' competitions have been hosted on the [[Art of Problem Solving]] message boards. They are generally made by one community member and then administered for any of the other community members to take. Sometimes, the administrator may ask other people to sign up to write problems for the contest.<br /> <br /> Mock AMCs are usually very popular in the months leading up to the actual [[AMC]] competition. There is no guarantee that community members will make Mock AMCs in any given year, but there probably will be one.<br /> <br /> Feel free to remove mock tests that are not high quality or to recommend others.<br /> <br /> Ongoing mock AMCs, as well as other mock contests, can be found in the [https://artofproblemsolving.com/community/c594864_aops_mock_contests AoPS Mock Contests forum].<br /> <br /> == Tips for Writing a Mock AMC ==<br /> Anyone can write a Mock AMC and administer it. If you are interested in writing one, here are some tips:<br /> <br /> * Look at past [[AMC]]/[[AHSME]] tests to get a feel for what kind of problems you should write and what difficulty level they should be.<br /> * Look at famous theorems and formulas and see if there's any way you can make a good problem out of them.<br /> * If you are running out of creative juice and decide to pull problems from contests, try using problems from obscure contests first, if possible. This way, even the more experienced test takers will hopefully find problems that they do not already know how to do.<br /> * Pair up with another user on AoPS and write it together. Two minds are much better than one. With just one person, the problems may be biased toward one subject, but with two people, the chances of this happening are smaller.<br /> <br /> == Past Mock AMCs ==<br /> <br /> Listed below are the [higher-quality] Mock AMCs that have been hosted over AoPS in the past. Feel free to add, remove, or recommend.<br /> <br /> Note that the &quot;level&quot; column represents the originally intended difficulty. In other words, if a person makes a mock [[AMC 12]], the level would be &quot;12&quot;, even if the problems themselves are much easier. Recommended mock tests are starred.<br /> <br /> === Mock AMC 12 ===<br /> <br /> {| class=&quot;wikitable&quot; style=&quot;text-align:center;width:100%&quot;<br /> |-<br /> |<br /> ! scope=&quot;col&quot; | '''Author'''<br /> ! scope=&quot;col&quot; | '''Year'''<br /> ! scope=&quot;col&quot; | '''Initial Discussion'''<br /> ! scope=&quot;col&quot; | '''Problems'''<br /> ! scope=&quot;col&quot; width=80 | '''Answers'''<br /> ! scope=&quot;col&quot; | '''Results/Discussion'''<br /> |-<br /> ! scope=&quot;row&quot; | '''Mock AMC #1'''<br /> | mathfanatic<br /> | 2003<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=9321 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=9353 Problems]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=9572 1-5]<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=9573 6-10]<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=9574 11-15]<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=9575 16-20]<br /> [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=9576 21-25]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=9365 Results]<br /> |-<br /> ! scope=&quot;row&quot; | '''Mock AMC #2'''<br /> | mathfanatic<br /> | 2004<br /> | [http://www.artofproblemsolving.com/community/c5h10497p67837 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=10497 Problems]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=10497 Solutions]<br /> | n/a<br /> |-<br /> ! scope=&quot;row&quot; | '''Mock AMC A'''<br /> | JSRosen3<br /> | 2004<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=14138 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=14361 Problems]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=102483#p102483 Answers] [http://www.artofproblemsolving.com/community/c5h14516 Solutions]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=14489 Results / Discussion]<br /> |-<br /> ! scope=&quot;row&quot; | '''Mock AMC B'''<br /> | beta<br /> | 2004<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=14735 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=14764 Problems]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=14894 Answers] [http://www.artofproblemsolving.com/community/c5h14884 Solutions]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=105741#p105741 Results / Discussion]<br /> |-<br /> ! scope=&quot;row&quot; | '''Mock AMC C'''<br /> | JGeneson<br /> | 2004<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=15001 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=15134 Problems]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=15251 Answers]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=15251 Results / Discussion]<br /> |-<br /> ! scope=&quot;row&quot; | '''Mock AMC D'''<br /> | joml88<br /> | 2004<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=16886 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=17888 Problems]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=17891 Answers]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=17891 Results / Discussion]<br /> |-<br /> ! scope=&quot;row&quot; | '''Mock AMC E'''<br /> | Silverfalcon<br /> | 2004<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=21997 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=22141 Problems]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=22344 Answers]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=22344 Results / Discussion]<br /> |-<br /> ! scope=&quot;row&quot; | '''Mock AMC F'''<br /> | joml88<br /> | 2005<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=22049 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=23163 Problems]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=23177 Answers]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=23177 Results / Discussion]<br /> |-<br /> ! scope=&quot;row&quot; | '''Mock AMC G'''<br /> | Lucky707<br /> | 2005<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=24355 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=24974 Problems]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=25087 Answers]<br /> | [http://www.artofproblemsolving.com/community/c5h25087p157983 Results / Discussion]<br /> |-<br /> ! scope=&quot;row&quot; | '''Mock AMC H'''<br /> | Silverfalcon<br /> | 2005<br /> | [http://www.artofproblemsolving.com/community/c5h24437p154514 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/community/c5h24437p154514 Problems]<br /> | throughout thread<br /> | n/a<br /> |-<br /> ! scope=&quot;row&quot; | '''Mock AMC I'''<br /> | white_horse_king88<br /> | 2005<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=21280 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=25181 Problems]<br /> | n/a<br /> | n/a<br /> |-<br /> ! scope=&quot;row&quot; | '''Mock AMC J'''<br /> | Silverfalcon<br /> | 2005<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=47625 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=48129 Problems]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=304246#p304246 Answers]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=48132 Results / Discussion]<br /> |-<br /> ! scope=&quot;row&quot; | '''Mock AMC K'''<br /> | amirhtlusa<br /> | 2005<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=49958 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=50515 Problems]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=321102#p321102 Answers]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=321102#p321102 Results / Discussion]<br /> |-<br /> ! scope=&quot;row&quot; | '''Mock AMC L'''<br /> | amirhtlusa<br /> | 2005<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=61330 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=63041 Problems]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=63258 Answers]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=63258 Results / Discussion]<br /> |-<br /> ! scope=&quot;row&quot; | '''Mock AMC M'''<br /> | Silverfalcon<br /> | 2006<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=63542 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=78982 Problems]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=79749 Answers]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=79749 Discussion]<br /> |-<br /> ! scope=&quot;row&quot; | '''Mock AMC N'''<br /> | chess64<br /> | 2006<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=98894 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=99307 Problems]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=99344 Answers]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=99344 Results] / [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=99566 Discussion]<br /> |-<br /> ! scope=&quot;row&quot; | '''Mock AMC O'''<br /> | mustafa<br /> | 2006<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=121312 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=122126 Problems]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=126071 Answers]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=126071 Results/Discussion]<br /> |-<br /> ! scope=&quot;row&quot; | '''Mock AMC P'''<br /> | Anirudh<br /> | 2006<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=125029 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=709655#p709655 Problems]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=716240#p716240 Answers]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=716262#p716262 Results]/[http://www.artofproblemsolving.com/Forum/viewtopic.php?t=125029 Discussion]<br /> |-<br /> ! scope=&quot;row&quot; | '''Mock AMC Q'''<br /> | calc rulz<br /> | 2006<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=125194 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=125886 Problems]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=128102 Answers]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=128102 Results/Discussion]<br /> |-<br /> ! scope=&quot;row&quot; | '''Mock AMC R'''<br /> | rnwang2<br /> | 2006<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=126107 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=715597#p715597 Problems]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=127421 Answers]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=127421 Results/Discussion]<br /> |-<br /> ! scope=&quot;row&quot; | '''Mock AMC S'''<br /> | mysmartmouth<br /> | 2007<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=127221 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=128689 Problems]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=131403 Answers]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=131403 Results/Discussion]<br /> |-<br /> ! scope=&quot;row&quot; | '''Mock AMC T'''<br /> | paladin8<br /> | 2007<br /> | [http://www.artofproblemsolving.com/community/c5h127979p726003 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=127979 Problems]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=129159 Answers]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=129159 Results/Discussion]<br /> |-<br /> ! scope=&quot;row&quot; | '''Mock AMC U'''<br /> | Silverfalcon<br /> | 2008<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=184067 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=185233 Problems]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=185238 Answers]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=185236 Results]/[http://www.artofproblemsolving.com/Forum/viewtopic.php?t=185238 Discussion]<br /> |-<br /> ! scope=&quot;row&quot; | '''Mock AMC V'''<br /> | gfour84<br /> | 2009<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=298452 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1634175#p1634175 Problems]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1637429#p1637429 Answers]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1637006#p1637006 Results]/[http://www.artofproblemsolving.com/Forum/viewtopic.php?t=302030 Discussion]<br /> |-<br /> ! scope=&quot;row&quot; | '''Mock AMC 12'''<br /> | MathTwo<br /> | 2010<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=319184 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1759276#p1759276 Problems]<br /> | n/a<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1778080#p1778080 Results]/[http://www.artofproblemsolving.com/Forum/viewtopic.php?t=328510 Discussion]<br /> |-<br /> ! scope=&quot;row&quot; | '''Mock AMC 2/11'''<br /> | Caelestor<br /> | 2011<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=390907 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2183293#p2183293 Problems]<br /> | n/a<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=390907 Discussion]<br /> |-<br /> ! scope=&quot;row&quot; | '''Mock AMC 1/12'''<br /> | Lord.of.AMC<br /> | 2012<br /> | [http://www.artofproblemsolving.com/community/c5h456321p2563731 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2605537#p2605537 Problems]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2610048#p2610048 Answers]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2610048#p2610048 Results]<br /> |-<br /> ! scope=&quot;row&quot; | '''Mock AMC 12'''<br /> | Diehard<br /> | 2012<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=459149 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=459710 Problems]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2580327#p2580327 Answers]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=459710 Discussion]<br /> |-<br /> ! scope=&quot;row&quot; | '''Mock AMC 12 2012'''<br /> | python123<br /> | 2012<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=456256 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2576205#p2576205 Problems]<br /> | [http://www.artofproblemsolving.com/community/c5h456256p2580610 Answers]<br /> | [http://www.artofproblemsolving.com/community/c5h456256p2580610 Results / Discussion]<br /> |-<br /> ! scope=&quot;row&quot; | '''Mock AMC A'''<br /> | Binomial-theorem<br /> | 2012<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=473867 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2692082#p2692082 Problems]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2717578#p2717578 Solutions]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2707954#p2707954 Results]<br /> |-<br /> ! scope = &quot;row&quot; | '''Almost 2016 Mock AMC 11.5'''<br /> | whatshisbucket<br /> | 2015<br /> | [http://www.artofproblemsolving.com/community/c5h1175087p5665743 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/community/c5h1175087p5665743 Problems]<br /> | [http://artofproblemsolving.com/community/c209194_almost_2016_mock_amc_11.5 Solutions]<br /> | [http://www.artofproblemsolving.com/community/c5h1175087p5665743 Results / Discussion]<br /> |-<br /> ! scope=&quot;row&quot; | '''hnkevin42 Mock AMC 12'''<br /> | hnkevin42<br /> | 2016<br /> | [http://www.artofproblemsolving.com/community/c5h1187484p5778754 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/community/c5h1187484p5778754 Problems]<br /> | [http://www.artofproblemsolving.com/community/c5h1187484p5913294 Answers]<br /> | [http://www.artofproblemsolving.com/community/c5h1187484p5778754 Results/Discussion]<br /> |-<br /> ! scope=&quot;row&quot; | '''Last-Minute Mock AMC 12'''<br /> | CountofMC<br /> | 2018<br /> | [https://artofproblemsolving.com/community/c594864h1582044_lastminute_mock_amc_12 Initial Discussion]<br /> | [https://artofproblemsolving.com/community/c594864h1582044_lastminute_mock_amc_12 Problems]<br /> | [https://artofproblemsolving.com/community/c594864h1582044_lastminute_mock_amc_12 Answers]<br /> | [https://artofproblemsolving.com/community/c4t334444f4_lastminute_mock_amc_12 Results/Discussion]<br /> |-<br /> !scope=&quot;row&quot; | Christmas Math Competitions Year 2*<br /> | CMC Committee<br /> | 2018-2019<br /> | [https://artofproblemsolving.com/community/c594864h1747367_aime_ii_released_christmas_mathematics_competition_cmc_year_2 Initial Discussion]<br /> | [https://drive.google.com/file/d/1uS3YqAK10jB1RkCQCLQgeBHUACydJph-/view CMC 12A] [https://drive.google.com/file/d/1txKy-MfZCPPnUNTLCIad8dbatV-EGu71/view CMC 12B]<br /> | [https://artofproblemsolving.com/community/c798404h1760024_cmc_10a12a_amp_10b12b_year_2__problems_and_answer_keys Answers]<br /> | [https://artofproblemsolving.com/community/c594864h1747367p11379904 Results] / [https://artofproblemsolving.com/community/c798404_christmas_mathematics_competitions_year_2 Discussion]<br /> |-<br /> !scope=&quot;row&quot; | 2019 Mock AMC 12B*<br /> | fidgetboss_4000<br /> | 2019<br /> | [https://artofproblemsolving.com/community/c594864h1897633p13160252 Initial Discussion]<br /> | [https://artofproblemsolving.com/community/c594864h1897633p13160252 Problems] (or click here: [[Mock AMC 12B Problems]]<br /> | [https://artofproblemsolving.com/community/c963448h1919463_answer_keys_d Answers]<br /> | N/A / N/A<br /> |-<br /> !scope=&quot;row&quot; | Christmas Math Competitions Year 3*<br /> | CMC Committee<br /> | 2019-2020<br /> | [https://artofproblemsolving.com/community/c594864h1967029_cime_ii_released_christmas_math_competition_cmc_year_3 Initial Discussion]<br /> | [http://cmc.ericshen.net/CMC-2020/CMC-2020-12A.pdf CMC 10A] [http://cmc.ericshen.net/CMC-2020/CMC-2020-12B-booklet.pdf CMC 10B]<br /> | [https://artofproblemsolving.com/community/c1035147h1980361p13760203 Answers]<br /> | [https://artofproblemsolving.com/community/c594864h1967029p13623910 Results] / [https://artofproblemsolving.com/community/c1035147_christmas_mathematics_competitions_year_3 Discussion]<br /> |-<br /> !scope=&quot;row&quot; | January 2020 Mock AMC 10/12<br /> | P_Groudon<br /> | 2020<br /> | <br /> [https://artofproblemsolving.com/community/c594864h1984167_concluded_january_2020_mock_amc_1012 Initial Discussion]<br /> | [https://artofproblemsolving.com/community/c594864h1984167_concluded_january_2020_mock_amc_1012 Problems]<br /> | [https://artofproblemsolving.com/community/c1035224h1984207_january_2020_answer_keys Answers]<br /> | [https://artofproblemsolving.com/community/c594864h1984167p13963794 Results]<br /> |-<br /> !scope=&quot;row&quot; | OTSS 2020<br /> | kevinmathz<br /> | 2020<br /> | <br /> [https://artofproblemsolving.com/community/c594864h2066002_otss_olympiad_test_spring_series Initial Discussion]<br /> | [https://drive.google.com/file/d/1JfuEvDGw9i99XeQYACPUDzHebxyrcmtE/edit TMC 10A] [https://drive.google.com/file/d/1FkKjAgehDwpmmNOj_mjvYuCCV1r7p08p/edit TMC 12A]<br /> | [https://artofproblemsolving.com/community/c1130807h2094651_tmc_1012_a_solutions Answers]<br /> | [https://artofproblemsolving.com/community/c1130807h2081571_links_to_mocks_solutions_and_leaderboards Results]<br /> |-<br /> !scope=&quot;row&quot; | FMC 2020<br /> | fidgetboss_4000<br /> | 2020<br /> | <br /> [https://artofproblemsolving.com/community/c594864h2069998 Initial Discussion]<br /> | [https://drive.google.com/file/d/17jscyJzVCFDlV6YL0OpPKczyk04Tltsx/view FMC 10A] [https://drive.google.com/file/d/1pMdXlGy9F5hbWSP8CGANAhAA9_ct7pdS/view FMC 12A]<br /> | [https://artofproblemsolving.com/community/c1177836_2020_fmc_public_discussion_forum_uwu Answers]<br /> | [https://artofproblemsolving.com/community/q2h2069998p15525666 Results]<br /> |-<br /> !scope=&quot;row&quot; | 2020 Mock AMC 12<br /> | scrabbler94<br /> | 2020<br /> | <br /> [https://artofproblemsolving.com/community/c594864h2159345_contest_over_2020_mock_amc_1012_scrabbler94 Initial Discussion]<br /> | [https://artofproblemsolving.com/community/c594864h2159345_contest_over_2020_mock_amc_1012_scrabbler94 Problems]<br /> | [https://artofproblemsolving.com/community/c594864h2159345p16460394 Answers]<br /> | [https://artofproblemsolving.com/community/c594864h2159345p16460394 Results]<br /> |-<br /> |}<br /> <br /> === Mock AMC 10 ===<br /> <br /> {| class=&quot;wikitable&quot; style=&quot;text-align:center;width:100%&quot;<br /> |-<br /> |<br /> ! scope=&quot;col&quot; | '''Author'''<br /> ! scope=&quot;col&quot; | '''Year'''<br /> ! scope=&quot;col&quot; | '''Initial Discussion'''<br /> ! scope=&quot;col&quot; | '''Problems'''<br /> ! scope=&quot;col&quot; | '''Answers'''<br /> ! scope=&quot;col&quot; | '''Results/Discussion'''<br /> |-<br /> ! scope=&quot;row&quot; | '''Mock AMC 10'''<br /> | #H34N1<br /> | 2008<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=212730 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=214081 Problems]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=213727 Answers]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=213732 Results/Discussion]<br /> |-<br /> ! scope=&quot;row&quot; | '''agent's Mock AMC 10'''<br /> | agentcx<br /> | 2009<br /> | [http://www.artofproblemsolving.com/community/c5h331823p1775678 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1781208#p1781208 Problems]<br /> | n/a<br /> | [http://www.artofproblemsolving.com/community/c5h331823p1782769 Results / Discussion]<br /> |-<br /> ! scope=&quot;row&quot; | '''Mock AMC 10 Set'''<br /> | AwesomeToad<br /> | 2010<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=311120 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1762829#p1762829 Problems]<br /> | n/a<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1778740#p1778740 Results]<br /> |-<br /> ! scope=&quot;row&quot; | '''Mock AMC 10/12'''<br /> | djmathman<br /> | 2013<br /> | [http://www.artofproblemsolving.com/community/c5h556673s1_first_mock_amcs_of_the_20132014_season Initial Discussion]<br /> | [http://www.artofproblemsolving.com/community/c5h556673p3294854 Problems]<br /> | [http://www.artofproblemsolving.com/community/c5h556673p3324794 Answers]<br /> | [http://www.artofproblemsolving.com/community/c5h556673p3324794 Results / Discussion]<br /> |-<br /> ! scope=&quot;row&quot; | '''Mock AMC 10 2014-2015*'''<br /> | AlcumusGuy<br /> | 2014<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=618080&amp;hilit=AlcumusGuy%27s+mock+amc+10 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=618080&amp;hilit=AlcumusGuy%27s+mock+amc+10 Problems]<br /> | [http://www.artofproblemsolving.com/community/c5h618080p3710795 Answers/Results]<br /> | [http://artofproblemsolving.com/community/c56018 Problem Discussion]<br /> |-<br /> ! scope=&quot;row&quot; | '''Kelvin the Frog v2015'''<br /> | BOGTRO<br /> | 2015<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=623373&amp;hilit=Kelvin+the+frog Initial Discussion]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=623373&amp;hilit=Kelvin+the+frog Problems]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3731642#p3731642 Answers]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3731642#p3731642 Results]<br /> |-<br /> ! scope=&quot;row&quot; | '''Mock AMC 10/12'''<br /> | joey8189681<br /> | 2015<br /> | [http://www.artofproblemsolving.com/community/c5h624714p3741743 Initial Discussion]<br /> | [https://www.dropbox.com/s/1c8hhl6yt2awpix/Mock%20AMC%2010.pdf?dl=0 Problems]<br /> | [http://www.artofproblemsolving.com/community/c5h624714p3754593 Answers]<br /> | [http://www.artofproblemsolving.com/community/q2h626092p3756574 Results]<br /> |-<br /> ! scope=&quot;row&quot; | '''May Mock AMC 10 Contest'''<br /> | azmath333<br /> | 2015<br /> | [http://www.artofproblemsolving.com/community/c5h1082684_mock_amc_1012_contest Initial Discussion]<br /> | [http://www.artofproblemsolving.com/community/c5h1082684_mock_amc_1012_contest Problems]<br /> | [http://www.artofproblemsolving.com/community/c5h1082684p4812575 Answers] [http://www.artofproblemsolving.com/community/c5h1082684p4825662 Solutions]<br /> | [http://www.artofproblemsolving.com/community/c5h1082684p4812575 Results / Discussion] <br /> |-<br /> ! scope=&quot;row&quot; | '''2015 Mock AMC 10*'''<br /> | droid347<br /> | 2015<br /> | [http://www.artofproblemsolving.com/community/c5h1094505p4893513 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/community/c5h1094505p4893513 Problems]<br /> | [http://www.artofproblemsolving.com/community/c5h1094505p5051006 Answers]<br /> | [http://www.artofproblemsolving.com/community/c5h1094505p5045295 Results / Discussion]<br /> |-<br /> ! scope=&quot;row&quot; | '''July Mock AMC 10 Contest'''<br /> | akshaygowrish<br /> | 2015<br /> | [http://artofproblemsolving.com/community/c5h1104125_july_mock_amc_10_contest Initial Discussion]<br /> | [https://docs.google.com/document/d/12OXd-mmS_T7SyMcqw1hfvykNlcj-4TR4mMXlRJ09uBg/edit?usp=sharing Problems]<br /> | [https://docs.google.com/document/d/1-Q_w5XNKcRaffvpCLSdoJ51TpRpcDAgYLmV9OlczBCs/edit?usp=sharing Answer Key]<br /> | [http://artofproblemsolving.com/community/c5h1104125p5154283 Results / Discussion]<br /> |-<br /> ! scope = &quot;row&quot; |'''August Mock AMC 10'''<br /> | azmath333<br /> | 2015<br /> | [http://www.artofproblemsolving.com/community/c5h1115462_august_mock_amc_10 Initial Discussion]<br /> | [http://latex.artofproblemsolving.com/miscpdf/augustmockamc10.pdf Problems]<br /> | [http://www.artofproblemsolving.com/community/c5h1115462p5270430 Answers]<br /> | [http://www.artofproblemsolving.com/community/c5h1115462p5270430 Results] / [http://www.artofproblemsolving.com/community/c121880_august_mock_amc_10_discussion_forum Discussion]<br /> |-<br /> ! scope = &quot;row&quot; | '''September Mock AMC 10'''<br /> | cpma213<br /> | 2015<br /> | [http://artofproblemsolving.com/community/c5h1137447p5320386 Initial Discussion]<br /> | [http://artofproblemsolving.com/community/c5h1137447p5320386 Problems]<br /> | n/a<br /> | [http://artofproblemsolving.com/community/c5h1137447p5479954 Results / Discussion]<br /> |-<br /> ! scope = &quot;row&quot; | '''December 2015 Mock AMC 1^3*(3+7)'''<br /> | mathisawesome2169<br /> | 2015<br /> | [http://www.artofproblemsolving.com/community/c5h1173465p5647504 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/community/c5h1173465p5647504 Problems]<br /> | n/a<br /> | [http://www.artofproblemsolving.com/community/c5h1173465p5707101 Results / Discussion]<br /> |-<br /> ! scope = &quot;row&quot; | '''New Year's Mock AMC10*'''<br /> | checkmatetang<br /> | 2015<br /> | [http://artofproblemsolving.com/community/c5h1171366p5627439 Initial Discussion]<br /> | [http://artofproblemsolving.com/community/c5h1171366p5627439 Problems]<br /> | [http://artofproblemsolving.com/community/c5h1171366p5711384 Answers]<br /> | [http://artofproblemsolving.com/community/c5h1171366p5711384 Results] / [http://artofproblemsolving.com/community/c202656_new_years_mock_amc10_discussion Discussion]<br /> |-<br /> ! scope = &quot;row&quot; | '''2015-2016 Mock AMC 10*'''<br /> | PersonPsychopath<br /> | 2015<br /> | [http://artofproblemsolving.com/community/c5h1168398p5595105 Initial Discussion]<br /> | [http://artofproblemsolving.com/community/c5h1168398p5595105 Problems]<br /> | [http://artofproblemsolving.com/community/c147536h1182005_mock_amc10_wrapup Answers]<br /> | [http://artofproblemsolving.com/community/c5h1168398p5595105 Results / Discussion]<br /> |-<br /> ! scope = &quot;row&quot; | '''Mock AMC 10 2015-2016*'''<br /> | AlcumusGuy<br /> | 2015<br /> | [http://www.artofproblemsolving.com/community/c5h1177413p5687990 Initial Discussion]<br /> | [http://artofproblemsolving.com/community/c5h1177413_mock_amc_10_20152016_released Problems]<br /> | [http://artofproblemsolving.com/community/c212844_mock_amc_10_20152016 Solutions]<br /> | [http://artofproblemsolving.com/community/c5h1177413p5769839 Results / Discussion]<br /> |-<br /> ! scope = &quot;row&quot; | '''January 2016 Mock AMC 10*'''<br /> | atmchallenge<br /> | 2015<br /> | [http://www.artofproblemsolving.com/community/c5h1178971p5700377 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/community/c5h1178971p5700377 Problems]<br /> | [http://www.artofproblemsolving.com/community/c5h1178971p5838468 Answers]<br /> | [http://www.artofproblemsolving.com/community/c5h1178971p5838468 Results/Discussion]<br /> |-<br /> ! scope = &quot;row&quot; | '''2015-2016 Mock AMC 10* #2'''<br /> | PersonPsychopath<br /> | 2015<br /> | [http://artofproblemsolving.com/community/c5h1188467 Initial Discussion]<br /> | [http://artofproblemsolving.com/community/c5h1188467 Problems]<br /> | [http://artofproblemsolving.com/community/c147536h1175930 Solutions]<br /> | [http://artofproblemsolving.com/community/c5h1188467 Results / Discussion]<br /> |-<br /> <br /> ! scope=&quot;row&quot; | '''MeepyMeepMeep Mock AMC 10'''<br /> | MeepyMeepMeep, speck<br /> | 2016<br /> | [http://artofproblemsolving.com/community/c5h1195432p5852002 Initial Discussion]<br /> | [https://www.dropbox.com/s/uw0herhuqk8zbp5/Mock%20AMC%2010.pdf?dl=0 Problems]<br /> | n/a<br /> | n/a<br /> |- <br /> <br /> !scope=&quot;row&quot; | eisirrational's Mock AMC 10<br /> | eisirrational, illogical_21, whatshisbucket<br /> | 2017<br /> | [https://artofproblemsolving.com/community/c5h1372991_eisirrationals_mock_amc_10 Initial Discussion]<br /> | [https://artofproblemsolving.com/community/c5h1372991_eisirrationals_mock_amc_10 Problems]<br /> | [http://artofproblemsolving.com/community/c418591h1392426p7811150 Answer Key]<br /> | [https://artofproblemsolving.com/community/c418591_mock_amc_10_discussion Discussion]<br /> |-<br /> <br /> !scope=&quot;row&quot; | Summer Mock AMC 10<br /> | Rowechen, OmicronGamma, FedeX333X, KenV, kvedula2004, ItzVineeth<br /> | 2017<br /> | [https://artofproblemsolving.com/community/c5h1471834_summer_mock_amc_series Initial Discussion]<br /> | [https://drive.google.com/file/d/0B3M-fxa6QG0_ODNpSnJZcEt3djQ/view Problems]<br /> | [http://latex.artofproblemsolving.com/miscpdf/hehdrwpi.pdf?t=1500270651030 Answers]<br /> | [https://artofproblemsolving.com/community/c481237_2017_summer_mock_amc_10_discussion_forum Statistics / Discussion]<br /> |-<br /> <br /> !scope=&quot;row&quot; | scrabbler94's Mock AMC 10<br /> | scrabbler94<br /> | 2018<br /> | [https://artofproblemsolving.com/community/c5h1569848_mock_2018_amc_10_test_scrabbler94 Initial Discussion]<br /> | [https://artofproblemsolving.com/community/c5h1569848_mock_2018_amc_10_test_scrabbler94 Problems]<br /> | [https://artofproblemsolving.com/community/c5h1569848_mock_2018_amc_10_test_scrabbler94 Answers]<br /> | [http://artofproblemsolving.com/community/c5h1569848p9704902 Results] / [https://artofproblemsolving.com/community/c593716_scrabbler94s_mock_amc_10_discussion Discussion]<br /> |-<br /> <br /> ! scope=&quot;row&quot; | '''2018 Mock AMC 10'''<br /> | blue8931<br /> | 2018<br /> | [https://artofproblemsolving.com/community/c594864h1606066 Initial Discussion]<br /> | [https://artofproblemsolving.com/downloads/printable_post_collections/628064 Problems]<br /> | [https://artofproblemsolving.com/community/c628116h1632436_official_answer_key Answer Key]<br /> | [https://artofproblemsolving.com/community/c628116_2018_mock_amc_10_discussion_forum Results / Discussion]<br /> |-<br /> <br /> !scope=&quot;row&quot; | 2018 Memorial Day Mock AMC 10<br /> | QIDb602<br /> | 2018<br /> | [https://artofproblemsolving.com/community/c594864h1645806_released_2018_memorial_day_mock_amc_10 Initial Discussion]<br /> | [https://drive.google.com/file/d/1paAHspy5PH1J9fMoNYTb7cIEH3Di72K5/view Problems]<br /> | [https://drive.google.com/file/d/1tipZuHE11jmmOfnSDnhkwKnvIOnbsbaL/view?usp=sharing Answers]<br /> | [https://artofproblemsolving.com/community/c594864h1645806_released_2018_memorial_day_mock_amc_10 Results] / [https://artofproblemsolving.com/community/c671484 Discussion]<br /> |-<br /> <br /> !scope=&quot;row&quot; | Autumn Mock AMC 10<br /> | Krypton36, AlastorMoody, alphaone001, dchenmathcounts, InternetPerson10, kootrapali, mathdragon2000, MathGeek2018, Stormersyle<br /> | 2018<br /> | [http://artofproblemsolving.com/community/c594864h1710983p11032970 Initial Discussion]<br /> | [http://artofproblemsolving.com/community/c594864h1710983p11032970 Problems]<br /> | [http://artofproblemsolving.com/community/c594864h1710983p11331788 Answers]<br /> | [http://artofproblemsolving.com/community/c594864h1710983p11332566 Results] / [https://artofproblemsolving.com/community/c761744_autumn_mock_amc_10_discussion_forum_d Discussion]<br /> |-<br /> <br /> !scope=&quot;row&quot; | 2018 December Mock AMC 10<br /> | mathchampion1, kcbhatraju, kootrapali, chocolatelover111<br /> | 2018<br /> | [https://artofproblemsolving.com/community/c594864h1743064_2018_december_mock_amc_10_released Initial Discussion]<br /> | [https://artofproblemsolving.com/community/c594864h1743064_2018_december_mock_amc_10_released Problems]<br /> | [https://artofproblemsolving.com/community/c784638_december_mock_amc_10_2018_discussion_forum Answers]<br /> | [https://artofproblemsolving.com/community/c784638_december_mock_amc_10_2018_discussion_forum Results] / [https://artofproblemsolving.com/community/c784638_december_mock_amc_10_2018_discussion_forum Discussion]<br /> |-<br /> <br /> !scope=&quot;row&quot; | Christmas Math Competitions Year 2*<br /> | CMC Committee<br /> | 2018-2019<br /> | [https://artofproblemsolving.com/community/c594864h1747367_aime_ii_released_christmas_mathematics_competition_cmc_year_2 Initial Discussion]<br /> | [https://drive.google.com/file/d/15heOR_6rnH70joxJRfZBhQXa8Wezb2AW/view CMC 10A] [https://drive.google.com/file/d/1ZNayjDykKYj4889nXtOb7k9KBu_bqXki/view CMC 10B]<br /> | [https://artofproblemsolving.com/community/c798404h1760024_cmc_10a12a_amp_10b12b_year_2__problems_and_answer_keys Answers]<br /> | [https://artofproblemsolving.com/community/c594864h1747367p11379904 Results] / [https://artofproblemsolving.com/community/c798404_christmas_mathematics_competitions_year_2 Discussion]<br /> |-<br /> <br /> !scope=&quot;row&quot; | 2019 Mock AMC 10C*<br /> | fidgetboss_4000<br /> | 2019<br /> | [https://artofproblemsolving.com/community/c594864h1786263_released_2019_amc_10c Initial Discussion]<br /> | [https://artofproblemsolving.com/community/c594864h1786263_released_2019_amc_10c Problems] (or click here: [[2019 AMC 10C Problems]]<br /> | [https://artofproblemsolving.com/community/c594864h1786263_released_2019_amc_10c Answers - In Thread]<br /> | N/A / N/A<br /> |-<br /> !scope=&quot;row&quot; | 2019 Mock AMC 10B*<br /> | fidgetboss_4000<br /> | 2019<br /> | [https://artofproblemsolving.com/community/c594864h1897633p13160252 Initial Discussion]<br /> | [https://artofproblemsolving.com/community/c594864h1897633p13160252 Problems] (or click here: [[Mock AMC 10B Problems]]<br /> | [https://artofproblemsolving.com/community/c963448h1919463_answer_keys_d Answers]<br /> | N/A / N/A<br /> |-<br /> !scope=&quot;row&quot; | Christmas Math Competitions Year 3*<br /> | CMC Committee<br /> | 2019-2020<br /> | [https://artofproblemsolving.com/community/c594864h1967029_cime_ii_released_christmas_math_competition_cmc_year_3 Initial Discussion]<br /> | [http://cmc.ericshen.net/CMC-2020/CMC-2020-10A.pdf CMC 10A] [http://cmc.ericshen.net/CMC-2020/CMC-2020-10B-booklet.pdf CMC 10B]<br /> | [https://artofproblemsolving.com/community/c1035147h1980361p13760203 Answers]<br /> | [https://artofproblemsolving.com/community/c594864h1967029p13623910 Results] / [https://artofproblemsolving.com/community/c1035147_christmas_mathematics_competitions_year_3 Discussion]<br /> |-<br /> !scope=&quot;row&quot; | Stormersyle's Mock AMC 10*<br /> | Stormersyle<br /> | 2019-2020<br /> | [https://artofproblemsolving.com/community/c594864h1974214p13694448 Initial Discussion]<br /> | [https://artofproblemsolving.com/community/c594864h1974214p13694448] [Problems]<br /> | [https://artofproblemsolving.com/community/c594864h1974214p13694448 Answers]<br /> | [https://artofproblemsolving.com/community/c594864h1974214p13694448 Results] / [https://artofproblemsolving.com/community/c594864h1974214p13694448 Discussion]<br /> |-<br /> !scope=&quot;row&quot; | January 2020 Mock AMC 10/12<br /> | P_Groudon<br /> | 2020<br /> | <br /> [https://artofproblemsolving.com/community/c594864h1984167_concluded_january_2020_mock_amc_1012 Initial Discussion]<br /> | [https://artofproblemsolving.com/community/c594864h1984167_concluded_january_2020_mock_amc_1012 Problems]<br /> | [https://artofproblemsolving.com/community/c1035224h1984207_january_2020_answer_keys Answers]<br /> | [https://artofproblemsolving.com/community/c594864h1984167p13963794 Results]<br /> |-<br /> !scope=&quot;row&quot; | 2020 Mock Combo AMC 10*<br /> | fidgetboss_4000<br /> | 2020<br /> | <br /> [https://artofproblemsolving.com/community/c594864h2005582_amc_1012_released_mock_combo_amc_1012_ii Initial Discussion]<br /> | [https://artofproblemsolving.com/wiki/index.php?title=2020_Mock_Combo_AMC_10&amp;action=view Problems]<br /> | [https://artofproblemsolving.com/community/c1121049h2053291_answer_key Answers]<br /> | [https://artofproblemsolving.com/community/c594864h2005582_amc_1012_released_mock_combo_amc_1012_ii Results]<br /> |-<br /> !scope=&quot;row&quot; | OTSS 2020<br /> | kevinmathz<br /> | 2020<br /> | <br /> [https://artofproblemsolving.com/community/c594864h2066002_otss_olympiad_test_spring_series Initial Discussion]<br /> | [https://drive.google.com/file/d/1JfuEvDGw9i99XeQYACPUDzHebxyrcmtE/edit TMC 10A] [https://drive.google.com/file/d/1FkKjAgehDwpmmNOj_mjvYuCCV1r7p08p/edit TMC 12A]<br /> | [https://artofproblemsolving.com/community/c1130807h2094651_tmc_1012_a_solutions Answers]<br /> | [https://artofproblemsolving.com/community/c1130807h2081571_links_to_mocks_solutions_and_leaderboards Results]<br /> |-<br /> !scope=&quot;row&quot; | FMC 2020<br /> | fidgetboss_4000<br /> | 2020<br /> | <br /> [https://artofproblemsolving.com/community/c594864h2069998 Initial Discussion]<br /> | [https://drive.google.com/file/d/17jscyJzVCFDlV6YL0OpPKczyk04Tltsx/view FMC 10A] [https://drive.google.com/file/d/1pMdXlGy9F5hbWSP8CGANAhAA9_ct7pdS/view FMC 12A]<br /> | [https://artofproblemsolving.com/community/c1177836_2020_fmc_public_discussion_forum_uwu Answers]<br /> | [https://artofproblemsolving.com/community/q2h2069998p15525666 Results]<br /> |-<br /> !scope=&quot;row&quot; | 2020 Mock AMC 10<br /> | scrabbler94<br /> | 2020<br /> | <br /> [https://artofproblemsolving.com/community/c594864h2159345_contest_over_2020_mock_amc_1012_scrabbler94 Initial Discussion]<br /> | [https://artofproblemsolving.com/community/c594864h2159345_contest_over_2020_mock_amc_1012_scrabbler94 Problems]<br /> | [https://artofproblemsolving.com/community/c594864h2159345p16460394 Answers]<br /> | [https://artofproblemsolving.com/community/c594864h2159345p16460394 Results]<br /> |-<br /> |}<br /> <br /> === Mock AMC 8 ===<br /> <br /> {| class=&quot;wikitable&quot; style=&quot;text-align:center;width:100%&quot;<br /> |-<br /> |<br /> ! scope=&quot;col&quot; | '''Author'''<br /> ! scope=&quot;col&quot; | '''Year'''<br /> ! scope=&quot;col&quot; | '''Initial Discussion'''<br /> ! scope=&quot;col&quot; | '''Problems'''<br /> ! scope=&quot;col&quot; width=80 | '''Answers'''<br /> ! scope=&quot;col&quot; | '''Results/Discussion'''<br /> |-<br /> ! scope=&quot;row&quot; | '''Mock AMC 8'''<br /> | mathfanatic<br /> | 2004<br /> | [http://www.artofproblemsolving.com/community/c5h15878p111575 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=15878 Problems]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=114070#p114070 Answers]<br /> | [http://www.artofproblemsolving.com/community/c5h15878p114404 Results / Discussion]<br /> |-<br /> ! scope=&quot;row&quot; | '''Mock AMC 8'''<br /> | 13375P34K43V312<br /> | 2006<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=107507 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=108455 Problems]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=630802#p630802 Answers]<br /> | [http://www.artofproblemsolving.com/community/c5h108455p623522 Results/Discussion]<br /> |-<br /> ! scope=&quot;row&quot; | '''Mock AMC 8'''<br /> | ahaanomegas<br /> | 2012<br /> | [http://www.artofproblemsolving.com/community/c5h459346p2577870 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2598111#p2598111 Problems]<br /> | n/a<br /> | n/a<br /> |-<br /> ! scope=&quot;row&quot; | '''Mock AMC 8'''<br /> | utahjazz<br /> | 2012<br /> | [http://www.artofproblemsolving.com/community/c5h485041p2717878 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2728186#p2728186 Problems]<br /> | n/a<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2731391#p2731391 Results]<br /> |-<br /> ! scope=&quot;row&quot; | '''Mock AMC 8'''<br /> | iNomOnCountdown<br /> | 2014<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=132&amp;t=590451&amp;hilit=iNomonCountdown%27s+mock+amc+8 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=132&amp;t=590451&amp;hilit=iNomonCountdown%27s+mock+amc+8 Problems]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=132&amp;t=590451&amp;start=0 Answers]<br /> | [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=132&amp;t=590451&amp;start=0 Results / Discussion]<br /> |-<br /> ! scope = &quot;row&quot; | '''Mock AMC 8'''<br /> | Tan<br /> | 2014<br /> | [http://www.artofproblemsolving.com/community/c5h613297p3648226 Initial Discussion]<br /> | [http://www.artofproblemsolving.com/community/c5h613297p3648227 Problems]<br /> | [https://docs.google.com/spreadsheets/d/1U1QJ9r4r2xFbRByk9gXOhEVV48ScjoLWZp4IFanbKF0/edit#gid=1475864782 Answers]<br /> | [https://docs.google.com/spreadsheets/d/1U1QJ9r4r2xFbRByk9gXOhEVV48ScjoLWZp4IFanbKF0/edit#gid=1475864782 Results / Discussion]<br /> |-<br /> ! scope = &quot;row&quot; | '''2015 Hard Mock AMC 8*'''<br /> | Not_a_Username, 8invalid8<br /> | 2015<br /> | [http://artofproblemsolving.com/community/q1h1128391p5227860 Initial Discussion]<br /> | [http://latex.artofproblemsolving.com/miscpdf/kashimyo.pdf Problems]<br /> | [http://www.artofproblemsolving.com/community/c5h1128391p5423312 Answers]<br /> | [http://www.artofproblemsolving.com/community/c5h1128391p5423312 Results / Discussion]<br /> |-<br /> ! scope = &quot;row&quot; | '''Mock AMC 8'''<br /> | PersonPsychopath<br /> | 2015<br /> | [http://artofproblemsolving.com/community/c3h1152332p5457736 Initial Discussion]<br /> | [http://artofproblemsolving.com/downloads/printable_post_collections/163667 Problems]<br /> | [http://artofproblemsolving.com/community/category-admin/165328 Forum]<br /> | [http://artofproblemsolving.com/community/c165328h1155882p5484131 Results / Discussion]<br /> |-<br /> ! scope = &quot;row&quot; | '''Mock AMC 8 #2'''<br /> | PersonPsychopath<br /> | 2015<br /> | [http://artofproblemsolving.com/community/c3h1167425p5586244 Initial Discussion]<br /> | [http://artofproblemsolving.com/downloads/printable_post_collections/165838 Problems]<br /> | [http://artofproblemsolving.com/community/c147536_the_mock_amc_8_forum Forum]<br /> | [http://artofproblemsolving.com/community/c3h1167425p5586244 Results / Discussion]<br /> |-<br /> ! scope = &quot;row&quot; | '''Mock AMC 8 2015'''<br /> | Alberty44<br /> | 2015<br /> | [http://artofproblemsolving.com/community/c194276_discusssion Initial Discussion]<br /> | [http://artofproblemsolving.com/community/c5h1168822p5674003 Problems]<br /> | n/a<br /> | [http://artofproblemsolving.com/community/c194276h1175928p5673996 Results/Discussion]<br /> |-<br /> ! scope = &quot;row&quot; | '''Christmas AMC 8'''<br /> | Mudkipswims42<br /> | 2015<br /> | [http://artofproblemsolving.com/community/c5h1177489p5688588 Initial Discussion]<br /> | [http://artofproblemsolving.com/community/c5h1177489p5688588 Problems]<br /> | [http://www.artofproblemsolving.com/community/c229455_christmas_amc8_discussion_forum Forum]<br /> | [http://artofproblemsolving.com/community/c5h1177489p5880200 Results/Discussion]<br /> |- <br /> <br /> ! scope = &quot;row&quot; | '''Mock AMC 8!'''<br /> | eisirrational, pretzel, AOPS12142015, Th3Numb3rThr33, e_power_pi_times_i<br /> | 2017<br /> | [https://artofproblemsolving.com/community/c5h1526187_mock_amc_8 Initial Discussion]<br /> | [http://services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvYi84LzFkNzUyMWYwZTViOTQ1MzRmZmU3ODE0NmI2MzIzMGUxNjA2MTcwLnBkZg==&amp;rn=TW9jayBBTUMgOCB2Ny5wZGY= Problems]<br /> | [http://artofproblemsolving.com/community/c3h1545004p9368333 Answer Key/Solutions]<br /> | [https://artofproblemsolving.com/community/c561318_mock_amc_8_2017_discussion Discussion]<br /> |-<br /> <br /> ! scope = &quot;row&quot; | '''Mock Summ(er)ation AMC 8!'''<br /> | mathchamp1, kevinmathz, Gali, mathdragon2000, reddragon644, ShreyJ, Quadrastic, Mathnerd1223334444<br /> | 2018<br /> | [https://artofproblemsolving.com/community/c594864h1672811_2018_summeration_mock_amc_8 Initial Discussion]<br /> | [https://math-adventures.weebly.com/uploads/1/1/8/8/118819793/amc_8_problems.pdf Problems]<br /> | n/a<br /> | n/a<br /> |-<br /> <br /> ! scope = &quot;row&quot; | '''popcorn1's AMC 8 2018'''<br /> | popcorn1<br /> | 2018<br /> | [https://artofproblemsolving.com/community/c594864h1715072_popcorn1s_amc_8_2018 Initial Discussion]<br /> | [http://artofproblemsolving.com/community/c594864h1715072p11074872 Problems]<br /> | Answer Key / Solutions - not released<br /> | Discussion - not released<br /> |-<br /> <br /> ! scope = &quot;row&quot; | ''' Stormersyle's Mock AMC 8<br /> | Stormersyle <br /> | 2018<br /> | [http://artofproblemsolving.com/community/c594864h1722790p11146826 Initial Discussion]<br /> | [http://artofproblemsolving.com/community/c594864h1722790p11146826 Problems]<br /> | [https://latex.artofproblemsolving.com/miscpdf/qgdafjlp.pdf?t=1544494957792 Answer Key/Solutions]<br /> | Discussion - not released<br /> |-<br /> <br /> ! scope = &quot;row&quot; | '''mathchampion1's Christmas Mock AMC 8'''<br /> | mathchampion1<br /> | 2018<br /> | [https://artofproblemsolving.com/community/c594864h1752494_released_mock_amc_8_christmas_edition Initial Discussion]<br /> | [https://artofproblemsolving.com/community/c594864h1752494_released_mock_amc_8_christmas_edition Problems]<br /> | Answer Key / Solutions - not released<br /> | Discussion - not released<br /> |-<br /> <br /> ! scope = &quot;row&quot; | '''June 2019 Mock AMC 8'''<br /> | fidgetboss_4000<br /> | 2019<br /> | [https://artofproblemsolving.com/community/c594864h1853912_june_2019_mock_amc_8 Initial Discussion]<br /> | [https://artofproblemsolving.com/community/c594864h1853912_june_2019_mock_amc_8 Problems]<br /> | Answer Key / Solutions - not released<br /> | Discussion - not released<br /> |-<br /> <br /> }<br /> <br /> == See also ==<br /> * [[American Mathematics Competitions]]<br /> * [[Math books]]<br /> * [[Mathematics competitions]]<br /> * [[Mock AIME]]<br /> * [[Mock MathCounts]]<br /> * [[Mock USAMO]]<br /> * [[Mock USAJMO]]<br /> * [[Resources for mathematics competitions]]<br /> * [[AoPS Past Contests]]</div> Scrabbler94 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10A_Problems/Problem_2&diff=127192 2019 AMC 10A Problems/Problem 2 2020-07-01T04:41:04Z <p>Scrabbler94: Solution 2 is wrong since divisibility by 100 doesn't imply the hundreds digit of their difference is 0.</p> <hr /> <div>== Problem ==<br /> What is the hundreds digit of &lt;math&gt;(20!-15!)?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }4\qquad\textbf{(E) }5&lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> The last three digits of &lt;math&gt;n!&lt;/math&gt; for all &lt;math&gt;n\geq15&lt;/math&gt; are &lt;math&gt;000&lt;/math&gt;, because there are at least three &lt;math&gt;2&lt;/math&gt;s and three &lt;math&gt;5&lt;/math&gt;s in its prime factorization. Because &lt;math&gt;0-0=0&lt;/math&gt;, the answer is &lt;math&gt;\boxed{\textbf{(A) }0}&lt;/math&gt;.<br /> <br /> ==Video Solution==<br /> https://youtu.be/V1fY0oLSHvo<br /> <br /> ~savannahsolver<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2019|ab=A|num-b=1|num-a=3}}<br /> {{MAA Notice}}</div> Scrabbler94 https://artofproblemsolving.com/wiki/index.php?title=User:Scrabbler94&diff=127104 User:Scrabbler94 2020-06-30T06:00:54Z <p>Scrabbler94: </p> <hr /> <div>My username is Scrabbler94.<br /> <br /> &lt;b&gt;Bio:&lt;/b&gt;<br /> *MIT '16, major 18-C (Mathematics with computer science)<br /> *Ph.D. candidate in computer science<br /> *I play in NASPA-sanctioned Scrabble tournaments (as you can probably guess by my username). Rated ~1700-1800.<br /> *Amateur bowler (average ~175)<br /> <br /> &lt;b&gt;Past Contests:&lt;/b&gt;<br /> *2nd MATHCOUNTS State (2008), two-time MATHCOUNTS National qualifier (2007-2008)<br /> *AMC 10 perfect scorer (2009)<br /> *Six-time AIME qualifier (2007-2012)<br /> *USAMO qualifier, score: 13 (2010)<br /> *Putnam score: 20 (2014) - 1 point away from top 500 :(<br /> <br /> &lt;b&gt;AoPS Mock Contests:&lt;/b&gt;<br /> Contests held over AoPS.<br /> *[https://artofproblemsolving.com/community/c594864h1569848 Mock 2018 AMC 10]<br /> *[https://artofproblemsolving.com/community/c594864h1579782_released_mock_state_mathcounts_scrabbler94 Mock 2018 MATHCOUNTS State Competition]<br /> *[https://artofproblemsolving.com/community/c594864h1759381_2nd_annual_mock_mathcounts_state_scrabbler94 Mock 2019 MATHCOUNTS State Competition]<br /> *[https://artofproblemsolving.com/community/c594864h2159345_2020_mock_amc_1012_scrabbler94 Mock 2020 AMC 10/12]</div> Scrabbler94 https://artofproblemsolving.com/wiki/index.php?title=User:Scrabbler94&diff=127103 User:Scrabbler94 2020-06-30T05:59:01Z <p>Scrabbler94: add 2020 AMC 10/12</p> <hr /> <div>My username is Scrabbler94.<br /> <br /> &lt;b&gt;Bio:&lt;/b&gt;<br /> *MIT '16, major 18-C (Mathematics with computer science)<br /> *Ph.D. student in computer science<br /> *I play in NASPA-sanctioned Scrabble tournaments (as you can probably guess by my username). Rated ~1700-1800.<br /> *Amateur bowler (average ~175)<br /> <br /> &lt;b&gt;Past Contests:&lt;/b&gt;<br /> *2nd MATHCOUNTS State (2008), two-time MATHCOUNTS National qualifier (2007-2008)<br /> *AMC 10 perfect scorer (2009)<br /> *Six-time AIME qualifier (2007-2012)<br /> *USAMO qualifier, score: 13 (2010)<br /> *Putnam score: 20 (2014) - 1 point away from top 500 :(<br /> <br /> &lt;b&gt;AoPS Mock Contests:&lt;/b&gt;<br /> Contests held over AoPS.<br /> *[https://artofproblemsolving.com/community/c594864h1569848 Mock 2018 AMC 10]<br /> *[https://artofproblemsolving.com/community/c594864h1579782_released_mock_state_mathcounts_scrabbler94 Mock 2018 MATHCOUNTS State Competition]<br /> *[https://artofproblemsolving.com/community/c594864h1759381_2nd_annual_mock_mathcounts_state_scrabbler94 Mock 2019 MATHCOUNTS State Competition]<br /> *[https://artofproblemsolving.com/community/c594864h2159345_2020_mock_amc_1012_scrabbler94 Mock 2020 AMC 10/12]</div> Scrabbler94 https://artofproblemsolving.com/wiki/index.php?title=LaTeX:Commands&diff=121177 LaTeX:Commands 2020-04-20T05:43:02Z <p>Scrabbler94: various improvements</p> <hr /> <div>{{LaTeX: Commands}}<br /> <br /> This page introduces various useful commands for rendering math in LaTeX, as well as instructions for building your own commands.<br /> <br /> ===Subscripts and Superscripts===<br /> Subscripts and superscripts (such as exponents) can be made using the underscore _ and carat ^ symbols respectively.<br /> {| class=&quot;latextable&quot;<br /> !Symbol !! Command!!Symbol!!Command<br /> |-<br /> |&lt;math&gt;2^{2}&lt;/math&gt;||2^2||&lt;math&gt;\textstyle a_i&lt;/math&gt;||a_i<br /> |-<br /> | &lt;math&gt;\textstyle 2^{23}&lt;/math&gt;||2^{23}||&lt;math&gt;\textstyle n_{i-1}&lt;/math&gt;||n_{i-1}<br /> |-<br /> | &lt;math&gt;a^{i+1}_3&lt;/math&gt;||a^{i+1}_3||&lt;math&gt;x^{3^2}&lt;/math&gt;||x^{3^2}<br /> |-<br /> | &lt;math&gt;2^{a_i}&lt;/math&gt;||2^{a_i}||&lt;math&gt;2^a_i&lt;/math&gt;||2^a_i<br /> |}<br /> Notice that we can apply both a subscript and a superscript at the same time. For subscripts or superscripts with more than one character, you must surround with curly braces. For example, &lt;code&gt;x^10&lt;/code&gt; produces &lt;math&gt;x^10&lt;/math&gt;, while &lt;code&gt;x^{10}&lt;/code&gt; produces &lt;math&gt;x^{10}&lt;/math&gt;.<br /> <br /> ==Math Commands==<br /> Here are some commonly used math commands in LaTeX:<br /> ===Fractions===<br /> {|class=&quot;latextable&quot;<br /> !Symbol!!Command<br /> |-<br /> |&lt;math&gt;\frac {1}{2}&lt;/math&gt;||\frac{1}{2} or \frac12<br /> |-<br /> | &lt;math&gt;\frac{2}{x+2}&lt;/math&gt;||\frac{2}{x+2}<br /> |-<br /> | &lt;math&gt;\frac{1+\frac{1}{x}}{3x + 2}&lt;/math&gt;||\frac{1+\frac{1}{x}}{3x + 2}<br /> |}<br /> <br /> <br /> Notice that with fractions with a 1-digit numerator and a 1-digit denominator, we can simply group the numerator and the denominator together as one number. However, for fractions with either a numerator or a denominator that requires more than one character, you need to surround everything in curly brackets.<br /> <br /> Use \cfrac for continued fractions.<br /> {| class=&quot;latextable&quot;<br /> !Expression !! Command<br /> |-<br /> |&lt;math&gt;\cfrac{2}{1+\cfrac{2}{1+\cfrac{2}{1+\cfrac{2}{1}}}}&lt;/math&gt;||\cfrac{2}{1+\cfrac{2}{1+\cfrac{2}{1+\cfrac{2}{1}}}}<br /> |}<br /> <br /> ===Radicals===<br /> {| class=&quot;latextable&quot;<br /> !Symbol !! Command<br /> |-<br /> |&lt;math&gt;\sqrt{2}&lt;/math&gt;||\sqrt{2}<br /> |-<br /> | &lt;math&gt;\sqrt{x+y}&lt;/math&gt;||\sqrt{x+y}<br /> |-<br /> | &lt;math&gt;\sqrt{x+\frac{1}{2}}&lt;/math&gt;||\sqrt{x+\frac{1}{2}}<br /> |-<br /> | &lt;math&gt;\sqrt{3}&lt;/math&gt;||\sqrt{3}<br /> |-<br /> | &lt;math&gt;\sqrt[n]{x}&lt;/math&gt;||\sqrt[n]{x}<br /> |}<br /> <br /> ===Sums, Products, Limits and Logarithms===<br /> Use the commands \sum, \prod, \lim, and \log respectively. To denote lower and upper bounds, or the base of the logarithm, use _ and ^ in the same way they are used for subscripts and superscripts. (Lower and upper bounds for integrals work the same way, as you'll see in the [[LaTeX:Commands#Calculus|calculus section]])<br /> {| class=&quot;latextable&quot;<br /> !Symbol !! Command<br /> |-<br /> |&lt;math&gt;\textstyle \sum_{i=1}^{\infty}\frac{1}{i}&lt;/math&gt;||\sum_{i=1}^{\infty}\frac{1}{i}<br /> |-<br /> | &lt;math&gt;\textstyle \prod_{n=1}^5\frac{n}{n-1}&lt;/math&gt;||\prod_{n=1}^5\frac{n}{n-1}<br /> |-<br /> | &lt;math&gt;\textstyle \lim_{x\to\infty}\frac{1}{x}&lt;/math&gt;||\lim_{x\to\infty}\frac{1}{x}<br /> |-<br /> | &lt;math&gt;\textstyle \lim\limits_{x\to\infty}\frac{1}{x}&lt;/math&gt;||\lim\limits_{x\to\infty}\frac{1}{x}<br /> |-<br /> |&lt;math&gt;\textstyle \log_n n^2&lt;/math&gt;||\log_n n^2<br /> |}<br /> Some of these are prettier in display mode:<br /> {| class=&quot;latextable&quot;<br /> !Symbol !! Command<br /> |-<br /> |&lt;math&gt;\sum_{i=1}^{\infty}\frac{1}{i}&lt;/math&gt;||\sum_{i=1}^{\infty}\frac{1}{i}<br /> |-<br /> | &lt;math&gt;\prod_{n=1}^5\frac{n}{n-1}&lt;/math&gt;||\prod_{n=1}^5\frac{n}{n-1}<br /> |-<br /> | &lt;math&gt;\lim_{x\to\infty}\frac{1}{x}&lt;/math&gt;||\lim_{x\to\infty}\frac{1}{x}<br /> |}<br /> Note that we can use sums, products, and logarithms without _ or ^ modifiers.<br /> {| class=&quot;latextable&quot;<br /> !Symbol !! Command<br /> |-<br /> |&lt;math&gt;\sum\frac{1}{i}&lt;/math&gt;||\sum\frac{1}{i}<br /> |-<br /> | &lt;math&gt;\frac{n}{n-1}&lt;/math&gt;||\frac{n}{n-1}<br /> |-<br /> | &lt;math&gt;\textstyle \log n^2&lt;/math&gt;||\log n^2<br /> |-<br /> | &lt;math&gt;\textstyle \ln e&lt;/math&gt;||\ln e<br /> |}<br /> <br /> ===Mods===<br /> {| class=&quot;latextable&quot;<br /> !Symbol !! Command<br /> |-<br /> |&lt;math&gt;9\equiv 3 \bmod{6}&lt;/math&gt;||9\equiv 3 \bmod{6}<br /> |-<br /> | &lt;math&gt;9\equiv 3 \pmod{6}&lt;/math&gt;||9\equiv 3 \pmod{6}<br /> |-<br /> | &lt;math&gt;9\equiv 3 \mod{6}&lt;/math&gt;||9\equiv 3 \mod{6}<br /> |-<br /> | &lt;math&gt;9\equiv 3\pod{6}&lt;/math&gt;||9\equiv 3 \pod{6}<br /> |}<br /> <br /> ===Combinations===<br /> {| class=&quot;latextable&quot;<br /> !Symbol !! Command<br /> |-<br /> |&lt;math&gt;\scriptstyle\binom{1}{1}&lt;/math&gt;||\binom{1}{1}<br /> |-<br /> | &lt;math&gt;\scriptstyle\binom{n-1}{r-1}&lt;/math&gt;||\binom{n-1}{r-1}<br /> |}<br /> These often look better in display mode:<br /> {| class=&quot;latextable&quot;<br /> !Symbol !! Command<br /> |-<br /> |&lt;math&gt;\dbinom{9}{3}&lt;/math&gt;||\dbinom{9}{3}<br /> |-<br /> | &lt;math&gt;\dbinom{n-1}{r-1}&lt;/math&gt;||\dbinom{n-1}{r-1}<br /> |}<br /> <br /> ===Trigonometric Functions===<br /> <br /> Most of these are just the abbreviation of the trigonometric function with simply a backslash added before the abbreviation. <br /> <br /> {| class=&quot;latextable&quot;<br /> <br /> !Symbol!!Command!!Symbol!!Command!!Symbol!!Command<br /> |-<br /> |&lt;math&gt;\textstyle \cos&lt;/math&gt;||\cos||&lt;math&gt;\textstyle \sin&lt;/math&gt;||\sin||&lt;math&gt;\textstyle \tan&lt;/math&gt;||\tan<br /> |-<br /> | &lt;math&gt;\sec&lt;/math&gt;||\sec||&lt;math&gt;\textstyle \textstyle \csc&lt;/math&gt;||\csc||&lt;math&gt;\textstyle \cot&lt;/math&gt;||\cot<br /> |-<br /> | &lt;math&gt;\textstyle \arccos&lt;/math&gt;||\arccos||&lt;math&gt;\textstyle \arcsin&lt;/math&gt;||\arcsin||&lt;math&gt;\textstyle \arctan&lt;/math&gt;||\arctan<br /> |-<br /> | &lt;math&gt;\textstyle \cosh&lt;/math&gt;||\cosh||&lt;math&gt;\textstyle \sinh&lt;/math&gt;||\sinh||&lt;math&gt;\textstyle \tanh&lt;/math&gt;||\tanh<br /> |-<br /> | &lt;math&gt;\textstyle \coth&lt;/math&gt;||\coth <br /> |}<br /> Here are a couple examples:<br /> {| class=&quot;latextable&quot;<br /> !Symbol !! Command<br /> |-<br /> |&lt;math&gt;\textstyle \cos^2 x +\sin^2 x = 1&lt;/math&gt;||\cos^2 x +\sin^2 x = 1<br /> |-<br /> | &lt;math&gt;\cos 90^\circ = 0&lt;/math&gt;||\cos 90^\circ = 0 <br /> |}<br /> <br /> ===Calculus===<br /> Below are examples of calculus expressions rendered in LaTeX. Most of these commands have been introduced before. Notice how definite integrals are rendered (and the difference between regular math and display mode for definite integrals). The \, in the integrals makes a small space before the dx.<br /> {| class=&quot;latextable&quot;<br /> !Symbol !! Command<br /> |-<br /> |&lt;math&gt;\frac{d}{dx}\left(x^2\right) = 2x&lt;/math&gt;||\frac{d}{dx}\left(x^2\right) = 2x<br /> |-<br /> | &lt;math&gt;\int 2x\,dx = x^2+C&lt;/math&gt;||\int 2x\,dx = x^2+C<br /> |-<br /> | &lt;math&gt;\int^5_1 2x\,dx = 24&lt;/math&gt;||\int^5_1 2x\,dx = 24<br /> |-<br /> | &lt;math&gt;\int^5_1 2x\,dx = 24&lt;/math&gt;||\int^5_1 2x\,dx = 24<br /> |-<br /> | &lt;math&gt;\frac{\partial^2U}{\partial x^2} + \frac{\partial^2U}{\partial y^2}&lt;/math&gt;||\frac{\partial^2U}{\partial x^2} + \frac{\partial^2U}{\partial y^2}<br /> |-<br /> | &lt;math&gt;\frac{1}{4\pi}\oint_\Sigma\frac{1}{r}\frac{\partial U}{\partial n} ds&lt;/math&gt;||\frac{1}{4\pi}\oint_\Sigma\frac{1}{r}\frac{\partial U}{\partial n} ds<br /> |}<br /> <br /> ===Overline and Underline===<br /> <br /> {| class=&quot;latextable&quot;<br /> !Symbol !! Command<br /> |-<br /> |&lt;math&gt;\overline{a+bi}&lt;/math&gt;||\overline{a+bi}<br /> |-<br /> | &lt;math&gt;\underline{747}&lt;/math&gt;||\underline{747}<br /> |}<br /> <br /> == LaTeX ==<br /> ===Other Functions===<br /> {| class=&quot;latextable&quot;<br /> !Symbol !! Command!!Symbol !! Command!!Symbol !! Command<br /> |-<br /> |&lt;math&gt;\arg&lt;/math&gt;||\arg||&lt;math&gt;\textstyle\deg&lt;/math&gt;||\deg||&lt;math&gt;\textstyle\det&lt;/math&gt;||\det<br /> |-<br /> | &lt;math&gt;\dim&lt;/math&gt;||\dim||&lt;math&gt;\textstyle\exp&lt;/math&gt;||\exp||&lt;math&gt;\textstyle\gcd&lt;/math&gt;||\gcd<br /> |-<br /> |&lt;math&gt;\hom&lt;/math&gt;||\hom||&lt;math&gt;\inf&lt;/math&gt;||\inf||&lt;math&gt;\ker&lt;/math&gt;||\ker<br /> |-<br /> | &lt;math&gt;\textstyle\lg&lt;/math&gt;||\lg||&lt;math&gt;\liminf&lt;/math&gt;||\liminf||&lt;math&gt;\limsup&lt;/math&gt;||\limsup<br /> |-<br /> | &lt;math&gt;\textstyle\max&lt;/math&gt;||\max||&lt;math&gt;\textstyle\min&lt;/math&gt;||\min||&lt;math&gt;\Pr&lt;/math&gt;||\Pr<br /> |-<br /> | &lt;math&gt;\sup&lt;/math&gt;||\sup<br /> |}<br /> Some of these commands take subscripts in the same way sums, products, and logarithms do. Some render differently in display mode and regular math mode.<br /> {| class=&quot;latextable&quot;<br /> !Symbol !! Command!!Symbol !! Command!!Symbol !! Command<br /> |-<br /> | &lt;math&gt;\dim_x&lt;/math&gt;||\dim_x||&lt;math&gt;\textstyle\gcd_x&lt;/math&gt;||\gcd_x||&lt;math&gt;\inf_x&lt;/math&gt;||\inf_x<br /> |-<br /> | &lt;math&gt;\liminf_x&lt;/math&gt;||\liminf_x||&lt;math&gt;\limsup_x&lt;/math&gt;||\limsup_x||&lt;math&gt;\textstyle\max_x&lt;/math&gt;||\max_x<br /> |-<br /> | &lt;math&gt;\textstyle\min_x&lt;/math&gt;||\min_x||&lt;math&gt;\Pr_x&lt;/math&gt;||\Pr_x||&lt;math&gt;\sup_x&lt;/math&gt;||\sup_x<br /> |}<br /> <br /> ==Matrices==<br /> We can build an array or matrix with the \begin{array}…\end{array} commands, and use \left and \right to properly size the delimiters around the matrix:<br /> &lt;pre&gt;&lt;nowiki&gt;<br /> The characteristic polynomial $f(\lambda)$ of the<br /> $3 \times 3$ matrix<br /> $<br /> \left(<br /> \begin{array}{ccc}<br /> a &amp; b &amp; c &lt;br /&gt;d &amp; e &amp; f &lt;br /&gt;g &amp; h &amp; i \end{array}<br /> \right)$<br /> is given by the equation<br /> $f(\lambda)<br /> = \left|<br /> \begin{array}{ccc}<br /> \lambda - a &amp; -b &amp; -c &lt;br /&gt;-d &amp; \lambda - e &amp; -f &lt;br /&gt;-g &amp; -h &amp; \lambda - i \end{array}<br /> \right|.$<br /> &lt;/nowiki&gt;<br /> &lt;/pre&gt;<br /> More simply, we can use the shortcut matrix environments in the amsmath package:<br /> &lt;pre&gt;&lt;nowiki&gt;<br /> The characteristic polynomial $f(\lambda)$ of the<br /> $3 \times 3$ matrix<br /> $<br /> \begin{pmatrix}<br /> a &amp; b &amp; c &lt;br /&gt;d &amp; e &amp; f &lt;br /&gt;g &amp; h &amp; i<br /> \end{pmatrix}$<br /> is given by the equation<br /> $f(\lambda)<br /> = \begin{vmatrix}<br /> \lambda - a &amp; -b &amp; -c &lt;br /&gt;-d &amp; \lambda - e &amp; -f &lt;br /&gt;-g &amp; -h &amp; \lambda - i<br /> \end{vmatrix}.$<br /> &lt;/nowiki&gt;&lt;/pre&gt;<br /> You can read more about how the array environment works [[LaTeX:Layout|here]] (it works the same as tabular).<br /> <br /> We can also use this environment to typeset any mathematics that calls for multiple columns, such as piecewise-defined functions like this one:<br /> &lt;pre&gt;&lt;nowiki&gt;<br /> $f(x) = \left\{ \begin{array}{ll}<br /> x+7 &amp; \mbox{if 5&lt; x};&lt;br /&gt;x^2-3 &amp; \mbox{if -3 \le x \le 5};&lt;br /&gt;-x &amp; \mbox{if x &lt; -3}.\end{array} \right.$<br /> &lt;/nowiki&gt;&lt;/pre&gt;<br /> <br /> But it would be better to use the cases environment and \text command that the amsmath package provides:<br /> &lt;pre&gt;&lt;nowiki&gt;<br /> $<br /> f(x) = \begin{cases}<br /> x+7 &amp; \text{if 5&lt; x}; &lt;br /&gt;x^2-3 &amp; \text{if -3 \le x \le 5};&lt;br /&gt;-x &amp; \text{if x &lt; -3}.<br /> \end{cases}<br />$<br /> &lt;/nowiki&gt;&lt;/pre&gt;<br /> <br /> ==Text Styles in Math Mode==<br /> You can render letters in various styles in math mode. Below are examples; you should be able to use these with any letters. The \mathbb requires the amsfonts package to be included in your document's preamble. Do not try to do \mathbb{year}. You'll get &lt;math&gt;\mathbb{year}&lt;/math&gt;, and that looks nothing like it!<br /> {| class=&quot;latextable&quot;<br /> !Symbol !! Command!!Symbol !! Command!!Symbol !! Command!!Symbol !! Command<br /> |-<br /> |&lt;math&gt;\mathbb{R}&lt;/math&gt;||\mathbb{R}||&lt;math&gt;\mathbf{R}&lt;/math&gt;||\mathbf{R}||&lt;math&gt;\mathcal{R}&lt;/math&gt;||\mathcal{R}||&lt;math&gt;\mathfrak{R}&lt;/math&gt;||\mathfrak{R}<br /> |-<br /> | [[Image:Mathbb1.gif]]||\mathbb{Z}||&lt;math&gt;\mathbf{Z}&lt;/math&gt;||\mathbf{Z}||&lt;math&gt;\mathcal{Z}&lt;/math&gt;||\mathcal{Z}||&lt;math&gt;\mathfrak{Z}&lt;/math&gt;||\mathfrak{Z}<br /> |-<br /> | &lt;math&gt;\mathbb{Q}&lt;/math&gt;||\mathbb{Q}||&lt;math&gt;\mathbf{Q}&lt;/math&gt;||\mathbf{Q}||&lt;math&gt;\mathcal{Q}&lt;/math&gt;||\mathcal{Q}||&lt;math&gt;\mathfrak{Q}&lt;/math&gt;||\mathfrak{Q}<br /> |}<br /> If you're persistent, you can dig a few more out of [ftp://ftp.ams.org/pub/tex/doc/amsfonts/amsfndoc.pdf this document].<br /> <br /> If you want to drop a little bit of text in the middle of math mode, you can use the \text command. The \text command is most useful in &lt;nowiki&gt;$$...$$&lt;/nowiki&gt; or &lt;nowiki&gt;$...$&lt;/nowiki&gt; mode, where breaking up the math mode would force the output on to a new line entirely.<br /> So<br /> &lt;pre&gt;&lt;nowiki&gt;<br /> $$n^2 + 5 = 30\text{ so we have }n=\pm5$$<br /> &lt;/nowiki&gt;&lt;/pre&gt;<br /> gives<br /> <br /> [[Image:Text1.gif]]<br /> <br /> ==How to Build Your Own Commands==<br /> The command \newcommand is used to create your own commands. We'll start with an example:<br /> &lt;pre&gt;&lt;nowiki&gt;<br /> \documentclass[11pt]{article}<br /> \usepackage{amsmath}<br /> <br /> \pdfpagewidth 8.5in<br /> \pdfpageheight 11in<br /> \newcommand{\reci}{\frac{1}{#1}}<br /> \newcommand{\hypot}{\sqrt{#1^2+#2^2}}<br /> \newcommand{\cbrt}{\sqrt{#1}}<br /> <br /> \begin{document}<br /> <br /> The reciprocal of 2 is $\reci{2}$.<br /> <br /> The hypotenuse has length $\hypot{3}{4}$.<br /> <br /> I'm sick of writing $\backslash$sqrt{2}$' all the time, just to get$\cbrt{2}$.<br /> <br /> \end{document}<br /> &lt;/nowiki&gt;&lt;/pre&gt;<br /> The \newcommand declarations are in the preamble. Each is of the form<br /> <br /> \newcommand{name of new command}[number of arguments]{definition}<br /> <br /> The name of the new command, which must begin with a \, is the name you'll use in the document to use the command. The number of arguments is how many inputs will be sent to the command. The definition is just normal LaTeX code, with #1, #2, #3, etc., placed where you want the inputs to go when the new command is called.<br /> <br /> New commands can be used for all sorts of purposes, not just for making math commands you'll use a lot easier to call. For example, try this:<br /> &lt;pre&gt;&lt;nowiki&gt;<br /> \documentclass[11pt]{article}<br /> \usepackage{amsmath}<br /> <br /> \pdfpagewidth 8.5in<br /> \pdfpageheight 11in<br /> \newcounter{prob_num}<br /> \setcounter{prob_num}{1}<br /> \newcommand{\prob}{\bigskip \bigskip\arabic{prob_num}.\stepcounter{prob_num} #1<br /> \par\nopagebreak\medskip A.\ #2\hfill B.\ #3\hfill<br /> C.\ #4\hfill D.\ #5\hfill E.\ NOTA}<br /> <br /> \begin{document}<br /> <br /> \prob{What is$2+2$?}{4}{5}{6}{7}<br /> <br /> \prob{What is$\sqrt{100}$?}{81}{10}{9}{1}<br /> <br /> \prob{Evaluate$\sum_{n=1}^\infty \frac{1}{n^2}$.}<br /> {$\frac{1}{e}$} {$\frac{2}{\pi}$}<br /> {$\frac{\pi^3}{8}$} {$\frac{\pi^2}{6}$}<br /> <br /> \end{document}<br /> &lt;/nowiki&gt;&lt;/pre&gt;<br /> In the example above, we create a new command called \prob. Each time we call \prob, we supply 5 arguments, one for the question and one for each of the multiple choices.<br /> <br /> In the preamble and the definition of \prob, you'll see a few new LaTeX commands:<br /> <br /> \newcounter{prob_num} creates a counter variable called prob_num<br /> <br /> \setcounter{prob_num}{1} setsprob_num to equal 1.<br /> <br /> In the definition of \prob, the \bigskip and \medskip commands create vertical space.<br /> <br /> \arabic{prob_num} prints out the current value of the counter prob_num as an arabic numeral.<br /> <br /> \stepcounter{prob_num} increments the counter prob_num by 1.<br /> <br /> \nopagebreak tells LaTeX not to break the page between the problem and the choices unless it really, really, really has to.<br /> <br /> The \hfill commands put roughly equal space between the choices.<br /> <br /> Once you build a body of custom commands that you will be using in many LaTeX documents, you should learn about [[LaTeX:Packages|creating your own package]] so you don't have to copy all your custom commands from document to document.<br /> <br /> ==See Also==<br /> *[[LaTeX:Packages | Next: Packages]]<br /> *[[LaTeX:Symbols | Previous: Symbols]]</div> Scrabbler94 https://artofproblemsolving.com/wiki/index.php?title=LaTeX:Commands&diff=121176 LaTeX:Commands 2020-04-20T05:34:20Z <p>Scrabbler94: /* Fractions */ remove spacing after \frac</p> <hr /> <div>{{LaTeX: Commands}}<br /> <br /> This page introduces various useful commands for rendering math in LaTeX, as well as instructions for building your own commands.<br /> <br /> ===Subscripts and Superscripts===<br /> Subscripts and superscripts (such as exponents) can be made using the underscore _ and carat ^ symbols respectively.<br /> {| class=&quot;latextable&quot;<br /> !Symbol !! Command!!Symbol!!Command<br /> |-<br /> |&lt;math&gt;2^{2}&lt;/math&gt;||2^2||&lt;math&gt;\textstyle a_i&lt;/math&gt;||a_i<br /> |-<br /> | &lt;math&gt;\textstyle 2^{23}&lt;/math&gt;||2^{23}||&lt;math&gt;\textstyle n_{i-1}&lt;/math&gt;||n_{i-1}<br /> |-<br /> | &lt;math&gt;a^{i+1}_3&lt;/math&gt;||a^{i+1}_3||&lt;math&gt;x^{3^2}&lt;/math&gt;||x^{3^2}<br /> |-<br /> | &lt;math&gt;2^{a_i}&lt;/math&gt;||2^{a_i}||&lt;math&gt;2^a_i&lt;/math&gt;||2^a_i<br /> |}<br /> Notice that we can apply both a subscript and a superscript at the same time. For subscripts or superscripts with more than one character, you must surround with curly braces. For example, &lt;code&gt;x^10&lt;/code&gt; produces &lt;math&gt;x^10&lt;/math&gt;, while &lt;code&gt;x^{10}&lt;/code&gt; produces &lt;math&gt;x^{10}&lt;/math&gt;.<br /> <br /> ==Math Commands==<br /> Here are some commonly used math commands in LaTeX:<br /> ===Fractions===<br /> {|class=&quot;latextable&quot;<br /> !Symbol!!Command<br /> |-<br /> |&lt;math&gt;\frac {1}{2}&lt;/math&gt;||\frac{1}{2} or \frac12<br /> |-<br /> | &lt;math&gt;\frac{2}{x+2}&lt;/math&gt;||\frac{2}{x+2}<br /> |-<br /> | &lt;math&gt;\frac{1+\frac{1}{x}}{3x + 2}&lt;/math&gt;||\frac{1+\frac{1}{x}}{3x + 2}<br /> |}<br /> <br /> <br /> Notice that with fractions with a 1-digit numerator and a 1-digit denominator, we can simply group the numerator and the denominator together as one number. However, for fractions with either a numerator or a denominator that requires more than one character, you need to surround everything in curly brackets.<br /> <br /> Use \cfrac for continued fractions.<br /> {| class=&quot;latextable&quot;<br /> !Expression !! Command<br /> |-<br /> |&lt;math&gt;\cfrac{2}{1+\cfrac{2}{1+\cfrac{2}{1+\cfrac{2}{1}}}}&lt;/math&gt;||\cfrac{2}{1+\cfrac{2}{1+\cfrac{2}{1+\cfrac{2}{1}}}}<br /> |}<br /> <br /> ===Radicals===<br /> {| class=&quot;latextable&quot;<br /> !Symbol !! Command<br /> |-<br /> |&lt;math&gt;\sqrt(2)&lt;/math&gt;||\sqrt(2)<br /> |-<br /> |&lt;math&gt;\sqrt{2}&lt;/math&gt;||\sqrt{2}<br /> |-<br /> | &lt;math&gt;\sqrt{x+y}&lt;/math&gt;||\sqrt{x+y}<br /> |-<br /> | &lt;math&gt;\sqrt{x+\frac{1}{2}}&lt;/math&gt;||\sqrt{x+\frac{1}{2}}<br /> |-<br /> | &lt;math&gt;\sqrt{3}&lt;/math&gt;||\sqrt{3}<br /> |-<br /> | &lt;math&gt;\sqrt[n]{x}&lt;/math&gt;||\sqrt[n]{x}<br /> |}<br /> <br /> ===Sums, Products, Limits and Logarithms===<br /> We use _ to get the 'bottom' parts of summations, products, and limits, as well as the subscripts of logarithms. We use ^ to get the 'top' parts of sums and products. (Integration symbols work the same way, as you'll see in the [[LaTeX:Commands#Calculus|calculus section]].) Click here for a few [[LaTeX:Commands#Other_Functions|other commands]] which take 'bottom' parts.<br /> {| class=&quot;latextable&quot;<br /> !Symbol !! Command<br /> |-<br /> |&lt;math&gt;\textstyle \sum_{i=1}^{\infty}\frac{1}{i}&lt;/math&gt;||\sum_{i=1}^{\infty}\frac{1}{i}<br /> |-<br /> | &lt;math&gt;\textstyle \prod_{n=1}^5\frac{n}{n-1}&lt;/math&gt;||\prod_{n=1}^5\frac{n}{n-1}<br /> |-<br /> | &lt;math&gt;\textstyle \lim_{x\to\infty}\frac{1}{x}&lt;/math&gt;||\lim_{x\to\infty}\frac{1}{x}<br /> |-<br /> |&lt;math&gt;\textstyle \log_n n^2&lt;/math&gt;||\log_n n^2<br /> |}<br /> Some of these are prettier in display mode:<br /> {| class=&quot;latextable&quot;<br /> !Symbol !! Command<br /> |-<br /> |&lt;math&gt;\sum_{i=1}^{\infty}\frac{1}{i}&lt;/math&gt;||\sum_{i=1}^{\infty}\frac{1}{i}<br /> |-<br /> | &lt;math&gt;\prod_{n=1}^5\frac{n}{n-1}&lt;/math&gt;||\prod_{n=1}^5\frac{n}{n-1}<br /> |-<br /> | &lt;math&gt;\lim_{x\to\infty}\frac{1}{x}&lt;/math&gt;||\lim_{x\to\infty}\frac{1}{x}<br /> |}<br /> Note that we can use sums, products, and logarithms without _ or ^ modifiers.<br /> {| class=&quot;latextable&quot;<br /> !Symbol !! Command<br /> |-<br /> |&lt;math&gt;\sum\frac{1}{i}&lt;/math&gt;||\sum\frac{1}{i}<br /> |-<br /> | &lt;math&gt;\frac{n}{n-1}&lt;/math&gt;||\frac{n}{n-1}<br /> |-<br /> | &lt;math&gt;\textstyle \log n^2&lt;/math&gt;||\log n^2<br /> |-<br /> | &lt;math&gt;\textstyle \ln e&lt;/math&gt;||\ln e<br /> |}<br /> <br /> ===Mods===<br /> {| class=&quot;latextable&quot;<br /> !Symbol !! Command<br /> |-<br /> |&lt;math&gt;9\equiv 3 \bmod{6}&lt;/math&gt;||9\equiv 3 \bmod{6}<br /> |-<br /> | &lt;math&gt;9\equiv 3 \pmod{6}&lt;/math&gt;||9\equiv 3 \pmod{6}<br /> |-<br /> | &lt;math&gt;9\equiv 3 \mod{6}&lt;/math&gt;||9\equiv 3 \mod{6}<br /> |-<br /> | &lt;math&gt;9\equiv 3\pod{6}&lt;/math&gt;||9\equiv 3 \pod{6}<br /> |}<br /> <br /> ===Combinations===<br /> {| class=&quot;latextable&quot;<br /> !Symbol !! Command<br /> |-<br /> |&lt;math&gt;\scriptstyle\binom{1}{1}&lt;/math&gt;||\binom{1}{1}<br /> |-<br /> | &lt;math&gt;\scriptstyle\binom{n-1}{r-1}&lt;/math&gt;||\binom{n-1}{r-1}<br /> |}<br /> These often look better in display mode:<br /> {| class=&quot;latextable&quot;<br /> !Symbol !! Command<br /> |-<br /> |&lt;math&gt;\dbinom{9}{3}&lt;/math&gt;||\dbinom{9}{3}<br /> |-<br /> | &lt;math&gt;\dbinom{n-1}{r-1}&lt;/math&gt;||\dbinom{n-1}{r-1}<br /> |}<br /> <br /> ===Trigonometric Functions===<br /> <br /> Most of these are just the abbreviation of the trigonometric function with simply a backslash added before the abbreviation. <br /> <br /> {| class=&quot;latextable&quot;<br /> <br /> !Symbol!!Command!!Symbol!!Command!!Symbol!!Command<br /> |-<br /> |&lt;math&gt;\textstyle \cos&lt;/math&gt;||\cos||&lt;math&gt;\textstyle \sin&lt;/math&gt;||\sin||&lt;math&gt;\textstyle \tan&lt;/math&gt;||\tan<br /> |-<br /> | &lt;math&gt;\sec&lt;/math&gt;||\sec||&lt;math&gt;\textstyle \textstyle \csc&lt;/math&gt;||\csc||&lt;math&gt;\textstyle \cot&lt;/math&gt;||\cot<br /> |-<br /> | &lt;math&gt;\textstyle \arccos&lt;/math&gt;||\arccos||&lt;math&gt;\textstyle \arcsin&lt;/math&gt;||\arcsin||&lt;math&gt;\textstyle \arctan&lt;/math&gt;||\arctan<br /> |-<br /> | &lt;math&gt;\textstyle \cosh&lt;/math&gt;||\cosh||&lt;math&gt;\textstyle \sinh&lt;/math&gt;||\sinh||&lt;math&gt;\textstyle \tanh&lt;/math&gt;||\tanh<br /> |-<br /> | &lt;math&gt;\textstyle \coth&lt;/math&gt;||\coth <br /> |}<br /> Here are a couple examples:<br /> {| class=&quot;latextable&quot;<br /> !Symbol !! Command<br /> |-<br /> |&lt;math&gt;\textstyle \cos^2 x +\sin^2 x = 1&lt;/math&gt;||\cos^2 x +\sin^2 x = 1<br /> |-<br /> | &lt;math&gt;\cos 90^\circ = 0&lt;/math&gt;||\cos 90^\circ = 0 <br /> |}<br /> <br /> ===Calculus===<br /> Below are examples of calculus rendered in LaTeX. Most of these commands have been introduced before. Notice how definite integrals are rendered (and the difference between regular math and display mode for definite integrals). The , in the integrals makes a small space before the dx.<br /> {| class=&quot;latextable&quot;<br /> !Symbol !! Command<br /> |-<br /> |&lt;math&gt;\frac{d}{dx}\left(x^2\right) = 2x&lt;/math&gt;||\frac{d}{dx}\left(x^2\right) = 2x<br /> |-<br /> | &lt;math&gt;\int 2x\,dx = x^2+C&lt;/math&gt;||\int 2x\,dx = x^2+C<br /> |-<br /> | &lt;math&gt;\int^5_1 2x\,dx = 24&lt;/math&gt;||\int^5_1 2x\,dx = 24<br /> |-<br /> | &lt;math&gt;\int^5_1 2x\,dx = 24&lt;/math&gt;||\int^5_1 2x\,dx = 24<br /> |-<br /> | &lt;math&gt;\frac{\partial^2U}{\partial x^2} + \frac{\partial^2U}{\partial y^2}&lt;/math&gt;||\frac{\partial^2U}{\partial x^2} + \frac{\partial^2U}{\partial y^2}<br /> |-<br /> | &lt;math&gt;\frac{1}{4\pi}\oint_\Sigma\frac{1}{r}\frac{\partial U}{\partial n} ds&lt;/math&gt;||\frac{1}{4\pi}\oint_\Sigma\frac{1}{r}\frac{\partial U}{\partial n} ds<br /> |}<br /> <br /> ===Overline and Underline===<br /> <br /> {| class=&quot;latextable&quot;<br /> !Symbol !! Command<br /> |-<br /> |&lt;math&gt;\overline{a+bi}&lt;/math&gt;||\overline{a+bi}<br /> |-<br /> | &lt;math&gt;\underline{747}&lt;/math&gt;||\underline{747}<br /> |}<br /> <br /> == LaTeX ==<br /> ===Other Functions===<br /> {| class=&quot;latextable&quot;<br /> !Symbol !! Command!!Symbol !! Command!!Symbol !! Command<br /> |-<br /> |&lt;math&gt;\arg&lt;/math&gt;||\arg||&lt;math&gt;\textstyle\deg&lt;/math&gt;||\deg||&lt;math&gt;\textstyle\det&lt;/math&gt;||\det<br /> |-<br /> | &lt;math&gt;\dim&lt;/math&gt;||\dim||&lt;math&gt;\textstyle\exp&lt;/math&gt;||\exp||&lt;math&gt;\textstyle\gcd&lt;/math&gt;||\gcd<br /> |-<br /> |&lt;math&gt;\hom&lt;/math&gt;||\hom||&lt;math&gt;\inf&lt;/math&gt;||\inf||&lt;math&gt;\ker&lt;/math&gt;||\ker<br /> |-<br /> | &lt;math&gt;\textstyle\lg&lt;/math&gt;||\lg||&lt;math&gt;\liminf&lt;/math&gt;||\liminf||&lt;math&gt;\limsup&lt;/math&gt;||\limsup<br /> |-<br /> | &lt;math&gt;\textstyle\max&lt;/math&gt;||\max||&lt;math&gt;\textstyle\min&lt;/math&gt;||\min||&lt;math&gt;\Pr&lt;/math&gt;||\Pr<br /> |-<br /> | &lt;math&gt;\sup&lt;/math&gt;||\sup<br /> |}<br /> Some of these functions take 'bottom' parts just like sums and limits. Some render differently in display mode and regular math mode.<br /> {| class=&quot;latextable&quot;<br /> !Symbol !! Command!!Symbol !! Command!!Symbol !! Command<br /> |-<br /> | &lt;math&gt;\dim_x&lt;/math&gt;||\dim_x||&lt;math&gt;\textstyle\gcd_x&lt;/math&gt;||\gcd_x||&lt;math&gt;\inf_x&lt;/math&gt;||\inf_x<br /> |-<br /> | &lt;math&gt;\liminf_x&lt;/math&gt;||\liminf_x||&lt;math&gt;\limsup_x&lt;/math&gt;||\limsup_x||&lt;math&gt;\textstyle\max_x&lt;/math&gt;||\max_x<br /> |-<br /> | &lt;math&gt;\textstyle\min_x&lt;/math&gt;||\min_x||&lt;math&gt;\Pr_x&lt;/math&gt;||\Pr_x||&lt;math&gt;\sup_x&lt;/math&gt;||\sup_x<br /> |}<br /> <br /> ==Matrices==<br /> We can build an array or matrix with the \begin{array} command, and use \left and \right to properly size the delimiters around the matrix:<br /> &lt;pre&gt;&lt;nowiki&gt;<br /> The characteristic polynomial$f(\lambda)$of the<br />$3 \times 3$matrix<br /> $<br /> \left(<br /> \begin{array}{ccc}<br /> a &amp; b &amp; c &lt;br /&gt;d &amp; e &amp; f &lt;br /&gt;g &amp; h &amp; i \end{array}<br /> \right)$<br /> is given by the equation<br /> $f(\lambda)<br /> = \left|<br /> \begin{array}{ccc}<br /> \lambda - a &amp; -b &amp; -c &lt;br /&gt;-d &amp; \lambda - e &amp; -f &lt;br /&gt;-g &amp; -h &amp; \lambda - i \end{array}<br /> \right|.$<br /> &lt;/nowiki&gt;<br /> &lt;/pre&gt;<br /> More simply, we can use the shortcut commands in the amsmath package:<br /> &lt;pre&gt;&lt;nowiki&gt;<br /> The characteristic polynomial$f(\lambda)$of the<br />$3 \times 3$matrix<br /> $<br /> \begin{pmatrix}<br /> a &amp; b &amp; c &lt;br /&gt;d &amp; e &amp; f &lt;br /&gt;g &amp; h &amp; i<br /> \end{pmatrix}$<br /> is given by the equation<br /> $f(\lambda)<br /> = \begin{vmatrix}<br /> \lambda - a &amp; -b &amp; -c &lt;br /&gt;-d &amp; \lambda - e &amp; -f &lt;br /&gt;-g &amp; -h &amp; \lambda - i<br /> \end{vmatrix}.$<br /> &lt;/nowiki&gt;&lt;/pre&gt;<br /> You can read more about how the array command works [[LaTeX:Layout|here]] (it works the same as tabular).<br /> <br /> We can also use this environment to typeset any mathematics that calls for multiple columns, such as funky function definitions like this one:<br /> &lt;pre&gt;&lt;nowiki&gt;<br /> $f(x) = \left\{ \begin{array}{ll}<br /> x+7 &amp; \mbox{if 5&lt; x};&lt;br /&gt;x^2-3 &amp; \mbox{if -3 \le x \le 5};&lt;br /&gt;-x &amp; \mbox{if x &lt; -3}.\end{array} \right.$<br /> &lt;/nowiki&gt;&lt;/pre&gt;<br /> <br /> But it would be better to use the cases environment and \text command that the amsmath package provides:<br /> &lt;pre&gt;&lt;nowiki&gt;<br /> $<br /> f(x) = \begin{cases}<br /> x+7 &amp; \text{if 5&lt; x}; &lt;br /&gt;x^2-3 &amp; \text{if -3 \le x \le 5};&lt;br /&gt;-x &amp; \text{if x &lt; -3}.<br /> \end{cases}<br />$<br /> &lt;/nowiki&gt;&lt;/pre&gt;<br /> <br /> ==Text Styles in Math Mode==<br /> You can render letters in various styles in math mode. Below are examples; you should be able to use these with any letters. The \mathbb requires the amsfonts package to be included in your document's preamble. Do not try to do \mathbb{year}. You'll get &lt;math&gt;\mathbb{year}&lt;/math&gt;, and that looks nothing like it!<br /> {| class=&quot;latextable&quot;<br /> !Symbol !! Command!!Symbol !! Command!!Symbol !! Command!!Symbol !! Command<br /> |-<br /> |&lt;math&gt;\mathbb{R}&lt;/math&gt;||\mathbb{R}||&lt;math&gt;\mathbf{R}&lt;/math&gt;||\mathbf{R}||&lt;math&gt;\mathcal{R}&lt;/math&gt;||\mathcal{R}||&lt;math&gt;\mathfrak{R}&lt;/math&gt;||\mathfrak{R}<br /> |-<br /> | [[Image:Mathbb1.gif]]||\mathbb{Z}||&lt;math&gt;\mathbf{Z}&lt;/math&gt;||\mathbf{Z}||&lt;math&gt;\mathcal{Z}&lt;/math&gt;||\mathcal{Z}||&lt;math&gt;\mathfrak{Z}&lt;/math&gt;||\mathfrak{Z}<br /> |-<br /> | &lt;math&gt;\mathbb{Q}&lt;/math&gt;||\mathbb{Q}||&lt;math&gt;\mathbf{Q}&lt;/math&gt;||\mathbf{Q}||&lt;math&gt;\mathcal{Q}&lt;/math&gt;||\mathcal{Q}||&lt;math&gt;\mathfrak{Q}&lt;/math&gt;||\mathfrak{Q}<br /> |}<br /> If you're persistent, you can dig a few more out of [ftp://ftp.ams.org/pub/tex/doc/amsfonts/amsfndoc.pdf this document].<br /> <br /> If you want to drop a little bit of text in the middle of math mode, you can use the \text command. The \text command is most useful in &lt;nowiki&gt;$$...$$&lt;/nowiki&gt; or &lt;nowiki&gt;$...$&lt;/nowiki&gt; mode, where breaking up the math mode would force the output on to a new line entirely.<br /> So<br /> &lt;pre&gt;&lt;nowiki&gt;<br /> $$n^2 + 5 = 30\text{ so we have }n=\pm5$$<br /> &lt;/nowiki&gt;&lt;/pre&gt;<br /> gives<br /> <br /> [[Image:Text1.gif]]<br /> <br /> ==How to Build Your Own Commands==<br /> The command \newcommand is used to create your own commands. We'll start with an example:<br /> &lt;pre&gt;&lt;nowiki&gt;<br /> \documentclass[11pt]{article}<br /> \usepackage{amsmath}<br /> <br /> \pdfpagewidth 8.5in<br /> \pdfpageheight 11in<br /> \newcommand{\reci}{\frac{1}{#1}}<br /> \newcommand{\hypot}{\sqrt{#1^2+#2^2}}<br /> \newcommand{\cbrt}{\sqrt{#1}}<br /> <br /> \begin{document}<br /> <br /> The reciprocal of 2 is$\reci{2}$.<br /> <br /> The hypotenuse has length$\hypot{3}{4}$.<br /> <br /> I'm sick of writing $\backslash$sqrt{2}$' all the time, just to get $\cbrt{2}$.<br /> <br /> \end{document}<br /> &lt;/nowiki&gt;&lt;/pre&gt;<br /> The \newcommand declarations are in the preamble. Each is of the form<br /> <br /> \newcommand{name of new command}[number of arguments]{definition}<br /> <br /> The name of the new command, which must begin with a \, is the name you'll use in the document to use the command. The number of arguments is how many inputs will be sent to the command. The definition is just normal LaTeX code, with #1, #2, #3, etc., placed where you want the inputs to go when the new command is called.<br /> <br /> New commands can be used for all sorts of purposes, not just for making math commands you'll use a lot easier to call. For example, try this:<br /> &lt;pre&gt;&lt;nowiki&gt;<br /> \documentclass[11pt]{article}<br /> \usepackage{amsmath}<br /> <br /> \pdfpagewidth 8.5in<br /> \pdfpageheight 11in<br /> \newcounter{prob_num}<br /> \setcounter{prob_num}{1}<br /> \newcommand{\prob}{\bigskip \bigskip\arabic{prob_num}.\stepcounter{prob_num} #1<br /> \par\nopagebreak\medskip A.\ #2\hfill B.\ #3\hfill<br /> C.\ #4\hfill D.\ #5\hfill E.\ NOTA}<br /> <br /> \begin{document}<br /> <br /> \prob{What is $2+2$?}{4}{5}{6}{7}<br /> <br /> \prob{What is $\sqrt{100}$?}{81}{10}{9}{1}<br /> <br /> \prob{Evaluate $\sum_{n=1}^\infty \frac{1}{n^2}$.}<br /> {$\frac{1}{e}$} {$\frac{2}{\pi}$}<br /> {$\frac{\pi^3}{8}$} {$\frac{\pi^2}{6}$}<br /> <br /> \end{document}<br /> &lt;/nowiki&gt;&lt;/pre&gt;<br /> In the example above, we create a new command called \prob. Each time we call \prob, we supply 5 arguments, one for the question and one for each of the multiple choices.<br /> <br /> In the preamble and the definition of \prob, you'll see a few new LaTeX commands:<br /> <br /> \newcounter{prob_num} creates a counter variable called prob_num<br /> <br /> \setcounter{prob_num}{1} setsprob_num to equal 1.<br /> <br /> In the definition of \prob, the \bigskip and \medskip commands create vertical space.<br /> <br /> \arabic{prob_num} prints out the current value of the counter prob_num as an arabic numeral.<br /> <br /> \stepcounter{prob_num} increments the counter prob_num by 1.<br /> <br /> \nopagebreak tells LaTeX not to break the page between the problem and the choices unless it really, really, really has to.<br /> <br /> The \hfill commands put roughly equal space between the choices.<br /> <br /> Once you build a body of custom commands that you will be using in many LaTeX documents, you should learn about [[LaTeX:Packages|creating your own package]] so you don't have to copy all your custom commands from document to document.<br /> <br /> ==See Also==<br /> *[[LaTeX:Packages | Next: Packages]]<br /> *[[LaTeX:Symbols | Previous: Symbols]]</div> Scrabbler94 https://artofproblemsolving.com/wiki/index.php?title=LaTeX:Commands&diff=121175 LaTeX:Commands 2020-04-20T05:33:43Z <p>Scrabbler94: /* Subscripts and Superscripts */</p> <hr /> <div>{{LaTeX: Commands}}<br /> <br /> This page introduces various useful commands for rendering math in LaTeX, as well as instructions for building your own commands.<br /> <br /> ===Subscripts and Superscripts===<br /> Subscripts and superscripts (such as exponents) can be made using the underscore _ and carat ^ symbols respectively.<br /> {| class=&quot;latextable&quot;<br /> !Symbol !! Command!!Symbol!!Command<br /> |-<br /> |&lt;math&gt;2^{2}&lt;/math&gt;||2^2||&lt;math&gt;\textstyle a_i&lt;/math&gt;||a_i<br /> |-<br /> | &lt;math&gt;\textstyle 2^{23}&lt;/math&gt;||2^{23}||&lt;math&gt;\textstyle n_{i-1}&lt;/math&gt;||n_{i-1}<br /> |-<br /> | &lt;math&gt;a^{i+1}_3&lt;/math&gt;||a^{i+1}_3||&lt;math&gt;x^{3^2}&lt;/math&gt;||x^{3^2}<br /> |-<br /> | &lt;math&gt;2^{a_i}&lt;/math&gt;||2^{a_i}||&lt;math&gt;2^a_i&lt;/math&gt;||2^a_i<br /> |}<br /> Notice that we can apply both a subscript and a superscript at the same time. For subscripts or superscripts with more than one character, you must surround with curly braces. For example, &lt;code&gt;x^10&lt;/code&gt; produces &lt;math&gt;x^10&lt;/math&gt;, while &lt;code&gt;x^{10}&lt;/code&gt; produces &lt;math&gt;x^{10}&lt;/math&gt;.<br /> <br /> ==Math Commands==<br /> Here are some commonly used math commands in LaTeX:<br /> ===Fractions===<br /> {|class=&quot;latextable&quot;<br /> !Symbol!!Command<br /> |-<br /> |&lt;math&gt;\frac {1}{2}&lt;/math&gt;||\frac {1}{2}<br /> |-<br /> | &lt;math&gt;\frac{2}{x+2}&lt;/math&gt;||\frac{2}{x+2}<br /> |-<br /> | &lt;math&gt;\frac{1+\frac{1}{x}}{3x + 2}&lt;/math&gt;||\frac{1+\frac{1}{x}}{3x + 2}<br /> |}<br /> <br /> <br /> Notice that with fractions with a 1-digit numerator and a 1-digit denominator, we can simply group the numerator and the denominator together as one number. However, for fractions with either a numerator or a denominator that requires more than one character, you need to surround everything in curly brackets.<br /> <br /> Use \cfrac for continued fractions.<br /> {| class=&quot;latextable&quot;<br /> !Expression !! Command<br /> |-<br /> |&lt;math&gt;\cfrac{2}{1+\cfrac{2}{1+\cfrac{2}{1+\cfrac{2}{1}}}}&lt;/math&gt;||\cfrac{2}{1+\cfrac{2}{1+\cfrac{2}{1+\cfrac{2}{1}}}}<br /> |}<br /> <br /> ===Radicals===<br /> {| class=&quot;latextable&quot;<br /> !Symbol !! Command<br /> |-<br /> |&lt;math&gt;\sqrt(2)&lt;/math&gt;||\sqrt(2)<br /> |-<br /> |&lt;math&gt;\sqrt{2}&lt;/math&gt;||\sqrt{2}<br /> |-<br /> | &lt;math&gt;\sqrt{x+y}&lt;/math&gt;||\sqrt{x+y}<br /> |-<br /> | &lt;math&gt;\sqrt{x+\frac{1}{2}}&lt;/math&gt;||\sqrt{x+\frac{1}{2}}<br /> |-<br /> | &lt;math&gt;\sqrt{3}&lt;/math&gt;||\sqrt{3}<br /> |-<br /> | &lt;math&gt;\sqrt[n]{x}&lt;/math&gt;||\sqrt[n]{x}<br /> |}<br /> <br /> ===Sums, Products, Limits and Logarithms===<br /> We use _ to get the 'bottom' parts of summations, products, and limits, as well as the subscripts of logarithms. We use ^ to get the 'top' parts of sums and products. (Integration symbols work the same way, as you'll see in the [[LaTeX:Commands#Calculus|calculus section]].) Click here for a few [[LaTeX:Commands#Other_Functions|other commands]] which take 'bottom' parts.<br /> {| class=&quot;latextable&quot;<br /> !Symbol !! Command<br /> |-<br /> |&lt;math&gt;\textstyle \sum_{i=1}^{\infty}\frac{1}{i}&lt;/math&gt;||\sum_{i=1}^{\infty}\frac{1}{i}<br /> |-<br /> | &lt;math&gt;\textstyle \prod_{n=1}^5\frac{n}{n-1}&lt;/math&gt;||\prod_{n=1}^5\frac{n}{n-1}<br /> |-<br /> | &lt;math&gt;\textstyle \lim_{x\to\infty}\frac{1}{x}&lt;/math&gt;||\lim_{x\to\infty}\frac{1}{x}<br /> |-<br /> |&lt;math&gt;\textstyle \log_n n^2&lt;/math&gt;||\log_n n^2<br /> |}<br /> Some of these are prettier in display mode:<br /> {| class=&quot;latextable&quot;<br /> !Symbol !! Command<br /> |-<br /> |&lt;math&gt;\sum_{i=1}^{\infty}\frac{1}{i}&lt;/math&gt;||\sum_{i=1}^{\infty}\frac{1}{i}<br /> |-<br /> | &lt;math&gt;\prod_{n=1}^5\frac{n}{n-1}&lt;/math&gt;||\prod_{n=1}^5\frac{n}{n-1}<br /> |-<br /> | &lt;math&gt;\lim_{x\to\infty}\frac{1}{x}&lt;/math&gt;||\lim_{x\to\infty}\frac{1}{x}<br /> |}<br /> Note that we can use sums, products, and logarithms without _ or ^ modifiers.<br /> {| class=&quot;latextable&quot;<br /> !Symbol !! Command<br /> |-<br /> |&lt;math&gt;\sum\frac{1}{i}&lt;/math&gt;||\sum\frac{1}{i}<br /> |-<br /> | &lt;math&gt;\frac{n}{n-1}&lt;/math&gt;||\frac{n}{n-1}<br /> |-<br /> | &lt;math&gt;\textstyle \log n^2&lt;/math&gt;||\log n^2<br /> |-<br /> | &lt;math&gt;\textstyle \ln e&lt;/math&gt;||\ln e<br /> |}<br /> <br /> ===Mods===<br /> {| class=&quot;latextable&quot;<br /> !Symbol !! Command<br /> |-<br /> |&lt;math&gt;9\equiv 3 \bmod{6}&lt;/math&gt;||9\equiv 3 \bmod{6}<br /> |-<br /> | &lt;math&gt;9\equiv 3 \pmod{6}&lt;/math&gt;||9\equiv 3 \pmod{6}<br /> |-<br /> | &lt;math&gt;9\equiv 3 \mod{6}&lt;/math&gt;||9\equiv 3 \mod{6}<br /> |-<br /> | &lt;math&gt;9\equiv 3\pod{6}&lt;/math&gt;||9\equiv 3 \pod{6}<br /> |}<br /> <br /> ===Combinations===<br /> {| class=&quot;latextable&quot;<br /> !Symbol !! Command<br /> |-<br /> |&lt;math&gt;\scriptstyle\binom{1}{1}&lt;/math&gt;||\binom{1}{1}<br /> |-<br /> | &lt;math&gt;\scriptstyle\binom{n-1}{r-1}&lt;/math&gt;||\binom{n-1}{r-1}<br /> |}<br /> These often look better in display mode:<br /> {| class=&quot;latextable&quot;<br /> !Symbol !! Command<br /> |-<br /> |&lt;math&gt;\dbinom{9}{3}&lt;/math&gt;||\dbinom{9}{3}<br /> |-<br /> | &lt;math&gt;\dbinom{n-1}{r-1}&lt;/math&gt;||\dbinom{n-1}{r-1}<br /> |}<br /> <br /> ===Trigonometric Functions===<br /> <br /> Most of these are just the abbreviation of the trigonometric function with simply a backslash added before the abbreviation. <br /> <br /> {| class=&quot;latextable&quot;<br /> <br /> !Symbol!!Command!!Symbol!!Command!!Symbol!!Command<br /> |-<br /> |&lt;math&gt;\textstyle \cos&lt;/math&gt;||\cos||&lt;math&gt;\textstyle \sin&lt;/math&gt;||\sin||&lt;math&gt;\textstyle \tan&lt;/math&gt;||\tan<br /> |-<br /> | &lt;math&gt;\sec&lt;/math&gt;||\sec||&lt;math&gt;\textstyle \textstyle \csc&lt;/math&gt;||\csc||&lt;math&gt;\textstyle \cot&lt;/math&gt;||\cot<br /> |-<br /> | &lt;math&gt;\textstyle \arccos&lt;/math&gt;||\arccos||&lt;math&gt;\textstyle \arcsin&lt;/math&gt;||\arcsin||&lt;math&gt;\textstyle \arctan&lt;/math&gt;||\arctan<br /> |-<br /> | &lt;math&gt;\textstyle \cosh&lt;/math&gt;||\cosh||&lt;math&gt;\textstyle \sinh&lt;/math&gt;||\sinh||&lt;math&gt;\textstyle \tanh&lt;/math&gt;||\tanh<br /> |-<br /> | &lt;math&gt;\textstyle \coth&lt;/math&gt;||\coth <br /> |}<br /> Here are a couple examples:<br /> {| class=&quot;latextable&quot;<br /> !Symbol !! Command<br /> |-<br /> |&lt;math&gt;\textstyle \cos^2 x +\sin^2 x = 1&lt;/math&gt;||\cos^2 x +\sin^2 x = 1<br /> |-<br /> | &lt;math&gt;\cos 90^\circ = 0&lt;/math&gt;||\cos 90^\circ = 0 <br /> |}<br /> <br /> ===Calculus===<br /> Below are examples of calculus rendered in LaTeX. Most of these commands have been introduced before. Notice how definite integrals are rendered (and the difference between regular math and display mode for definite integrals). The , in the integrals makes a small space before the dx.<br /> {| class=&quot;latextable&quot;<br /> !Symbol !! Command<br /> |-<br /> |&lt;math&gt;\frac{d}{dx}\left(x^2\right) = 2x&lt;/math&gt;||\frac{d}{dx}\left(x^2\right) = 2x<br /> |-<br /> | &lt;math&gt;\int 2x\,dx = x^2+C&lt;/math&gt;||\int 2x\,dx = x^2+C<br /> |-<br /> | &lt;math&gt;\int^5_1 2x\,dx = 24&lt;/math&gt;||\int^5_1 2x\,dx = 24<br /> |-<br /> | &lt;math&gt;\int^5_1 2x\,dx = 24&lt;/math&gt;||\int^5_1 2x\,dx = 24<br /> |-<br /> | &lt;math&gt;\frac{\partial^2U}{\partial x^2} + \frac{\partial^2U}{\partial y^2}&lt;/math&gt;||\frac{\partial^2U}{\partial x^2} + \frac{\partial^2U}{\partial y^2}<br /> |-<br /> | &lt;math&gt;\frac{1}{4\pi}\oint_\Sigma\frac{1}{r}\frac{\partial U}{\partial n} ds&lt;/math&gt;||\frac{1}{4\pi}\oint_\Sigma\frac{1}{r}\frac{\partial U}{\partial n} ds<br /> |}<br /> <br /> ===Overline and Underline===<br /> <br /> {| class=&quot;latextable&quot;<br /> !Symbol !! Command<br /> |-<br /> |&lt;math&gt;\overline{a+bi}&lt;/math&gt;||\overline{a+bi}<br /> |-<br /> | &lt;math&gt;\underline{747}&lt;/math&gt;||\underline{747}<br /> |}<br /> <br /> == LaTeX ==<br /> ===Other Functions===<br /> {| class=&quot;latextable&quot;<br /> !Symbol !! Command!!Symbol !! Command!!Symbol !! Command<br /> |-<br /> |&lt;math&gt;\arg&lt;/math&gt;||\arg||&lt;math&gt;\textstyle\deg&lt;/math&gt;||\deg||&lt;math&gt;\textstyle\det&lt;/math&gt;||\det<br /> |-<br /> | &lt;math&gt;\dim&lt;/math&gt;||\dim||&lt;math&gt;\textstyle\exp&lt;/math&gt;||\exp||&lt;math&gt;\textstyle\gcd&lt;/math&gt;||\gcd<br /> |-<br /> |&lt;math&gt;\hom&lt;/math&gt;||\hom||&lt;math&gt;\inf&lt;/math&gt;||\inf||&lt;math&gt;\ker&lt;/math&gt;||\ker<br /> |-<br /> | &lt;math&gt;\textstyle\lg&lt;/math&gt;||\lg||&lt;math&gt;\liminf&lt;/math&gt;||\liminf||&lt;math&gt;\limsup&lt;/math&gt;||\limsup<br /> |-<br /> | &lt;math&gt;\textstyle\max&lt;/math&gt;||\max||&lt;math&gt;\textstyle\min&lt;/math&gt;||\min||&lt;math&gt;\Pr&lt;/math&gt;||\Pr<br /> |-<br /> | &lt;math&gt;\sup&lt;/math&gt;||\sup<br /> |}<br /> Some of these functions take 'bottom' parts just like sums and limits. Some render differently in display mode and regular math mode.<br /> {| class=&quot;latextable&quot;<br /> !Symbol !! Command!!Symbol !! Command!!Symbol !! Command<br /> |-<br /> | &lt;math&gt;\dim_x&lt;/math&gt;||\dim_x||&lt;math&gt;\textstyle\gcd_x&lt;/math&gt;||\gcd_x||&lt;math&gt;\inf_x&lt;/math&gt;||\inf_x<br /> |-<br /> | &lt;math&gt;\liminf_x&lt;/math&gt;||\liminf_x||&lt;math&gt;\limsup_x&lt;/math&gt;||\limsup_x||&lt;math&gt;\textstyle\max_x&lt;/math&gt;||\max_x<br /> |-<br /> | &lt;math&gt;\textstyle\min_x&lt;/math&gt;||\min_x||&lt;math&gt;\Pr_x&lt;/math&gt;||\Pr_x||&lt;math&gt;\sup_x&lt;/math&gt;||\sup_x<br /> |}<br /> <br /> ==Matrices==<br /> We can build an array or matrix with the \begin{array} command, and use \left and \right to properly size the delimiters around the matrix:<br /> &lt;pre&gt;&lt;nowiki&gt;<br /> The characteristic polynomial $f(\lambda)$ of the<br /> $3 \times 3$ matrix<br /> $<br /> \left(<br /> \begin{array}{ccc}<br /> a &amp; b &amp; c &lt;br /&gt;d &amp; e &amp; f &lt;br /&gt;g &amp; h &amp; i \end{array}<br /> \right)$<br /> is given by the equation<br /> $f(\lambda)<br /> = \left|<br /> \begin{array}{ccc}<br /> \lambda - a &amp; -b &amp; -c &lt;br /&gt;-d &amp; \lambda - e &amp; -f &lt;br /&gt;-g &amp; -h &amp; \lambda - i \end{array}<br /> \right|.$<br /> &lt;/nowiki&gt;<br /> &lt;/pre&gt;<br /> More simply, we can use the shortcut commands in the amsmath package:<br /> &lt;pre&gt;&lt;nowiki&gt;<br /> The characteristic polynomial $f(\lambda)$ of the<br /> $3 \times 3$ matrix<br /> $<br /> \begin{pmatrix}<br /> a &amp; b &amp; c &lt;br /&gt;d &amp; e &amp; f &lt;br /&gt;g &amp; h &amp; i<br /> \end{pmatrix}$<br /> is given by the equation<br /> $f(\lambda)<br /> = \begin{vmatrix}<br /> \lambda - a &amp; -b &amp; -c &lt;br /&gt;-d &amp; \lambda - e &amp; -f &lt;br /&gt;-g &amp; -h &amp; \lambda - i<br /> \end{vmatrix}.$<br /> &lt;/nowiki&gt;&lt;/pre&gt;<br /> You can read more about how the array command works [[LaTeX:Layout|here]] (it works the same as tabular).<br /> <br /> We can also use this environment to typeset any mathematics that calls for multiple columns, such as funky function definitions like this one:<br /> &lt;pre&gt;&lt;nowiki&gt;<br /> $f(x) = \left\{ \begin{array}{ll}<br /> x+7 &amp; \mbox{if 5&lt; x};&lt;br /&gt;x^2-3 &amp; \mbox{if -3 \le x \le 5};&lt;br /&gt;-x &amp; \mbox{if x &lt; -3}.\end{array} \right.$<br /> &lt;/nowiki&gt;&lt;/pre&gt;<br /> <br /> But it would be better to use the cases environment and \text command that the amsmath package provides:<br /> &lt;pre&gt;&lt;nowiki&gt;<br /> $<br /> f(x) = \begin{cases}<br /> x+7 &amp; \text{if 5&lt; x}; &lt;br /&gt;x^2-3 &amp; \text{if -3 \le x \le 5};&lt;br /&gt;-x &amp; \text{if x &lt; -3}.<br /> \end{cases}<br />$<br /> &lt;/nowiki&gt;&lt;/pre&gt;<br /> <br /> ==Text Styles in Math Mode==<br /> You can render letters in various styles in math mode. Below are examples; you should be able to use these with any letters. The \mathbb requires the amsfonts package to be included in your document's preamble. Do not try to do \mathbb{year}. You'll get &lt;math&gt;\mathbb{year}&lt;/math&gt;, and that looks nothing like it!<br /> {| class=&quot;latextable&quot;<br /> !Symbol !! Command!!Symbol !! Command!!Symbol !! Command!!Symbol !! Command<br /> |-<br /> |&lt;math&gt;\mathbb{R}&lt;/math&gt;||\mathbb{R}||&lt;math&gt;\mathbf{R}&lt;/math&gt;||\mathbf{R}||&lt;math&gt;\mathcal{R}&lt;/math&gt;||\mathcal{R}||&lt;math&gt;\mathfrak{R}&lt;/math&gt;||\mathfrak{R}<br /> |-<br /> | [[Image:Mathbb1.gif]]||\mathbb{Z}||&lt;math&gt;\mathbf{Z}&lt;/math&gt;||\mathbf{Z}||&lt;math&gt;\mathcal{Z}&lt;/math&gt;||\mathcal{Z}||&lt;math&gt;\mathfrak{Z}&lt;/math&gt;||\mathfrak{Z}<br /> |-<br /> | &lt;math&gt;\mathbb{Q}&lt;/math&gt;||\mathbb{Q}||&lt;math&gt;\mathbf{Q}&lt;/math&gt;||\mathbf{Q}||&lt;math&gt;\mathcal{Q}&lt;/math&gt;||\mathcal{Q}||&lt;math&gt;\mathfrak{Q}&lt;/math&gt;||\mathfrak{Q}<br /> |}<br /> If you're persistent, you can dig a few more out of [ftp://ftp.ams.org/pub/tex/doc/amsfonts/amsfndoc.pdf this document].<br /> <br /> If you want to drop a little bit of text in the middle of math mode, you can use the \text command. The \text command is most useful in &lt;nowiki&gt;$$...$$&lt;/nowiki&gt; or &lt;nowiki&gt;$...$&lt;/nowiki&gt; mode, where breaking up the math mode would force the output on to a new line entirely.<br /> So<br /> &lt;pre&gt;&lt;nowiki&gt;<br /> $$n^2 + 5 = 30\text{ so we have }n=\pm5$$<br /> &lt;/nowiki&gt;&lt;/pre&gt;<br /> gives<br /> <br /> [[Image:Text1.gif]]<br /> <br /> ==How to Build Your Own Commands==<br /> The command \newcommand is used to create your own commands. We'll start with an example:<br /> &lt;pre&gt;&lt;nowiki&gt;<br /> \documentclass[11pt]{article}<br /> \usepackage{amsmath}<br /> <br /> \pdfpagewidth 8.5in<br /> \pdfpageheight 11in<br /> \newcommand{\reci}{\frac{1}{#1}}<br /> \newcommand{\hypot}{\sqrt{#1^2+#2^2}}<br /> \newcommand{\cbrt}{\sqrt{#1}}<br /> <br /> \begin{document}<br /> <br /> The reciprocal of 2 is $\reci{2}$.<br /> <br /> The hypotenuse has length $\hypot{3}{4}$.<br /> <br /> I'm sick of writing $\backslash$sqrt{2}$' all the time, just to get$\cbrt{2}$.<br /> <br /> \end{document}<br /> &lt;/nowiki&gt;&lt;/pre&gt;<br /> The \newcommand declarations are in the preamble. Each is of the form<br /> <br /> \newcommand{name of new command}[number of arguments]{definition}<br /> <br /> The name of the new command, which must begin with a \, is the name you'll use in the document to use the command. The number of arguments is how many inputs will be sent to the command. The definition is just normal LaTeX code, with #1, #2, #3, etc., placed where you want the inputs to go when the new command is called.<br /> <br /> New commands can be used for all sorts of purposes, not just for making math commands you'll use a lot easier to call. For example, try this:<br /> &lt;pre&gt;&lt;nowiki&gt;<br /> \documentclass[11pt]{article}<br /> \usepackage{amsmath}<br /> <br /> \pdfpagewidth 8.5in<br /> \pdfpageheight 11in<br /> \newcounter{prob_num}<br /> \setcounter{prob_num}{1}<br /> \newcommand{\prob}{\bigskip \bigskip\arabic{prob_num}.\stepcounter{prob_num} #1<br /> \par\nopagebreak\medskip A.\ #2\hfill B.\ #3\hfill<br /> C.\ #4\hfill D.\ #5\hfill E.\ NOTA}<br /> <br /> \begin{document}<br /> <br /> \prob{What is$2+2$?}{4}{5}{6}{7}<br /> <br /> \prob{What is$\sqrt{100}$?}{81}{10}{9}{1}<br /> <br /> \prob{Evaluate$\sum_{n=1}^\infty \frac{1}{n^2}$.}<br /> {$\frac{1}{e}$} {$\frac{2}{\pi}$}<br /> {$\frac{\pi^3}{8}$} {$\frac{\pi^2}{6}}<br /> <br /> \end{document}<br /> &lt;/nowiki&gt;&lt;/pre&gt;<br /> In the example above, we create a new command called \prob. Each time we call \prob, we supply 5 arguments, one for the question and one for each of the multiple choices.<br /> <br /> In the preamble and the definition of \prob, you'll see a few new LaTeX commands:<br /> <br /> \newcounter{prob_num} creates a counter variable called prob_num<br /> <br /> \setcounter{prob_num}{1} setsprob_num to equal 1.<br /> <br /> In the definition of \prob, the \bigskip and \medskip commands create vertical space.<br /> <br /> \arabic{prob_num} prints out the current value of the counter prob_num as an arabic numeral.<br /> <br /> \stepcounter{prob_num} increments the counter prob_num by 1.<br /> <br /> \nopagebreak tells LaTeX not to break the page between the problem and the choices unless it really, really, really has to.<br /> <br /> The \hfill commands put roughly equal space between the choices.<br /> <br /> Once you build a body of custom commands that you will be using in many LaTeX documents, you should learn about [[LaTeX:Packages|creating your own package]] so you don't have to copy all your custom commands from document to document.<br /> <br /> ==See Also==<br /> *[[LaTeX:Packages | Next: Packages]]<br /> *[[LaTeX:Symbols | Previous: Symbols]]</div> Scrabbler94 https://artofproblemsolving.com/wiki/index.php?title=LaTeX:Layout&diff=121174 LaTeX:Layout 2020-04-20T05:24:21Z <p>Scrabbler94: /* Referencing */ use \eqref instead of \ref for equations</p> <hr /> <div>{{Latex}}<br /> This article outlines some of the basics of layout in LaTeX.<br /> <br /> Note: Rather than typing up all the examples, you can copy-paste the examples into your TeXnicCenter files. We highly recommend opening up your TeXnicCenter and trying out each of the examples as you go. It takes almost no time at all to just copy-paste, compile, and view the results.<br /> <br /> ==Source File Format==<br /> The source file of a LaTeX broadly consists of two parts, the '''preamble''' and the '''document''' itself. The '''preamble''' consists of everything before the \begin{document} command. Things like margin settings, document style definitions, paragraph spacing settings, custom function definition and page numeration style are items that are set in the preamble. Often, much of the preamble is placed in a separate file and included using the \usepackage statement. This allows you to use the same code in many source files by just including a single line in each source file.<br /> <br /> Our next three sections deal primarily with preamble items, while the rest cover tools you might use within your document.<br /> <br /> ==The Preamble==<br /> <br /> ===Document Class===<br /> <br /> The first line of the file sets the '''document class''' with the &lt;code&gt;\documentclass[''options'']{''class''}&lt;/code&gt; command. For typical use the document class should be &lt;code&gt;article&lt;/code&gt;. (Other supported classes are &lt;code&gt;book&lt;/code&gt;, &lt;code&gt;report&lt;/code&gt;, &lt;code&gt;letter&lt;/code&gt;, as well as &lt;code&gt;beamer&lt;/code&gt; for presentation slides.) There are several options that can be set with &lt;code&gt;\documentclass&lt;/code&gt;, but generally they can be left to their defaults. One option that might need modification is the font size, which is 10pt by default but can be increased to 11pt or 12pt. A reference on other options for this command can be found [http://www.nada.kth.se/~carsten/latex/class.html here].<br /> <br /> ===Packages===<br /> <br /> [[LaTeX:Packages | Packages]] are included after the \documentclass statement using &lt;code&gt;\usepackage[''options'']{''packagename''}&lt;/code&gt;. Two common packages are amsmath, which allows you to write math formulas in your document, and graphicx with the pdftex option, which allows you to include images in your document.<br /> <br /> ===Page Setup===<br /> <br /> LaTeX automatically sets reasonable values for the page dimensions, orientation, etc. However, in some cases customization may be required. There are two ways to do this: the easy way, using several packages which do all the work for you, and the hard way, which involves doing all the work yourself.<br /> <br /> ====The Easy Way====<br /> <br /> The easy way involves using certain packages to do the heavy lifting. For example, to set the margins using the geometry package, use the line &lt;code&gt;\usepackage[margin=2.5cm]{geometry}&lt;/code&gt;. Check out the [http://tug.ctan.org/tex-archive/macros/latex/contrib/geometry/geometry.pdf geometry package user manual] for more detailed possibilities. As another example, if you want a layout where paragraphs are separated by some space, but not indented (like in most HTML pages), use the parskip package: &lt;code&gt;\usepackage{parskip}&lt;/code&gt;.<br /> <br /> ====The Hard Way====<br /> <br /> The hard way involves setting all the desired values manually. Here are some values that can be set:<br /> <br /> {| class=&quot;latextable&quot;<br /> |-valign=&quot;top&quot;<br /> | \pdfpageheight, \pdfpagewidth || Dimensions of the PDF file.<br /> |-valign=&quot;top&quot;<br /> |\topmargin || Length of margin at top of page above all printing. 1 inch is added to this value.<br /> |-valign=&quot;top&quot;<br /> |\evensidemargin || Left margin on even numbered pages. 1 inch is added to this value.<br /> |-valign=&quot;top&quot;<br /> |\oddsidemargin || Left margin on odd numbered pages. 1 inch is added to this value.<br /> |-valign=&quot;top&quot;<br /> |\headheight || Height of the page header.<br /> |-valign=&quot;top&quot;<br /> |\headsep || Distance from bottom of header to the body of text on a page.<br /> |-valign=&quot;top&quot;<br /> |\topskip || Distance from top of main text box to the baseline of the first line of text in the main text box.<br /> |-valign=&quot;top&quot;<br /> |\textheight, \textwidth || Height and width of main text box.<br /> |-valign=&quot;top&quot;<br /> |\footskip || Distance from bottom of body to the bottom of the footer.<br /> |-valign=&quot;top&quot;<br /> |\parskip || Distance between paragraphs.<br /> |-valign=&quot;top&quot;<br /> |\parindent || Amount of indentation at the first line of a paragraph.<br /> |}<br /> <br /> The &lt;code&gt;\setlength{''length''}{''value''}&lt;/code&gt; command is used to set these lengths. Units include inches (in), centimeters (cm) and points (pt). Note that these lengths can be negative. Here is an example:<br /> <br /> &lt;pre&gt;<br /> \setlength{\topmargin}{0in}<br /> \setlength{\headheight}{0in}<br /> \setlength{\headsep}{0in}<br /> \setlength{\textheight}{7.7in}<br /> \setlength{\textwidth}{6.5in}<br /> \setlength{\oddsidemargin}{0in}<br /> \setlength{\evensidemargin}{0in}<br /> \setlength{\parindent}{0.25in}<br /> \setlength{\parskip}{0.25in}<br /> &lt;/pre&gt;<br /> <br /> Note that the right margin of a page is automatically set based on the left margin (with &lt;code&gt;\oddsidemargin&lt;/code&gt; and &lt;code&gt;\evensidemargin&lt;/code&gt;) and the width of the text (with &lt;code&gt;\textwidth&lt;/code&gt;).<br /> <br /> There are many other parameters that you can set in the preamble, such as the title of the document, the header style, the footer style, page numbering, etc. You can consult books or Google for more information on these areas.<br /> <br /> ==Document Formatting==<br /> <br /> The document begins after the preamble. Use the command &lt;code&gt;\begin{document}&lt;/code&gt; to start the document and &lt;code&gt;\end{document}&lt;/code&gt; at the end.<br /> <br /> This section will cover techniques to format material in the document you create.<br /> <br /> ===Paragraphs===<br /> <br /> Any time LaTeX sees a blank line, it treats the next line as the start of a new paragraph. For example, try the following text:<br /> <br /> &lt;pre&gt;<br /> Fourscore and seven years ago our fathers brought forth on this continent a new nation, <br /> conceived in liberty and dedicated to the proposition that all men are created equal.<br /> <br /> Now we are engaged in a great civil war, testing whether that nation or any nation so<br /> conceived and so dedicated can long endure. We are met on a great battlefield of <br /> that war. We have come to dedicate a portion of it as a final resting place for those<br /> who died here that the nation might live. This we may, in all propriety do.<br /> But in a larger sense, we cannot dedicate, we cannot consecrate, we cannot hallow <br /> this ground. The brave men, living and dead who struggled here have hallowed it<br /> far above our poor power to add or detract. The world will little note nor long <br /> remember what we say here, but it can never forget what they did here.<br /> <br /> It is rather for us the living, we here be dedicated to the great task remaining before <br /> us--that from these honored dead we take increased devotion to that cause for <br /> which they here gave the last full measure of devotion--that we here highly <br /> resolve that these dead shall not have died in vain, that this nation shall <br /> have a new birth of freedom, and that government of the people, by the people, <br /> for the people shall not perish from the earth.<br /> &lt;/pre&gt;<br /> <br /> When you typeset this, you should find that each of the three sections above is indented and is its own paragraph. Moreover, you should see that if you don't have a full empty line between two lines in your source file, there is not only no new paragraph, but there is no line break, either. To end one paragraph and start another it is not enough to simply hit return and start typing on the next line - you must hit return twice and create an empty line for LaTeX to know to start a new paragraph.<br /> <br /> You can also create indents wherever you want in text by adding the \indent command, and you can suppress the automatic indent caused by a new paragraph by using \noindent. For example, try typesetting this:<br /> <br /> &lt;pre&gt;Fourscore and seven years ago our fathers brought forth on this continent a <br /> new nation, conceived in liberty and dedicated to the proposition that all men are <br /> created equal.<br /> <br /> \indent \indent Now \indent we \indent are engaged in a great civil war, testing <br /> whether that nation or any nation so conceived and so dedicated can long endure. <br /> We are met on a great battlefield of that war. We have come to dedicate a portion of <br /> it as a final resting place for those who died here that the nation might live. <br /> This we may, in all propriety do. But in a larger sense, we cannot dedicate, <br /> we cannot consecrate, we cannot hallow this ground. The brave men, living and <br /> dead who struggled here have hallowed it far above our poor power to add or <br /> detract. The world will little note nor long remember what we say here, but it <br /> can never forget what they did here.<br /> <br /> \noindent It is rather for us the living, we here be dedicated to the great task <br /> remaining before us--that from these honored dead we take increased devotion <br /> to that cause for which they here gave the last full measure of devotion--that <br /> we here highly resolve that these dead shall not have died in vain, that this <br /> nation shall have a new birth of freedom, and that government of the people, by <br /> the people, for the people shall not perish from the earth.<br /> &lt;/pre&gt;<br /> <br /> Note the effects of \indent and \noindent. Finally, understanding paragraphing rules in LaTeX is very important when using display math. Notice the difference in the following:<br /> <br /> &lt;pre&gt;&lt;nowiki&gt;<br /> Now, if we can prove<br /> $\Big(2\sqrt{\frac{a}{b}}\Big)^m + \Big(2\sqrt{\frac{b}{a}}\Big)^m<br /> \ge 2^{m+1}$<br /> then we will be done. Dividing both sides of this inequality by2^m$yields<br /> $\Big(\frac{a}{b}\Big)^m + \Big(\frac{b}{a}\Big)^m \ge 2,$<br /> which we can verify easily by AM-GM on the reciprocals that make up the LHS,<br /> thus we have our desired<br /> $\Big(1 + \frac{a}{b}\Big)^m + \Big(1 + \frac{b}{a}\Big)^m \ge<br /> \Big(2\sqrt{\frac{a}{b}}\Big)^m + \Big(2\sqrt{\frac{b}{a}}\Big)^m \ge 2^{m+1}.$<br /> &lt;/nowiki&gt;&lt;/pre&gt;<br /> and<br /> &lt;pre&gt;&lt;nowiki&gt;<br /> Now, if we can prove<br /> $\Big(2\sqrt{\frac{a}{b}}\Big)^m + \Big(2\sqrt{\frac{b}{a}}\Big)^m<br /> \ge 2^{m+1}$<br /> <br /> then we will be done. Dividing both sides of this inequality by$2^m$yields<br /> $\Big(\frac{a}{b}\Big)^m + \Big(\frac{b}{a}\Big)^m \ge 2,$<br /> <br /> which we can verify easily by AM-GM on the reciprocals that make up the LHS,<br /> thus we have our desired<br /> $\Big(1 + \frac{a}{b}\Big)^m + \Big(1 + \frac{b}{a}\Big)^m \ge<br /> \Big(2\sqrt{\frac{a}{b}}\Big)^m + \Big(2\sqrt{\frac{b}{a}}\Big)^m \ge 2^{m+1}.$<br /> &lt;/nowiki&gt;&lt;/pre&gt;<br /> In the latter, there are indents after each display math, which we clearly don't want. These are caused by the blank lines (and could be suppressed with \noindent).<br /> <br /> ===Sections===<br /> For longer documents that you want to split into parts, you can use LaTeX sectioning commands. Here's an example illustrating them. If you use the book document class, you can also use \chapter, but for most documents short of books, these should be sufficient.<br /> &lt;pre&gt;<br /> \section{In This First Section}<br /> <br /> This is the first section.<br /> <br /> \subsection{We Have This First Subsection}<br /> <br /> This is the first subsection.<br /> <br /> \subsubsection{And This Subsubsection}<br /> <br /> A subsubsection.<br /> <br /> \paragraph{And This Paragraph}<br /> <br /> A notable paragraph.<br /> <br /> \subparagraph{And This Subparagraph}<br /> <br /> A notable subparagraph.<br /> <br /> \subsubsection{And Then This Subsubsection}<br /> \section{The This Second Section}<br /> \subsection{We Have This Subsection}<br /> &lt;/pre&gt;<br /> <br /> ===Font Sizes and Styles===<br /> To change the font size, use any one of the following commands. To change it for just a portion of the page, enclose that potion in { } and have the relevant font size command occur right at the beginning of the text inside the curly braces. In order from smallest to largest, the font sizes you can use are:<br /> {| class=&quot;latextable&quot; style=&quot;margin: 1em auto 1em auto&quot;<br /> |-<br /> |\tiny<br /> |-<br /> |\scriptsize<br /> |-<br /> |\footnotesize<br /> |-<br /> |\small<br /> |-<br /> |\normalsize<br /> |-<br /> |\large<br /> |-<br /> |\Large<br /> |-<br /> |\LARGE<br /> |-<br /> |\huge<br /> |-<br /> |\Huge<br /> |}<br /> Try this out; the effects should be pretty clear:<br /> &lt;pre&gt;<br /> When I was born, I was {\small small}. Actually, {\scriptsize I was<br /> very small}. When I got older, I thought some day {\Large I would be<br /> large}, {\Huge maybe even gigantic}. But instead, I'm not even<br /> normalsize. {\small I'm still small.}<br /> &lt;/pre&gt;<br /> Here is a simple example that will probably show you all you need to know about bold, italics, and underlining.<br /> &lt;pre&gt;<br /> When something is \emph{really}, \textbf{really} important, you can<br /> \underline{underline it}, \emph{italicize it}, \textbf{bold it}. If<br /> you \underline{\textbf{\emph{must do all three}}}, then you can nest<br /> them.<br /> &lt;/pre&gt;<br /> Here is another example that demonstrates font families:<br /> &lt;pre&gt;<br /> You may want to write things \textsf{in a sans-serif font}, or<br /> \texttt{in a typewriter font}, or \textsl{in a slanted font} (which<br /> is \emph{slightly different} than italics). Sometimes it pays<br /> \textsc{to write things in small capitals}. You can next go to<br /> \textbf{bold and then \textsl{bold and slanted} and then back to just<br /> bold} again.<br /> &lt;/pre&gt;<br /> <br /> ===Spacing===<br /> There are a few spacing items you'll find useful in LaTeX. First, you can force a normal-size space (as between words) by using a single backslash followed by a space. This is particularly useful after periods: LaTeX interprets periods as ends of sentences, so it puts extra space after them, but if a period doesn't in fact end a sentence, you don't want that extra space. Try this to see an example.<br /> &lt;pre&gt;<br /> When Mr. Rogers read this, he was confused because the first sentence<br /> was only two words long. Mrs.\ Rogers wasn't confused at all.<br /> &lt;/pre&gt;<br /> In math mode, it's a little different. LaTeX ignores normal spaces in math mode, so all three of the following will come out the same:<br /> &lt;pre&gt;&lt;nowiki&gt;<br /> Spacing in math mode:<br /> <br />$x + y$<br /> <br />$x + y$<br /> <br />$x+y$<br /> &lt;/nowiki&gt;&lt;/pre&gt;<br /> Notice that the three math expressions come out all exactly the same. In general, you can trust math mode to space things out right rather than forcing any special spacing. This means that you should write formulas in your source document to be easily readable (by you), and trust LaTeX to do the right spacing.<br /> <br /> However, if you do need to tweak the spacing in math mode, there are some special commands:<br /> {|class=&quot;latextable&quot;<br /> |-<br /> | \,<br /> | a small space<br /> |-<br /> | \:<br /> | a medium space<br /> |-<br /> | \;<br /> | a large space<br /> |-<br /> | \quad<br /> | a really large space<br /> |-<br /> | \qquad<br /> | a huge space<br /> |-<br /> | \!<br /> | a negative space (moves things back to the left)<br /> |}<br /> <br /> Here are examples of these in action:<br /> &lt;pre&gt;&lt;nowiki&gt;<br />$x+y$<br /> <br />$x+\,y$<br /> <br />$x+\:y$<br /> <br />$x+\;y$<br /> <br />$x+\quad y$<br /> <br />$x+\qquad y$<br /> <br />$x+\!y<br /> &lt;/nowiki&gt;&lt;/pre&gt;<br /> Two spacing tools that can be bluntly used to move things are \hspace and \vspace.<br /> &lt;pre&gt;<br /> I \hspace{2in} like \hspace{1in} space.<br /> <br /> \vspace{5in}<br /> <br /> In every direction.<br /> &lt;/pre&gt;<br /> Sometimes, \hspace and \vspace will not produce the space you want; this happens most frequently at the beginning or the end of a line or of a page. If you want to force space there, use \hspace* or \vspace* instead.<br /> <br /> Two more that can be used even more bluntly are \hfill and \vfill, which cause LaTeX to create as much horizontal or vertical space as possible (within the boundaries of a page as defined in the preamble):<br /> &lt;pre&gt;<br /> I\hfill like \hfill space.<br /> \vfill<br /> In every direction.<br /> &lt;/pre&gt;<br /> Finally, you can force line breaks and prevent them in LaTeX. To force a linebreak somewhere in a block of text, simply use \\. To prevent a linebreak, you can use ~. The ~ in the text below prevents LaTeX from breaking a line between Professor and Reiter:<br /> &lt;pre&gt;<br /> This website is largely a result of Professor Harold Reiter getting<br /> me involved in math education again.\\ Professor~Reiter's dedication<br /> to educating eager math students is an inspiration. Students,<br /> parents, and educators who like this site should thank<br /> Professor~Reiter. If you don't already know Professor~Reiter, don't<br /> worry, you'll eventually meet him. Professor~Reiter meets everybody.<br /> &lt;/pre&gt;<br /> Try taking out the ~ symbols, and you'll probably see Professor and Reiter get split up at least once. Notice also that the \\ forces LaTeX to start a new line, but not a new paragraph. \par can be used to introduce a new paragraph.<br /> <br /> ===Justification (Centering, etc).===<br /> We can center text using<br /> <br /> \begin{center} text \end{center}<br /> <br /> and can justify it left or right with<br /> <br /> \begin{flushleft} text \end{flushleft}<br /> \begin{flushright} text \end{flushright}<br /> <br /> &lt;pre&gt;<br /> \begin{center}<br /> Fourscore and seven years ago our fathers brought forth on this<br /> continent a new nation, conceived in liberty and dedicated to the<br /> proposition that all men are created equal.<br /> \end{center}<br /> <br /> \begin{flushleft}<br /> Now we are engaged in a great civil war, testing whether that nation<br /> or any nation so conceived and so dedicated can long endure. We are<br /> met on a great battlefield of that war. We have come to dedicate a<br /> portion of it as a final resting place for those who died here that<br /> the nation might live. This we may, in all propriety do. But in a<br /> larger sense, we cannot dedicate, we cannot consecrate, we cannot<br /> hallow this ground. The brave men, living and dead who struggled here<br /> have hallowed it far above our poor power to add or detract. The<br /> world will little note nor long remember what we say here, but it can<br /> never forget what they did here.<br /> \end{flushleft}<br /> <br /> \begin{flushright}<br /> It is rather for us the living, we here be dedicated to the great<br /> task remaining before us--that from these honored dead we take<br /> increased devotion to that cause for which they here gave the last<br /> full measure of devotion--that we here highly resolve that these dead<br /> shall not have died in vain, that this nation shall have a new birth<br /> of freedom, and that government of the people, by the people, for the<br /> people shall not perish from the earth.<br /> \end{flushright}<br /> &lt;/pre&gt;<br /> You can also use \raggedleft to replace \begin{flushright} (and omit the \end{flushright}) and use \raggedright to replace \begin{flushleft} (and omit the \end{flushleft}).<br /> <br /> ===Tables===<br /> The primary way to build a table is to use the tabular environment. Here's an example:<br /> &lt;pre&gt;<br /> \begin{tabular}[t]{|l|ccccc|c|}<br /> \multicolumn{7}{c}{USAMTS Scores Round 1}\\\hline<br /> Name&amp;\#1&amp;\#2&amp;\#3&amp;\#4&amp;\#5&amp;Total\\\hline<br /> John Doe&amp;5&amp;5&amp;3&amp;2&amp;1&amp;16\\<br /> Jane Doe&amp;5&amp;5&amp;5&amp;4&amp;5&amp;24\\<br /> Richard Feynman&amp;5&amp;5&amp;5&amp;5&amp;5&amp;25\\\hline<br /> \end{tabular}<br /> &lt;/pre&gt;<br /> When you typeset that code, you should see a simple table&lt;!-- like [http://www.artofproblemsolving.com/LaTeX/Images/tabletex1.pdf this one]--&gt;. Read through the following general description of the tabular environment to understand how the code above produced the table.<br /> <br /> General form of the tabular environment:<br /> {|class=&quot;wikitable&quot; style=&quot;margin: 1em auto 1em auto&quot;<br /> |-<br /> |\begin{tabular}[''alignment'']{''columns''}<br /> |-<br /> |''rows''<br /> |-<br /> |\end{tabular}<br /> |}<br /> The italics show where you have to put code to create your table.<br /> <br /> ''alignment'' - put either b or t, or omit this completely. This determines how your table is vertically positioned with the text around it. This entry is not too important - experiment using different values (or omitting it) when you have a table in the midst of a document to get a better feel for it.<br /> <br /> ''columns'' - this describes the number of columns and the alignment of each column. Put r for a right-justified column, c for a centered column, and l for a left-justified column. Put a | if you want a vertical line between columns. For example, the column declaration {||rr|cc|l} will produce a table that has 2 vertical lines on the left, then two columns that are right-justified, then a vertical line, then 2 columns that are centered, then another vertical line, then a left-justified column. There are more complicating things that you can do, and even more complicated things if you include the array packing in your document (check a good LaTeX book for more details), but for most tables, the options we've described here are sufficient.<br /> <br /> ''rows'' - You can have as many rows as you like. For each row, you need an entry for each column. Each of these entries is separated by an &amp;. Use \\ to indicate that your input for that row is finished. Hence, if your column declaration was {cccc}, a possible row entry could be<br /> {| class=&quot;wikitable&quot; style=&quot;margin: 1em auto 1em auto&quot;<br /> |-<br /> |5&amp;5&amp;5&amp;5\\<br /> |}<br /> If you wish for one row to have fewer columns (i.e. one column takes up several of the usual table columns), use the command \multicolumn. In the example above, we had as our first row<br /> &lt;pre&gt;<br /> \multicolumn{7}{c}{USAMTS Scores Round 1}\\<br /> &lt;/pre&gt;<br /> The first { } indicates how many regular columns this entry will take up. The second { } indicates whether the text in this entry is right (r), left (l) or center (c) justified. The final { } contains our entry. As with the regular column declaration, use | if you want a vertical line before or after the entry of \multicolumn.<br /> <br /> In general, you can use \vline to introduce a vertical line anywhere in a table (try putting one between John and Nash in the example below and see what happens).<br /> <br /> Finally, at the end of some of the rows in our example, we have the command \hline. This produces a horizontal line after the row it follows. If you want a horizontal line atop a table, use \hline right before the first row. If you only want a horizontal line under a portion of the row, use \cline{start column-end column} as indicated in the example below:<br /> &lt;pre&gt;<br /> \begin{tabular}[t]{|l|l|cccc|c|}\hline<br /> \multicolumn{7}{|c|}{USAMTS Final Scores by Round}\\\hline<br /> Medal&amp;Name&amp;\#1&amp;\#2&amp;\#3&amp;\#4&amp;Total\\\hline\hline<br /> &amp;Richard Feynman&amp;25&amp;25&amp;25&amp;25&amp;100\\\cline{3-7}<br /> Gold&amp;Albert Einstein&amp;25&amp;25&amp;25&amp;25&amp;100\\\cline{3-7}<br /> &amp;Marie Curie&amp;25&amp;24&amp;24&amp;25&amp;98\\\hline<br /> Silver&amp;John Nash&amp;20&amp;20&amp;25&amp;24&amp;89\\\hline<br /> &amp;Jane Doe&amp;23&amp;\multicolumn{2}{c}{None}&amp;25&amp;48\\\cline{3-7}<br /> None&amp;John Doe&amp;\multicolumn{2}{c}{None}&amp;25&amp;20&amp;45\\\cline{3-7}<br /> &amp;Lazy Person&amp;5&amp;\multicolumn{3}{c|}{None}&amp;5\\\hline<br /> \end{tabular}<br /> &lt;/pre&gt;<br /> &lt;!--When you typeset this, you should get output [http://www.artofproblemsolving.com/LaTeX/Images/tabletex2.pdf like this]. --&gt;Note how we made a double horizontal line after the table headings.<br /> <br /> Finally, sometimes you'll want to create a table that consists solely of items in math mode. For such a table, use the array environment. The array environment works exactly like tabular, except that all its entries are rendered in math mode:<br /> &lt;pre&gt;<br /> $\begin{array}[b]{ccc}<br /> x&amp;y&amp;z \\<br /> y&amp;x&amp;z \\<br /> 1&amp;2&amp;3<br /> \end{array}$<br /> &lt;/pre&gt;<br /> Change both array declarations to tabular and delete the &lt;nowiki&gt;$&lt;/nowiki&gt; and &lt;nowiki&gt;$&lt;/nowiki&gt; and see what happens. You can do a number of other things with the array and tabular environments, but the above should cover most of what you'll want to do with them.<br /> <br /> If you build a table in which some entries are text that will take up multiple lines, you'll probably want to learn about boxes (below).<br /> <br /> ===Boxes===<br /> To create boxes of text that behave differently from the rest of the text, we can use<br /> <br /> \makebox[width][pos]{text}<br /> <br /> The width sets the width the of the box. The pos sets the positioning of the text - either r (right justified text), l (left justified), or s (stretched to fill the box). If the pos parameter is left out, as in \makebox[1in]{centerme}, the text is centered. The text is placed in the box. If you want to draw a box around the text, use \framebox just as you would use \makebox.<br /> <br /> \mbox{text} and \fbox{text} are quick versions of \makebox and \framebox, which create a box to fit the size of the text.<br /> <br /> Try the following example to see some of these features:<br /> &lt;pre&gt;<br /> I can create basic boxes for text \makebox[2in]{like this}. Notice<br /> that there's a 2in wide space with like this' in the middle of it.<br /> <br /> If I want to put a box around the text, I can use a frame box. The<br /> result looks \framebox[2in]{like this}.<br /> <br /> I can also justify the text to the right within a box<br /> \makebox[1.5in][r]{like so} or \framebox[2.5in][l]{like so}.<br /> <br /> We can also use quick versions of these. We can just \mbox{do this}<br /> or \fbox{this} to create a quick box that's exactly the size of what we put in it.<br /> &lt;/pre&gt;<br /> In some of the boxes above, you'll see the limitations of these boxes. In particular, LaTeX won't do a line break in the middle of a box, so your box might run off the right side of the page. For boxes that span many lines, we can use<br /> <br /> \parbox[pos]{width}{text}<br /> <br /> The pos is used to align the box with the text outside the box - set it to either b or t to align the bottom or top edge of box with the current baseline (try both and see what happens). You can also just omit this parameter and the box is centered vertically on the current line of text. Look at the following example to see how pos affects the box.<br /> <br /> The parameter width sets the width of the box and the text is what is in the box.<br /> <br /> Here's an example:<br /> &lt;pre&gt;<br /> \parbox[b]{2in}{I like using parbox to create funny little boxes of<br /> text all over my page. This one has its bottom edge aligned with the<br /> current line}<br /> CURRENT LINE.<br /> \parbox[t]{2in}{The top of my text is aligned with that current<br /> line.}<br /> <br /> \parbox{1.5in}{I'm just centered on the current line.} CURRENT LINE<br /> \parbox{2.5in}{You probably got a few Overfull or Underfull warnings<br /> when you typeset this. Sometimes narrow boxes will do that; if you're<br /> happy with the output, don't sweat it.}<br /> &lt;/pre&gt;<br /> \parbox is very useful with tables, as it can be used to include multirow input on a single row. Without it, long input in a cell will sometimes cause a table to stretch out into the margins. Note how boxes are used to create the tables below, and notice the effect of not using the \parbox in the second table.<br /> &lt;pre&gt;<br /> \begin{tabular}[t]{ccccc}<br /> \makebox[2.0in]{}&amp;&amp;\makebox[1.5in]{}&amp;&amp;<br /> \makebox[1.5in]{}\\\cline{1-1}\cline{3-3}\cline{5-5}<br /> Student Name&amp;&amp;\parbox[t][0.2in]{1.5in}{Art of Problem Solving<br /> Community Username}&amp;&amp;Student Email\\<br /> \end{tabular}<br /> <br /> \begin{tabular}[t]{ccccc}<br /> \makebox[2.0in]{}&amp;&amp;\makebox[1.5in]{}&amp;&amp;<br /> \makebox[1.5in]{}\\\cline{1-1}\cline{3-3}\cline{5-5}<br /> Student Name&amp;&amp;Art of Problem Solving Community Username&amp;&amp;Student<br /> Email\\<br /> \end{tabular}<br /> &lt;/pre&gt;<br /> <br /> ===Lists===<br /> There are three simple versions of lists you can make with LaTeX.<br /> <br /> To separate list items with bullet points, use itemize:<br /> &lt;pre&gt;<br /> \noindent Here's my list:<br /> <br /> \begin{itemize}<br /> \item Item 1.<br /> \item Item 2.<br /> \item Item 3.<br /> \end{itemize}<br /> &lt;/pre&gt;<br /> The \noindent is there to exhibit how the list is indented.<br /> <br /> If you want a numbered list, use enumerate:<br /> &lt;pre&gt;<br /> \noindent Here's my list:<br /> <br /> \begin{enumerate}<br /> \item Item 1.<br /> \item Item 2.<br /> \item Item 3.<br /> \end{enumerate}<br /> &lt;/pre&gt;<br /> If you want to make your own headings for each item, use description:<br /> &lt;pre&gt;<br /> \noindent Here's my list:<br /> <br /> \begin{description}<br /> \item[Heading 1.] Item 1.<br /> \item[Heading 2.] Item 2.<br /> \item[Heading 3.] Item 3.<br /> \end{description}<br /> &lt;/pre&gt;<br /> Lists can be nested up to four levels. Here's one nested 2 levels deep:<br /> &lt;pre&gt;<br /> \noindent Here's my list:<br /> <br /> \begin{itemize}<br /> \item Item 1.<br /> \begin{itemize}<br /> \item List 2, Item 1<br /> \item List 2, Item 2<br /> \end{itemize}<br /> \item Item 2.<br /> \item Item 3.<br /> \end{itemize}<br /> &lt;/pre&gt;<br /> You can use the command \renewcommand to change the labels for items. The labels for itemize have the names \labelitemi, \labelitemii, \labelitemiii, \labelitemiv (one for each level of nesting, with \labelitemi being the first):<br /> &lt;pre&gt;&lt;nowiki&gt;<br /> \renewcommand{\labelitemi}{\heartsuit$}<br /> \renewcommand{\labelitemii}{$\spadesuit}<br /> <br /> \noindent Here's my list:<br /> <br /> \begin{itemize}<br /> \item Item 1.<br /> \begin{itemize}<br /> \item List 2, Item 1<br /> \item List 2, Item 2<br /> \end{itemize}<br /> \item Item 2.<br /> \item Item 3.<br /> \end{itemize}<br /> &lt;/nowiki&gt;&lt;/pre&gt;<br /> As expected, you can also get pretty fancy with enumerate:<br /> &lt;pre&gt;<br /> \renewcommand{\labelenumi}{\Roman{enumi}.}<br /> \renewcommand{\labelenumii}{\Roman{enumi}. \alph{enumii}}<br /> <br /> \noindent Here's my list:<br /> <br /> \begin{enumerate}<br /> \item Item 1.<br /> \begin{enumerate}<br /> \item List 2, Item 1<br /> \item List 2, Item 2<br /> \end{enumerate}<br /> \item Item 2.<br /> \item Item 3.<br /> \end{enumerate}<br /> &lt;/pre&gt;<br /> The \labelenumi and \labelenumii are our labels for the two levels of nesting (as with itemize, there are four levels possible). The<br /> <br /> \renewcommand{\labelenumi}{\Roman{enumi}.}<br /> <br /> causes each item in the primary list to have a capitalized Roman numeral followed by a period. The enumi is the counter for the main items. The<br /> <br /> \renewcommand{\labelenumii}{\Roman{enumi}. \alph{enumii}}<br /> <br /> causes each item in the secondary list to have a capitalized Roman numeral of the primary list followed by a period, then a lowercase letter corresponding to the item of the secondary list. As you probably guessed, enumii is the counter that tells us which number of the secondary list we're on.<br /> <br /> In addition to \Roman and \alph, you can use \arabic, \roman, and \Alph to format counter numbers such as enumii to numbers, lowercase Roman numerals, or uppercase letters. Try them in place of \Roman and \alph above.<br /> <br /> ===Cross referencing===<br /> Sometimes you'll want to refer to past equations, or even to past pages. Rather than having to manually number equations then change your text if the equation labels change, or having to manually put in page numbers which might change later, you can use the referencing built into LaTeX. To tell LaTeX to note where something is so you can refer to it later, use<br /> <br /> \label{label name}<br /> <br /> To refer to it later, use<br /> <br /> \ref{label name}<br /> <br /> or<br /> <br /> \pageref{label name}<br /> <br /> The latter gives you the page number, and the former gives you the Theorem number or equation number or whatever other special number might be attached to the item labeled with label name. For referencing equations, it is preferable to use \eqref instead of \ref so that the equation is displayed in parentheses.<br /> <br /> When using \label, you'll have to compile your LaTeX code '''twice'''. The first compilation will produce a .aux file that contains all the label information based on your \label commands. The second compilation reads and uses that information to fill out your \ref and \pageref commands.<br /> <br /> Here's an example with several uses of \label and \ref.<br /> &lt;pre&gt;<br /> \section{Some Sums}\label{sec:formulas}<br /> <br /> Here are a few sums I know.<br /> <br /> \begin{eqnarray}<br /> 1+2+3+\cdots+n&amp;=&amp;\frac{n(n+1)}{2}\label{eqn:linear}\\<br /> 1^2+2^2+3^2+\cdots+n^2&amp;=&amp; \frac{n(n+1)(2n+1)}{6}\label{eqn:squares}\\<br /> 1^3+2^3+3^3+\cdots+n^3&amp;=&amp; \frac{n^2(n+1)^2}{4}\label{eqn:cubes}<br /> \end{eqnarray}<br /> <br /> I can find the sum of the first 10 squares easily with formula~\eqref{eqn:squares} above.<br /> <br /> \pagebreak<br /> <br /> \section{A Cool Relationship}<br /> <br /> Take a look at formulas~\eqref{eqn:linear} and~\eqref{eqn:cubes} on<br /> page~\pageref{eqn:linear} of section~\ref{sec:formulas}. Notice that the<br /> right side of~\eqref{eqn:cubes} is the square of the right side<br /> of~\eqref{eqn:linear}.<br /> &lt;/pre&gt;<br /> Remember, you have to compile the previous example '''twice''' in order the see the references. After the first compilation, LaTeX will warn you about &quot;undefined references&quot;, and you'll see &quot;??&quot; in your document where the references should be. Compile it a second time, and the references will work.<br /> <br /> The \label commands in the \begin{eqnarray} allow us to use \ref later to refer back to the equations. The \label{sec:formulas} allows us to use \ref to refer to the section. Note that we can use the same label for the \pageref as we use for the \ref. Also, note the ~ before each of the references; these aren't required, but are used for style concerns. They prevent a line break right before the reference. Generally, it's poor style to start off lines with numbers or references; these ~ symbols prevent that if possible.<br /> <br /> ==See Also==<br /> *[[LaTeX:Symbols | Next: Symbols]]<br /> *[[LaTeX:Pictures | Previous: Pictures]]</div> Scrabbler94 https://artofproblemsolving.com/wiki/index.php?title=LaTeX:Layout&diff=121173 LaTeX:Layout 2020-04-20T05:17:41Z <p>Scrabbler94: /* Document Class */</p> <hr /> <div>{{Latex}}<br /> This article outlines some of the basics of layout in LaTeX.<br /> <br /> Note: Rather than typing up all the examples, you can copy-paste the examples into your TeXnicCenter files. We highly recommend opening up your TeXnicCenter and trying out each of the examples as you go. It takes almost no time at all to just copy-paste, compile, and view the results.<br /> <br /> ==Source File Format==<br /> The source file of a LaTeX broadly consists of two parts, the '''preamble''' and the '''document''' itself. The '''preamble''' consists of everything before the \begin{document} command. Things like margin settings, document style definitions, paragraph spacing settings, custom function definition and page numeration style are items that are set in the preamble. Often, much of the preamble is placed in a separate file and included using the \usepackage statement. This allows you to use the same code in many source files by just including a single line in each source file.<br /> <br /> Our next three sections deal primarily with preamble items, while the rest cover tools you might use within your document.<br /> <br /> ==The Preamble==<br /> <br /> ===Document Class===<br /> <br /> The first line of the file sets the '''document class''' with the &lt;code&gt;\documentclass[''options'']{''class''}&lt;/code&gt; command. For typical use the document class should be &lt;code&gt;article&lt;/code&gt;. (Other supported classes are &lt;code&gt;book&lt;/code&gt;, &lt;code&gt;report&lt;/code&gt;, &lt;code&gt;letter&lt;/code&gt;, as well as &lt;code&gt;beamer&lt;/code&gt; for presentation slides.) There are several options that can be set with &lt;code&gt;\documentclass&lt;/code&gt;, but generally they can be left to their defaults. One option that might need modification is the font size, which is 10pt by default but can be increased to 11pt or 12pt. A reference on other options for this command can be found [http://www.nada.kth.se/~carsten/latex/class.html here].<br /> <br /> ===Packages===<br /> <br /> [[LaTeX:Packages | Packages]] are included after the \documentclass statement using &lt;code&gt;\usepackage[''options'']{''packagename''}&lt;/code&gt;. Two common packages are amsmath, which allows you to write math formulas in your document, and graphicx with the pdftex option, which allows you to include images in your document.<br /> <br /> ===Page Setup===<br /> <br /> LaTeX automatically sets reasonable values for the page dimensions, orientation, etc. However, in some cases customization may be required. There are two ways to do this: the easy way, using several packages which do all the work for you, and the hard way, which involves doing all the work yourself.<br /> <br /> ====The Easy Way====<br /> <br /> The easy way involves using certain packages to do the heavy lifting. For example, to set the margins using the geometry package, use the line &lt;code&gt;\usepackage[margin=2.5cm]{geometry}&lt;/code&gt;. Check out the [http://tug.ctan.org/tex-archive/macros/latex/contrib/geometry/geometry.pdf geometry package user manual] for more detailed possibilities. As another example, if you want a layout where paragraphs are separated by some space, but not indented (like in most HTML pages), use the parskip package: &lt;code&gt;\usepackage{parskip}&lt;/code&gt;.<br /> <br /> ====The Hard Way====<br /> <br /> The hard way involves setting all the desired values manually. Here are some values that can be set:<br /> <br /> {| class=&quot;latextable&quot;<br /> |-valign=&quot;top&quot;<br /> | \pdfpageheight, \pdfpagewidth || Dimensions of the PDF file.<br /> |-valign=&quot;top&quot;<br /> |\topmargin || Length of margin at top of page above all printing. 1 inch is added to this value.<br /> |-valign=&quot;top&quot;<br /> |\evensidemargin || Left margin on even numbered pages. 1 inch is added to this value.<br /> |-valign=&quot;top&quot;<br /> |\oddsidemargin || Left margin on odd numbered pages. 1 inch is added to this value.<br /> |-valign=&quot;top&quot;<br /> |\headheight || Height of the page header.<br /> |-valign=&quot;top&quot;<br /> |\headsep || Distance from bottom of header to the body of text on a page.<br /> |-valign=&quot;top&quot;<br /> |\topskip || Distance from top of main text box to the baseline of the first line of text in the main text box.<br /> |-valign=&quot;top&quot;<br /> |\textheight, \textwidth || Height and width of main text box.<br /> |-valign=&quot;top&quot;<br /> |\footskip || Distance from bottom of body to the bottom of the footer.<br /> |-valign=&quot;top&quot;<br /> |\parskip || Distance between paragraphs.<br /> |-valign=&quot;top&quot;<br /> |\parindent || Amount of indentation at the first line of a paragraph.<br /> |}<br /> <br /> The &lt;code&gt;\setlength{''length''}{''value''}&lt;/code&gt; command is used to set these lengths. Units include inches (in), centimeters (cm) and points (pt). Note that these lengths can be negative. Here is an example:<br /> <br /> &lt;pre&gt;<br /> \setlength{\topmargin}{0in}<br /> \setlength{\headheight}{0in}<br /> \setlength{\headsep}{0in}<br /> \setlength{\textheight}{7.7in}<br /> \setlength{\textwidth}{6.5in}<br /> \setlength{\oddsidemargin}{0in}<br /> \setlength{\evensidemargin}{0in}<br /> \setlength{\parindent}{0.25in}<br /> \setlength{\parskip}{0.25in}<br /> &lt;/pre&gt;<br /> <br /> Note that the right margin of a page is automatically set based on the left margin (with &lt;code&gt;\oddsidemargin&lt;/code&gt; and &lt;code&gt;\evensidemargin&lt;/code&gt;) and the width of the text (with &lt;code&gt;\textwidth&lt;/code&gt;).<br /> <br /> There are many other parameters that you can set in the preamble, such as the title of the document, the header style, the footer style, page numbering, etc. You can consult books or Google for more information on these areas.<br /> <br /> ==Document Formatting==<br /> <br /> The document begins after the preamble. Use the command &lt;code&gt;\begin{document}&lt;/code&gt; to start the document and &lt;code&gt;\end{document}&lt;/code&gt; at the end.<br /> <br /> This section will cover techniques to format material in the document you create.<br /> <br /> ===Paragraphs===<br /> <br /> Any time LaTeX sees a blank line, it treats the next line as the start of a new paragraph. For example, try the following text:<br /> <br /> &lt;pre&gt;<br /> Fourscore and seven years ago our fathers brought forth on this continent a new nation, <br /> conceived in liberty and dedicated to the proposition that all men are created equal.<br /> <br /> Now we are engaged in a great civil war, testing whether that nation or any nation so<br /> conceived and so dedicated can long endure. We are met on a great battlefield of <br /> that war. We have come to dedicate a portion of it as a final resting place for those<br /> who died here that the nation might live. This we may, in all propriety do.<br /> But in a larger sense, we cannot dedicate, we cannot consecrate, we cannot hallow <br /> this ground. The brave men, living and dead who struggled here have hallowed it<br /> far above our poor power to add or detract. The world will little note nor long <br /> remember what we say here, but it can never forget what they did here.<br /> <br /> It is rather for us the living, we here be dedicated to the great task remaining before <br /> us--that from these honored dead we take increased devotion to that cause for <br /> which they here gave the last full measure of devotion--that we here highly <br /> resolve that these dead shall not have died in vain, that this nation shall <br /> have a new birth of freedom, and that government of the people, by the people, <br /> for the people shall not perish from the earth.<br /> &lt;/pre&gt;<br /> <br /> When you typeset this, you should find that each of the three sections above is indented and is its own paragraph. Moreover, you should see that if you don't have a full empty line between two lines in your source file, there is not only no new paragraph, but there is no line break, either. To end one paragraph and start another it is not enough to simply hit return and start typing on the next line - you must hit return twice and create an empty line for LaTeX to know to start a new paragraph.<br /> <br /> You can also create indents wherever you want in text by adding the \indent command, and you can suppress the automatic indent caused by a new paragraph by using \noindent. For example, try typesetting this:<br /> <br /> &lt;pre&gt;Fourscore and seven years ago our fathers brought forth on this continent a <br /> new nation, conceived in liberty and dedicated to the proposition that all men are <br /> created equal.<br /> <br /> \indent \indent Now \indent we \indent are engaged in a great civil war, testing <br /> whether that nation or any nation so conceived and so dedicated can long endure. <br /> We are met on a great battlefield of that war. We have come to dedicate a portion of <br /> it as a final resting place for those who died here that the nation might live. <br /> This we may, in all propriety do. But in a larger sense, we cannot dedicate, <br /> we cannot consecrate, we cannot hallow this ground. The brave men, living and <br /> dead who struggled here have hallowed it far above our poor power to add or <br /> detract. The world will little note nor long remember what we say here, but it <br /> can never forget what they did here.<br /> <br /> \noindent It is rather for us the living, we here be dedicated to the great task <br /> remaining before us--that from these honored dead we take increased devotion <br /> to that cause for which they here gave the last full measure of devotion--that <br /> we here highly resolve that these dead shall not have died in vain, that this <br /> nation shall have a new birth of freedom, and that government of the people, by <br /> the people, for the people shall not perish from the earth.<br /> &lt;/pre&gt;<br /> <br /> Note the effects of \indent and \noindent. Finally, understanding paragraphing rules in LaTeX is very important when using display math. Notice the difference in the following:<br /> <br /> &lt;pre&gt;&lt;nowiki&gt;<br /> Now, if we can prove<br /> $\Big(2\sqrt{\frac{a}{b}}\Big)^m + \Big(2\sqrt{\frac{b}{a}}\Big)^m<br /> \ge 2^{m+1}$<br /> then we will be done. Dividing both sides of this inequality by2^m$yields<br /> $\Big(\frac{a}{b}\Big)^m + \Big(\frac{b}{a}\Big)^m \ge 2,$<br /> which we can verify easily by AM-GM on the reciprocals that make up the LHS,<br /> thus we have our desired<br /> $\Big(1 + \frac{a}{b}\Big)^m + \Big(1 + \frac{b}{a}\Big)^m \ge<br /> \Big(2\sqrt{\frac{a}{b}}\Big)^m + \Big(2\sqrt{\frac{b}{a}}\Big)^m \ge 2^{m+1}.$<br /> &lt;/nowiki&gt;&lt;/pre&gt;<br /> and<br /> &lt;pre&gt;&lt;nowiki&gt;<br /> Now, if we can prove<br /> $\Big(2\sqrt{\frac{a}{b}}\Big)^m + \Big(2\sqrt{\frac{b}{a}}\Big)^m<br /> \ge 2^{m+1}$<br /> <br /> then we will be done. Dividing both sides of this inequality by$2^m$yields<br /> $\Big(\frac{a}{b}\Big)^m + \Big(\frac{b}{a}\Big)^m \ge 2,$<br /> <br /> which we can verify easily by AM-GM on the reciprocals that make up the LHS,<br /> thus we have our desired<br /> $\Big(1 + \frac{a}{b}\Big)^m + \Big(1 + \frac{b}{a}\Big)^m \ge<br /> \Big(2\sqrt{\frac{a}{b}}\Big)^m + \Big(2\sqrt{\frac{b}{a}}\Big)^m \ge 2^{m+1}.$<br /> &lt;/nowiki&gt;&lt;/pre&gt;<br /> In the latter, there are indents after each display math, which we clearly don't want. These are caused by the blank lines (and could be suppressed with \noindent).<br /> <br /> ===Sections===<br /> For longer documents that you want to split into parts, you can use LaTeX sectioning commands. Here's an example illustrating them. If you use the book document class, you can also use \chapter, but for most documents short of books, these should be sufficient.<br /> &lt;pre&gt;<br /> \section{In This First Section}<br /> <br /> This is the first section.<br /> <br /> \subsection{We Have This First Subsection}<br /> <br /> This is the first subsection.<br /> <br /> \subsubsection{And This Subsubsection}<br /> <br /> A subsubsection.<br /> <br /> \paragraph{And This Paragraph}<br /> <br /> A notable paragraph.<br /> <br /> \subparagraph{And This Subparagraph}<br /> <br /> A notable subparagraph.<br /> <br /> \subsubsection{And Then This Subsubsection}<br /> \section{The This Second Section}<br /> \subsection{We Have This Subsection}<br /> &lt;/pre&gt;<br /> <br /> ===Font Sizes and Styles===<br /> To change the font size, use any one of the following commands. To change it for just a portion of the page, enclose that potion in { } and have the relevant font size command occur right at the beginning of the text inside the curly braces. In order from smallest to largest, the font sizes you can use are:<br /> {| class=&quot;latextable&quot; style=&quot;margin: 1em auto 1em auto&quot;<br /> |-<br /> |\tiny<br /> |-<br /> |\scriptsize<br /> |-<br /> |\footnotesize<br /> |-<br /> |\small<br /> |-<br /> |\normalsize<br /> |-<br /> |\large<br /> |-<br /> |\Large<br /> |-<br /> |\LARGE<br /> |-<br /> |\huge<br /> |-<br /> |\Huge<br /> |}<br /> Try this out; the effects should be pretty clear:<br /> &lt;pre&gt;<br /> When I was born, I was {\small small}. Actually, {\scriptsize I was<br /> very small}. When I got older, I thought some day {\Large I would be<br /> large}, {\Huge maybe even gigantic}. But instead, I'm not even<br /> normalsize. {\small I'm still small.}<br /> &lt;/pre&gt;<br /> Here is a simple example that will probably show you all you need to know about bold, italics, and underlining.<br /> &lt;pre&gt;<br /> When something is \emph{really}, \textbf{really} important, you can<br /> \underline{underline it}, \emph{italicize it}, \textbf{bold it}. If<br /> you \underline{\textbf{\emph{must do all three}}}, then you can nest<br /> them.<br /> &lt;/pre&gt;<br /> Here is another example that demonstrates font families:<br /> &lt;pre&gt;<br /> You may want to write things \textsf{in a sans-serif font}, or<br /> \texttt{in a typewriter font}, or \textsl{in a slanted font} (which<br /> is \emph{slightly different} than italics). Sometimes it pays<br /> \textsc{to write things in small capitals}. You can next go to<br /> \textbf{bold and then \textsl{bold and slanted} and then back to just<br /> bold} again.<br /> &lt;/pre&gt;<br /> <br /> ===Spacing===<br /> There are a few spacing items you'll find useful in LaTeX. First, you can force a normal-size space (as between words) by using a single backslash followed by a space. This is particularly useful after periods: LaTeX interprets periods as ends of sentences, so it puts extra space after them, but if a period doesn't in fact end a sentence, you don't want that extra space. Try this to see an example.<br /> &lt;pre&gt;<br /> When Mr. Rogers read this, he was confused because the first sentence<br /> was only two words long. Mrs.\ Rogers wasn't confused at all.<br /> &lt;/pre&gt;<br /> In math mode, it's a little different. LaTeX ignores normal spaces in math mode, so all three of the following will come out the same:<br /> &lt;pre&gt;&lt;nowiki&gt;<br /> Spacing in math mode:<br /> <br />$x + y$<br /> <br />$x + y$<br /> <br />$x+y$<br /> &lt;/nowiki&gt;&lt;/pre&gt;<br /> Notice that the three math expressions come out all exactly the same. In general, you can trust math mode to space things out right rather than forcing any special spacing. This means that you should write formulas in your source document to be easily readable (by you), and trust LaTeX to do the right spacing.<br /> <br /> However, if you do need to tweak the spacing in math mode, there are some special commands:<br /> {|class=&quot;latextable&quot;<br /> |-<br /> | \,<br /> | a small space<br /> |-<br /> | \:<br /> | a medium space<br /> |-<br /> | \;<br /> | a large space<br /> |-<br /> | \quad<br /> | a really large space<br /> |-<br /> | \qquad<br /> | a huge space<br /> |-<br /> | \!<br /> | a negative space (moves things back to the left)<br /> |}<br /> <br /> Here are examples of these in action:<br /> &lt;pre&gt;&lt;nowiki&gt;<br />$x+y$<br /> <br />$x+\,y$<br /> <br />$x+\:y$<br /> <br />$x+\;y$<br /> <br />$x+\quad y$<br /> <br />$x+\qquad y$<br /> <br />$x+\!y<br /> &lt;/nowiki&gt;&lt;/pre&gt;<br /> Two spacing tools that can be bluntly used to move things are \hspace and \vspace.<br /> &lt;pre&gt;<br /> I \hspace{2in} like \hspace{1in} space.<br /> <br /> \vspace{5in}<br /> <br /> In every direction.<br /> &lt;/pre&gt;<br /> Sometimes, \hspace and \vspace will not produce the space you want; this happens most frequently at the beginning or the end of a line or of a page. If you want to force space there, use \hspace* or \vspace* instead.<br /> <br /> Two more that can be used even more bluntly are \hfill and \vfill, which cause LaTeX to create as much horizontal or vertical space as possible (within the boundaries of a page as defined in the preamble):<br /> &lt;pre&gt;<br /> I\hfill like \hfill space.<br /> \vfill<br /> In every direction.<br /> &lt;/pre&gt;<br /> Finally, you can force line breaks and prevent them in LaTeX. To force a linebreak somewhere in a block of text, simply use \\. To prevent a linebreak, you can use ~. The ~ in the text below prevents LaTeX from breaking a line between Professor and Reiter:<br /> &lt;pre&gt;<br /> This website is largely a result of Professor Harold Reiter getting<br /> me involved in math education again.\\ Professor~Reiter's dedication<br /> to educating eager math students is an inspiration. Students,<br /> parents, and educators who like this site should thank<br /> Professor~Reiter. If you don't already know Professor~Reiter, don't<br /> worry, you'll eventually meet him. Professor~Reiter meets everybody.<br /> &lt;/pre&gt;<br /> Try taking out the ~ symbols, and you'll probably see Professor and Reiter get split up at least once. Notice also that the \\ forces LaTeX to start a new line, but not a new paragraph. \par can be used to introduce a new paragraph.<br /> <br /> ===Justification (Centering, etc).===<br /> We can center text using<br /> <br /> \begin{center} text \end{center}<br /> <br /> and can justify it left or right with<br /> <br /> \begin{flushleft} text \end{flushleft}<br /> \begin{flushright} text \end{flushright}<br /> <br /> &lt;pre&gt;<br /> \begin{center}<br /> Fourscore and seven years ago our fathers brought forth on this<br /> continent a new nation, conceived in liberty and dedicated to the<br /> proposition that all men are created equal.<br /> \end{center}<br /> <br /> \begin{flushleft}<br /> Now we are engaged in a great civil war, testing whether that nation<br /> or any nation so conceived and so dedicated can long endure. We are<br /> met on a great battlefield of that war. We have come to dedicate a<br /> portion of it as a final resting place for those who died here that<br /> the nation might live. This we may, in all propriety do. But in a<br /> larger sense, we cannot dedicate, we cannot consecrate, we cannot<br /> hallow this ground. The brave men, living and dead who struggled here<br /> have hallowed it far above our poor power to add or detract. The<br /> world will little note nor long remember what we say here, but it can<br /> never forget what they did here.<br /> \end{flushleft}<br /> <br /> \begin{flushright}<br /> It is rather for us the living, we here be dedicated to the great<br /> task remaining before us--that from these honored dead we take<br /> increased devotion to that cause for which they here gave the last<br /> full measure of devotion--that we here highly resolve that these dead<br /> shall not have died in vain, that this nation shall have a new birth<br /> of freedom, and that government of the people, by the people, for the<br /> people shall not perish from the earth.<br /> \end{flushright}<br /> &lt;/pre&gt;<br /> You can also use \raggedleft to replace \begin{flushright} (and omit the \end{flushright}) and use \raggedright to replace \begin{flushleft} (and omit the \end{flushleft}).<br /> <br /> ===Tables===<br /> The primary way to build a table is to use the tabular environment. Here's an example:<br /> &lt;pre&gt;<br /> \begin{tabular}[t]{|l|ccccc|c|}<br /> \multicolumn{7}{c}{USAMTS Scores Round 1}\\\hline<br /> Name&amp;\#1&amp;\#2&amp;\#3&amp;\#4&amp;\#5&amp;Total\\\hline<br /> John Doe&amp;5&amp;5&amp;3&amp;2&amp;1&amp;16\\<br /> Jane Doe&amp;5&amp;5&amp;5&amp;4&amp;5&amp;24\\<br /> Richard Feynman&amp;5&amp;5&amp;5&amp;5&amp;5&amp;25\\\hline<br /> \end{tabular}<br /> &lt;/pre&gt;<br /> When you typeset that code, you should see a simple table&lt;!-- like [http://www.artofproblemsolving.com/LaTeX/Images/tabletex1.pdf this one]--&gt;. Read through the following general description of the tabular environment to understand how the code above produced the table.<br /> <br /> General form of the tabular environment:<br /> {|class=&quot;wikitable&quot; style=&quot;margin: 1em auto 1em auto&quot;<br /> |-<br /> |\begin{tabular}[''alignment'']{''columns''}<br /> |-<br /> |''rows''<br /> |-<br /> |\end{tabular}<br /> |}<br /> The italics show where you have to put code to create your table.<br /> <br /> ''alignment'' - put either b or t, or omit this completely. This determines how your table is vertically positioned with the text around it. This entry is not too important - experiment using different values (or omitting it) when you have a table in the midst of a document to get a better feel for it.<br /> <br /> ''columns'' - this describes the number of columns and the alignment of each column. Put r for a right-justified column, c for a centered column, and l for a left-justified column. Put a | if you want a vertical line between columns. For example, the column declaration {||rr|cc|l} will produce a table that has 2 vertical lines on the left, then two columns that are right-justified, then a vertical line, then 2 columns that are centered, then another vertical line, then a left-justified column. There are more complicating things that you can do, and even more complicated things if you include the array packing in your document (check a good LaTeX book for more details), but for most tables, the options we've described here are sufficient.<br /> <br /> ''rows'' - You can have as many rows as you like. For each row, you need an entry for each column. Each of these entries is separated by an &amp;. Use \\ to indicate that your input for that row is finished. Hence, if your column declaration was {cccc}, a possible row entry could be<br /> {| class=&quot;wikitable&quot; style=&quot;margin: 1em auto 1em auto&quot;<br /> |-<br /> |5&amp;5&amp;5&amp;5\\<br /> |}<br /> If you wish for one row to have fewer columns (i.e. one column takes up several of the usual table columns), use the command \multicolumn. In the example above, we had as our first row<br /> &lt;pre&gt;<br /> \multicolumn{7}{c}{USAMTS Scores Round 1}\\<br /> &lt;/pre&gt;<br /> The first { } indicates how many regular columns this entry will take up. The second { } indicates whether the text in this entry is right (r), left (l) or center (c) justified. The final { } contains our entry. As with the regular column declaration, use | if you want a vertical line before or after the entry of \multicolumn.<br /> <br /> In general, you can use \vline to introduce a vertical line anywhere in a table (try putting one between John and Nash in the example below and see what happens).<br /> <br /> Finally, at the end of some of the rows in our example, we have the command \hline. This produces a horizontal line after the row it follows. If you want a horizontal line atop a table, use \hline right before the first row. If you only want a horizontal line under a portion of the row, use \cline{start column-end column} as indicated in the example below:<br /> &lt;pre&gt;<br /> \begin{tabular}[t]{|l|l|cccc|c|}\hline<br /> \multicolumn{7}{|c|}{USAMTS Final Scores by Round}\\\hline<br /> Medal&amp;Name&amp;\#1&amp;\#2&amp;\#3&amp;\#4&amp;Total\\\hline\hline<br /> &amp;Richard Feynman&amp;25&amp;25&amp;25&amp;25&amp;100\\\cline{3-7}<br /> Gold&amp;Albert Einstein&amp;25&amp;25&amp;25&amp;25&amp;100\\\cline{3-7}<br /> &amp;Marie Curie&amp;25&amp;24&amp;24&amp;25&amp;98\\\hline<br /> Silver&amp;John Nash&amp;20&amp;20&amp;25&amp;24&amp;89\\\hline<br /> &amp;Jane Doe&amp;23&amp;\multicolumn{2}{c}{None}&amp;25&amp;48\\\cline{3-7}<br /> None&amp;John Doe&amp;\multicolumn{2}{c}{None}&amp;25&amp;20&amp;45\\\cline{3-7}<br /> &amp;Lazy Person&amp;5&amp;\multicolumn{3}{c|}{None}&amp;5\\\hline<br /> \end{tabular}<br /> &lt;/pre&gt;<br /> &lt;!--When you typeset this, you should get output [http://www.artofproblemsolving.com/LaTeX/Images/tabletex2.pdf like this]. --&gt;Note how we made a double horizontal line after the table headings.<br /> <br /> Finally, sometimes you'll want to create a table that consists solely of items in math mode. For such a table, use the array environment. The array environment works exactly like tabular, except that all its entries are rendered in math mode:<br /> &lt;pre&gt;<br /> $\begin{array}[b]{ccc}<br /> x&amp;y&amp;z \\<br /> y&amp;x&amp;z \\<br /> 1&amp;2&amp;3<br /> \end{array}$<br /> &lt;/pre&gt;<br /> Change both array declarations to tabular and delete the &lt;nowiki&gt;$&lt;/nowiki&gt; and &lt;nowiki&gt;$&lt;/nowiki&gt; and see what happens. You can do a number of other things with the array and tabular environments, but the above should cover most of what you'll want to do with them.<br /> <br /> If you build a table in which some entries are text that will take up multiple lines, you'll probably want to learn about boxes (below).<br /> <br /> ===Boxes===<br /> To create boxes of text that behave differently from the rest of the text, we can use<br /> <br /> \makebox[width][pos]{text}<br /> <br /> The width sets the width the of the box. The pos sets the positioning of the text - either r (right justified text), l (left justified), or s (stretched to fill the box). If the pos parameter is left out, as in \makebox[1in]{centerme}, the text is centered. The text is placed in the box. If you want to draw a box around the text, use \framebox just as you would use \makebox.<br /> <br /> \mbox{text} and \fbox{text} are quick versions of \makebox and \framebox, which create a box to fit the size of the text.<br /> <br /> Try the following example to see some of these features:<br /> &lt;pre&gt;<br /> I can create basic boxes for text \makebox[2in]{like this}. Notice<br /> that there's a 2in wide space with like this' in the middle of it.<br /> <br /> If I want to put a box around the text, I can use a frame box. The<br /> result looks \framebox[2in]{like this}.<br /> <br /> I can also justify the text to the right within a box<br /> \makebox[1.5in][r]{like so} or \framebox[2.5in][l]{like so}.<br /> <br /> We can also use quick versions of these. We can just \mbox{do this}<br /> or \fbox{this} to create a quick box that's exactly the size of what we put in it.<br /> &lt;/pre&gt;<br /> In some of the boxes above, you'll see the limitations of these boxes. In particular, LaTeX won't do a line break in the middle of a box, so your box might run off the right side of the page. For boxes that span many lines, we can use<br /> <br /> \parbox[pos]{width}{text}<br /> <br /> The pos is used to align the box with the text outside the box - set it to either b or t to align the bottom or top edge of box with the current baseline (try both and see what happens). You can also just omit this parameter and the box is centered vertically on the current line of text. Look at the following example to see how pos affects the box.<br /> <br /> The parameter width sets the width of the box and the text is what is in the box.<br /> <br /> Here's an example:<br /> &lt;pre&gt;<br /> \parbox[b]{2in}{I like using parbox to create funny little boxes of<br /> text all over my page. This one has its bottom edge aligned with the<br /> current line}<br /> CURRENT LINE.<br /> \parbox[t]{2in}{The top of my text is aligned with that current<br /> line.}<br /> <br /> \parbox{1.5in}{I'm just centered on the current line.} CURRENT LINE<br /> \parbox{2.5in}{You probably got a few Overfull or Underfull warnings<br /> when you typeset this. Sometimes narrow boxes will do that; if you're<br /> happy with the output, don't sweat it.}<br /> &lt;/pre&gt;<br /> \parbox is very useful with tables, as it can be used to include multirow input on a single row. Without it, long input in a cell will sometimes cause a table to stretch out into the margins. Note how boxes are used to create the tables below, and notice the effect of not using the \parbox in the second table.<br /> &lt;pre&gt;<br /> \begin{tabular}[t]{ccccc}<br /> \makebox[2.0in]{}&amp;&amp;\makebox[1.5in]{}&amp;&amp;<br /> \makebox[1.5in]{}\\\cline{1-1}\cline{3-3}\cline{5-5}<br /> Student Name&amp;&amp;\parbox[t][0.2in]{1.5in}{Art of Problem Solving<br /> Community Username}&amp;&amp;Student Email\\<br /> \end{tabular}<br /> <br /> \begin{tabular}[t]{ccccc}<br /> \makebox[2.0in]{}&amp;&amp;\makebox[1.5in]{}&amp;&amp;<br /> \makebox[1.5in]{}\\\cline{1-1}\cline{3-3}\cline{5-5}<br /> Student Name&amp;&amp;Art of Problem Solving Community Username&amp;&amp;Student<br /> Email\\<br /> \end{tabular}<br /> &lt;/pre&gt;<br /> <br /> ===Lists===<br /> There are three simple versions of lists you can make with LaTeX.<br /> <br /> To separate list items with bullet points, use itemize:<br /> &lt;pre&gt;<br /> \noindent Here's my list:<br /> <br /> \begin{itemize}<br /> \item Item 1.<br /> \item Item 2.<br /> \item Item 3.<br /> \end{itemize}<br /> &lt;/pre&gt;<br /> The \noindent is there to exhibit how the list is indented.<br /> <br /> If you want a numbered list, use enumerate:<br /> &lt;pre&gt;<br /> \noindent Here's my list:<br /> <br /> \begin{enumerate}<br /> \item Item 1.<br /> \item Item 2.<br /> \item Item 3.<br /> \end{enumerate}<br /> &lt;/pre&gt;<br /> If you want to make your own headings for each item, use description:<br /> &lt;pre&gt;<br /> \noindent Here's my list:<br /> <br /> \begin{description}<br /> \item[Heading 1.] Item 1.<br /> \item[Heading 2.] Item 2.<br /> \item[Heading 3.] Item 3.<br /> \end{description}<br /> &lt;/pre&gt;<br /> Lists can be nested up to four levels. Here's one nested 2 levels deep:<br /> &lt;pre&gt;<br /> \noindent Here's my list:<br /> <br /> \begin{itemize}<br /> \item Item 1.<br /> \begin{itemize}<br /> \item List 2, Item 1<br /> \item List 2, Item 2<br /> \end{itemize}<br /> \item Item 2.<br /> \item Item 3.<br /> \end{itemize}<br /> &lt;/pre&gt;<br /> You can use the command \renewcommand to change the labels for items. The labels for itemize have the names \labelitemi, \labelitemii, \labelitemiii, \labelitemiv (one for each level of nesting, with \labelitemi being the first):<br /> &lt;pre&gt;&lt;nowiki&gt;<br /> \renewcommand{\labelitemi}{\heartsuit$}<br /> \renewcommand{\labelitemii}{$\spadesuit$}<br /> <br /> \noindent Here's my list:<br /> <br /> \begin{itemize}<br /> \item Item 1.<br /> \begin{itemize}<br /> \item List 2, Item 1<br /> \item List 2, Item 2<br /> \end{itemize}<br /> \item Item 2.<br /> \item Item 3.<br /> \end{itemize}<br /> &lt;/nowiki&gt;&lt;/pre&gt;<br /> As expected, you can also get pretty fancy with enumerate:<br /> &lt;pre&gt;<br /> \renewcommand{\labelenumi}{\Roman{enumi}.}<br /> \renewcommand{\labelenumii}{\Roman{enumi}. \alph{enumii}}<br /> <br /> \noindent Here's my list:<br /> <br /> \begin{enumerate}<br /> \item Item 1.<br /> \begin{enumerate}<br /> \item List 2, Item 1<br /> \item List 2, Item 2<br /> \end{enumerate}<br /> \item Item 2.<br /> \item Item 3.<br /> \end{enumerate}<br /> &lt;/pre&gt;<br /> The \labelenumi and \labelenumii are our labels for the two levels of nesting (as with itemize, there are four levels possible). The<br /> <br /> \renewcommand{\labelenumi}{\Roman{enumi}.}<br /> <br /> causes each item in the primary list to have a capitalized Roman numeral followed by a period. The enumi is the counter for the main items. The<br /> <br /> \renewcommand{\labelenumii}{\Roman{enumi}. \alph{enumii}}<br /> <br /> causes each item in the secondary list to have a capitalized Roman numeral of the primary list followed by a period, then a lowercase letter corresponding to the item of the secondary list. As you probably guessed, enumii is the counter that tells us which number of the secondary list we're on.<br /> <br /> In addition to \Roman and \alph, you can use \arabic, \roman, and \Alph to format counter numbers such as enumii to numbers, lowercase Roman numerals, or uppercase letters. Try them in place of \Roman and \alph above.<br /> <br /> ===Referencing===<br /> Sometimes you'll want to refer to past equations, or even to past pages. Rather than having to manually number equations then change your text if the equation labels change, or having to manually put in page numbers which might change later, you can use the referencing built into LaTeX. To tell LaTeX to note where something is so you can refer to it later, use<br /> <br /> \label{label name}<br /> <br /> To refer to it later, use<br /> <br /> \ref{label name}<br /> <br /> or<br /> <br /> \pageref{label name}<br /> <br /> The latter gives you the page number, and the former gives you the Theorem number or equation number or whatever other special number might be attached to the item labeled with label name.<br /> <br /> When using \label, you'll have to compile your LaTeX code '''twice'''. The first compilation will produce a .aux file that contains all the label information based on your \label commands. The second compilation reads and uses that information to fill out your \ref and \pageref commands.<br /> <br /> Here's an example with several uses of \label and \ref.<br /> &lt;pre&gt;<br /> \section{Some Sums}<br /> <br /> Here are a few sums I know.\label{sec:formulas}<br /> <br /> \begin{eqnarray}<br /> 1+2+3+\cdots+n&amp;=&amp;\frac{n(n+1)}{2}\label{linear}\\<br /> 1^2+2^2+3^2+\cdots+n^2&amp;=&amp; \frac{n(n+1)(2n+1)}{6}\label{squares}\\<br /> 1^3+2^3+3^3+\cdots+n^3&amp;=&amp; \frac{n^2(n+1)^2}{4}\label{cubes}<br /> \end{eqnarray}<br /> <br /> I can find the sum of the first 10 squares easily with formula~(\ref{squares}) above.<br /> <br /> \pagebreak<br /> <br /> \section{A Cool Relationship}<br /> <br /> Take a look at formulas~(\ref{linear}) and~(\ref{cubes}) on<br /> page~\pageref{linear} of section~\ref{sec:formulas}. Notice that the<br /> right side of~(\ref{cubes}) is the square of the right side<br /> of~(\ref{linear}).<br /> &lt;/pre&gt;<br /> Remember, you have to compile the previous example '''twice''' in order the see the references. After the first compilation, LaTeX will warn you about &quot;undefined references&quot;, and you'll see &quot;??&quot; in your document where the references should be. Compile it a second time, and the references will work.<br /> <br /> The \label commands in the \begin{eqnarray} allow us to use \ref later to refer back to the equations. The \label{sec:formulas} allows us to use \ref to refer to the section. Note that we can use the same label for the \pageref as we use for the \ref. Also, note the ~ before each of the references; these aren't required, but are used for style concerns. They prevent a line break right before the reference. Generally, it's poor style to start off lines with numbers or references; these ~ symbols prevent that if possible.<br /> <br /> ==See Also==<br /> *[[LaTeX:Symbols | Next: Symbols]]<br /> *[[LaTeX:Pictures | Previous: Pictures]]</div> Scrabbler94 https://artofproblemsolving.com/wiki/index.php?title=LaTeX:Math&diff=121172 LaTeX:Math 2020-04-20T05:13:54Z <p>Scrabbler94: /* Display Math */ expand and improve section on equations, labeling</p> <hr /> <div>{{Latex}}<br /> <br /> This article will detail how to work with math mode in LaTeX and how to display equations, formulas, and mathematical expressions in general.<br /> <br /> ==Math Mode==<br /> LaTeX uses a special math mode to display mathematics. To place something written in TeX in '''math mode''', use &lt;nowiki&gt;$&lt;/nowiki&gt; signs to enclose the math you want to display. For example, open a new source file in TeXnicCenter and type or copy/paste the following: <br /> <br /> <br /> &lt;pre&gt;&lt;nowiki&gt;<br /> \documentclass{article}<br /> \begin{document}<br /> The solution to $\sqrt{x}=5$ is $x=25$.<br /> \end{document} <br /> &lt;/nowiki&gt;&lt;/pre&gt;<br /> Save the document (press Ctrl-S or click File, then Save) as 'mymath' (don't include the quote marks in the name) in a folder of your choice. The file will appear in your folder as 'mymath.tex.'<br /> <br /> Compile the document just as you compiled your [[LaTeX:Basics|first document]]. When you view the output file, you should see<br /> <br /> [[Image:Mathsamp1.gif]]<br /> <br /> If you remove the &lt;nowiki&gt;$&lt;/nowiki&gt; symbols from your source file then try to compile, you should get 'Missing &lt;nowiki&gt;$&lt;/nowiki&gt; inserted' error messages in the Output window of TeXnicCenter (try it and see - you may have to scroll up in the Output window to see the errors).<br /> <br /> Nearly all mathematical items, such as variables, expressions, equations, etc., should be written in math mode. In fact, most math will generate errors if you don't remember to put it in math mode.<br /> <br /> == Display Math ==<br /> As we saw above, when using &lt;nowiki&gt;$&lt;/nowiki&gt;...&lt;nowiki&gt;$&lt;/nowiki&gt; to typeset math, the resulting math expression appears in-line. Sometimes, we may wish to display a mathematical expression on its own line. To do so, we use &lt;nowiki&gt;$&lt;/nowiki&gt;math stuff here&lt;nowiki&gt;$&lt;/nowiki&gt; or &lt;nowiki&gt;$$math stuff here$$&lt;/nowiki&gt; (the former is usually preferred now) to put the expression in display math mode:<br /> &lt;pre&gt;&lt;nowiki&gt;<br /> \documentclass{article} <br /> \begin{document}<br /> The solution to $\sqrt{x} = 5$ is $x=25.$<br /> \end{document} <br /> &lt;/nowiki&gt;&lt;/pre&gt;<br /> After you compile this and view it, you should see:<br /> <br /> [[Image:Mathsamp2.gif]]<br /> <br /> &lt;nowiki&gt;$&lt;/nowiki&gt;Math stuff here&lt;nowiki&gt;$&lt;/nowiki&gt; no longer works on the AoPS forums.<br /> <br /> Notice that the equations are on their own lines and are centered. As a matter of style, usually we put this '''display math''' on their own lines in the source file, like this:<br /> &lt;pre&gt;&lt;nowiki&gt;<br /> \documentclass{article}<br /> \begin{document}<br /> The solution to<br /> $<br /> \sqrt{x} = 5<br />$<br /> is<br /> $<br /> x=25.<br />$<br /> \end{document} <br /> &lt;/nowiki&gt;&lt;/pre&gt;<br /> <br /> We can also use<br /> <br /> \begin{equation} ... \end{equation}<br /> <br /> to display mathematics. This also creates a label, which we can refer to throughout the document using \label and \ref (or \eqref, using the amsmath package). See the following example:<br /> &lt;pre&gt;&lt;nowiki&gt;<br /> \documentclass{article}<br /> \usepackage{amsmath}<br /> <br /> \begin{document}<br /> The quadratic formula is shown in Equation~\eqref{eqn:quadratic}.<br /> <br /> \begin{equation} \label{eqn:quadratic}<br /> x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}<br /> \end{equation}<br /> <br /> Using Equation~\eqref{eqn:quadratic}, we obtain...<br /> \end{document} <br /> &lt;/nowiki&gt;&lt;/pre&gt;<br /> <br /> Notice the (1) to the right of the equation. Once again, rather than typing (1) in your source file to refer to this equation, use LaTeX [[LaTeX:Layout|referencing]] commands. It is also considered good style to add ~ before commands such as \ref, \eqref, \cite; the ~ symbol represents an unbreakable space (so that you do not have a reference on the following line).<br /> <br /> == Display Style (\&lt;!-- --&gt;displaystyle) ==<br /> Sometimes we have complicated expressions that we don't want to put on their own lines, but that doesn't render well with &lt;nowiki&gt;$&lt;/nowiki&gt;...&lt;nowiki&gt;$&lt;/nowiki&gt; mode. For example:<br /> &lt;pre&gt;&lt;nowiki&gt;<br /> \documentclass{article}<br /> \begin{document}<br /> Evaluate the sum $\sum_{i=0}^n i^3$.<br /> \end{document} <br /> &lt;/nowiki&gt;&lt;/pre&gt;<br /> gives us<br /> <br /> [[Image:Mathsamp3.gif]]<br /> <br /> That summation symbol is a little ugly. We can make it prettier by using \&lt;!-- --&gt;displaystyle:<br /> &lt;pre&gt;&lt;nowiki&gt;<br /> \documentclass{article}<br /> \begin{document}<br /> Evaluate the sum $\&lt;/nowiki&gt;&lt;nowiki&gt;displaystyle\sum\limits_{i=0}^n i^3$.<br /> \end{document} <br /> &lt;/nowiki&gt;&lt;/pre&gt;<br /> This gives us:<br /> <br /> [[Image:Mathsamp4.gif]]<br /> <br /> Notice that the summation symbol looks much nicer now - adding the \&lt;!-- --&gt;displaystyle at the beginning of your math (inside the &lt;nowiki&gt;$&lt;/nowiki&gt;...&lt;nowiki&gt;$&lt;/nowiki&gt;) will often make complicated math render more nicely. Note that it is not necessary to use \&lt;!-- --&gt;displaystyle when using display mode (&lt;nowiki&gt;$and$&lt;/nowiki&gt; or \begin{equation} and \end{equation}).<br /> <br /> == Aligning Equations (align) ==<br /> A pair of very useful tools for displaying equations well are the &quot;align&quot; and &quot;align*&quot; environments. They allow you to neatly align a string of equations:<br /> &lt;pre&gt;<br /> \documentclass{article}<br /> \usepackage{amsmath}<br /> \begin{document}<br /> \begin{align*}<br /> 2x^2 + 3(x-1)(x-2) &amp; = 2x^2 + 3(x^2-3x+2)\\&amp;= 2x^2 + 3x^2 - 9x + 6\\&amp;= 5x^2 - 9x + 6<br /> \end{align*}<br /> \end{document} <br /> &lt;/pre&gt;<br /> Compiling this should give:<br /> &lt;cmath&gt;<br /> \begin{align*}<br /> 2x^2 + 3(x-1)(x-2) &amp; = 2x^2 + 3(x^2-3x+2)\\<br /> &amp;= 2x^2 + 3x^2 - 9x + 6\\<br /> &amp;= 5x^2 - 9x + 6<br /> \end{align*} &lt;/cmath&gt;<br /> <br /> There are a few things to notice here. First, the align command requires that you use the package amsmath (and there's no reason to ''not'' use this package). Second, the * after align prevents line numbers from popping up after each line - try removing both of the *s from the source file and compile to see equation numbers. Next, notice that each line is of the form<br /> &lt;pre&gt;Math stuff &amp; more math stuff \\&lt;/pre&gt;<br /> The &amp; symbols separate the columns. There must be two columns (i.e. one &amp; symbol). The \\ tells LaTeX that you are finished with this line and are on to the next. Notice that there's no \\ on the last line; the \end{align*} tells LaTeX that you're finished. As you see above, you can leave some columns blank. As a style issue, notice that we start a new line in our source file after each \\. We could run all the lines together, but that makes editing very difficult.<br /> <br /> Typically, we use relational symbols like =, &gt;, or &lt; immediately following the &amp;; align ensures that these symbols are arranged into a vertical column as you see above. That's why we like align.<br /> <br /> Finally, notice that there are no &lt;dollar/&gt; symbols, &lt;nowiki&gt;$$...$$&lt;/nowiki&gt;, or &lt;nowiki&gt;$...$&lt;/nowiki&gt;, yet everything is rendered in math mode. This happens because align automatically puts everything in math mode - you don't need &lt;dollar/&gt;s or &lt;nowiki&gt;$...$&lt;/nowiki&gt; tags.<br /> <br /> Also note that in an align environment, you can use the \nonumber command if you want only some lines to be numbered. For example,<br /> &lt;pre&gt;<br /> \documentclass{article}<br /> \usepackage{amsmath}<br /> \begin{document}<br /> \begin{align}<br /> 2x^2 + 3(x-1)(x-2) &amp; = 2x^2 + 3(x^2-3x+2)\\ \nonumber &amp;= 2x^2 + 3x^2 - 9x + 6\\ &amp;= 5x^2 - 9x + 6<br /> \end{align}<br /> \end{document} <br /> &lt;/pre&gt;<br /> compiles to this:<br /> &lt;cmath&gt;\begin{align}<br /> 2x^2 + 3(x-1)(x-2) &amp; = 2x^2 + 3(x^2-3x+2)\\<br /> \nonumber &amp;= 2x^2 + 3x^2 - 9x + 6\\<br /> &amp;= 5x^2 - 9x + 6<br /> \end{align}&lt;/cmath&gt;<br /> <br /> ==Additional Packages==<br /> The basic LaTeX program does not include all the math you'll want to use. In order to access all the math functions and symbols we will introduce in the guide pages, you'll have to include a number of packages. We include these packages by using the \usepackage command between the \documentclass line and the \begin{document} line, such as:<br /> &lt;pre&gt;&lt;nowiki&gt;<br /> \documentclass{article}<br /> \usepackage{amsmath}<br /> \begin{document}<br /> We can write more than just $x$ now.<br /> Now we can write things like $\binom{5}{3}$.<br /> \end{document}<br /> &lt;/nowiki&gt;&lt;/pre&gt;<br /> <br /> The package used above is part of the basic MiKTeX installation, so you don't have to download anything new to include them. Later, you may want to read more about [[LaTeX:Layout|how to include more packages]] and [[LaTeX:Packages|how you can create packages of your own]].<br /> <br /> Finally, one last point of style - notice in that last example that we put the x in math mode by writing &lt;nowiki&gt;$x$&lt;/nowiki&gt; instead of just x. Try compiling with and without the x in math mode and you'll see why. Always put your math in math mode!<br /> <br /> If you find you want to do some math typesetting that you can't find on this page, or among our discussions of [[LaTeX:Symbols|symbols]] or [[LaTeX:Commands|commands]], try reading the [http://texdoc.net/texmf-dist/doc/latex/amsmath/amsldoc.pdf user's guide for the amsmath package], which contains some of the really fancy applications of the ams packages.<br /> <br /> ==See Also==<br /> *[[LaTeX:Examples | Next: Examples]]<br /> *[[LaTeX:Basics | Previous: Basics]]</div> Scrabbler94 https://artofproblemsolving.com/wiki/index.php?title=LaTeX:Symbols&diff=121171 LaTeX:Symbols 2020-04-20T05:01:44Z <p>Scrabbler94: /* Bracketing Symbols */ cases environment is more preferable than using \left\{ ....\right. esp. for piecewise functions</p> <hr /> <div>{{Latex}}<br /> <br /> This article will provide a short list of commonly used LaTeX symbols. <br /> <br /> == Finding Other Symbols ==<br /> <br /> Here are some external resources for finding less commonly used symbols:<br /> &lt;ul&gt;<br /> &lt;li&gt;<br /> [http://detexify.kirelabs.org/classify.html Detexify] is an app which allows you to draw the symbol you'd like and shows you the &lt;math&gt;\text{\LaTeX}&lt;/math&gt; code for it!<br /> &lt;br/&gt;&lt;br/&gt;&lt;/li&gt;<br /> <br /> &lt;li&gt;<br /> MathJax (what allows us to use &lt;math&gt;\text{\LaTeX}&lt;/math&gt; on the web, (technically an AJAX library simulating it.)) maintains a [http://docs.mathjax.org/en/latest/tex.html#supported-latex-commands list of supported commands].<br /> &lt;br/&gt;&lt;br/&gt;&lt;/li&gt;<br /> <br /> &lt;li&gt;<br /> [http://mirrors.ctan.org/info/symbols/comprehensive/symbols-a4.pdf The Comprehensive LaTeX Symbol List].<br /> &lt;br/&gt;&lt;br/&gt;&lt;/li&gt;<br /> &lt;/ul&gt;<br /> <br /> -------------------------------------------------------------------------------------------------------------<br /> <br /> ==Operators==<br /> {| class=&quot;latextable&quot;<br /> !Symbol!!Command!!Symbol!!Command!!Symbol!!Command<br /> |-<br /> |&lt;math&gt;\pm&lt;/math&gt;||\pm||&lt;math&gt;\mp&lt;/math&gt;||\mp||&lt;math&gt;\times&lt;/math&gt;||\times<br /> |-<br /> |&lt;math&gt;\div&lt;/math&gt;||\div||&lt;math&gt;\cdot&lt;/math&gt;||\cdot||&lt;math&gt;\ast&lt;/math&gt;||\ast<br /> |-<br /> |&lt;math&gt;\star&lt;/math&gt;||\star||&lt;math&gt;\dagger&lt;/math&gt;||\dagger||&lt;math&gt;\ddagger&lt;/math&gt;||\ddagger<br /> |-<br /> |&lt;math&gt;\amalg&lt;/math&gt;||\amalg||&lt;math&gt;\cap&lt;/math&gt;||\cap||&lt;math&gt;\cup&lt;/math&gt;||\cup<br /> |-<br /> |&lt;math&gt;\uplus&lt;/math&gt;||\uplus||&lt;math&gt;\sqcap&lt;/math&gt;||\sqcap||&lt;math&gt;\sqcup&lt;/math&gt;||\sqcup<br /> |-<br /> |&lt;math&gt;\vee&lt;/math&gt;||\vee||&lt;math&gt;\wedge&lt;/math&gt;||\wedge||&lt;math&gt;\oplus&lt;/math&gt;||\oplus<br /> |-<br /> |&lt;math&gt;\ominus&lt;/math&gt;||\ominus||&lt;math&gt;\otimes&lt;/math&gt;||\otimes||&lt;math&gt;\circ&lt;/math&gt;||\circ<br /> |-<br /> |&lt;math&gt;\bullet&lt;/math&gt;||\bullet||&lt;math&gt;\diamond&lt;/math&gt;||\diamond||&lt;math&gt;\lhd&lt;/math&gt;||\lhd<br /> |-<br /> |&lt;math&gt;\rhd&lt;/math&gt;||\rhd||&lt;math&gt;\unlhd&lt;/math&gt;||\unlhd||&lt;math&gt;\unrhd&lt;/math&gt;||\unrhd<br /> |-<br /> |&lt;math&gt;\oslash&lt;/math&gt;||\oslash||&lt;math&gt;\odot&lt;/math&gt;||\odot||&lt;math&gt;\bigcirc&lt;/math&gt;||\bigcirc<br /> |-<br /> |&lt;math&gt;\triangleleft&lt;/math&gt;||\triangleleft||&lt;math&gt;\Diamond&lt;/math&gt;||\Diamond||&lt;math&gt;\bigtriangleup&lt;/math&gt;||\bigtriangleup<br /> |-<br /> |&lt;math&gt;\bigtriangledown&lt;/math&gt;||\bigtriangledown||&lt;math&gt;\Box&lt;/math&gt;||\Box||&lt;math&gt;\triangleright&lt;/math&gt;||\triangleright<br /> |-<br /> |&lt;math&gt;\setminus&lt;/math&gt;||\setminus||&lt;math&gt;\wr&lt;/math&gt;||\wr||&lt;math&gt;\sqrt{x}&lt;/math&gt;||\sqrt{x}<br /> |-<br /> |&lt;math&gt;x^{\circ}&lt;/math&gt;||x^{\circ}||&lt;math&gt;\triangledown&lt;/math&gt;||\triangledown||&lt;math&gt;\sqrt[n]{x}&lt;/math&gt;||\sqrt[n]{x}<br /> |-<br /> |&lt;math&gt;a^x&lt;/math&gt;||a^x||&lt;math&gt;a^{xyz}&lt;/math&gt;||a^{xyz}||&lt;math&gt;a_x&lt;/math&gt;||a_x<br /> |}<br /> <br /> ==Relations==<br /> {| class=&quot;latextable&quot;<br /> !Symbol !! Command !!Symbol !! Command!!Symbol !! Command<br /> |-<br /> | &lt;math&gt;\le&lt;/math&gt;||\le||&lt;math&gt;\ge&lt;/math&gt;||\ge||&lt;math&gt;\neq&lt;/math&gt;||\neq<br /> |-<br /> | &lt;math&gt;\sim&lt;/math&gt;||\sim||&lt;math&gt;\ll&lt;/math&gt;||\ll||&lt;math&gt;\gg&lt;/math&gt;||\gg<br /> |-<br /> | &lt;math&gt;\doteq&lt;/math&gt;||\doteq||&lt;math&gt;\simeq&lt;/math&gt;||\simeq||&lt;math&gt;\subset&lt;/math&gt;||\subset<br /> |-<br /> | &lt;math&gt;\supset&lt;/math&gt;||\supset||&lt;math&gt;\approx&lt;/math&gt;||\approx||&lt;math&gt;\asymp&lt;/math&gt;||\asymp<br /> |-<br /> | &lt;math&gt;\subseteq&lt;/math&gt;||\subseteq||&lt;math&gt;\supseteq&lt;/math&gt;||\supseteq||&lt;math&gt;\cong&lt;/math&gt;||\cong<br /> |-<br /> | &lt;math&gt;\smile&lt;/math&gt;||\smile||&lt;math&gt;\sqsubset&lt;/math&gt;||\sqsubset||&lt;math&gt;\sqsupset&lt;/math&gt;||\sqsupset<br /> |-<br /> | &lt;math&gt;\equiv&lt;/math&gt;||\equiv||&lt;math&gt;\frown&lt;/math&gt;||\frown||&lt;math&gt;\sqsubseteq&lt;/math&gt;||\sqsubseteq<br /> |-<br /> | &lt;math&gt;\sqsupseteq&lt;/math&gt;||\sqsupseteq||&lt;math&gt;\propto&lt;/math&gt;||\propto||&lt;math&gt;\bowtie&lt;/math&gt;||\bowtie<br /> |-<br /> | &lt;math&gt;\in&lt;/math&gt;||\in||&lt;math&gt;\ni&lt;/math&gt;||\ni||&lt;math&gt;\prec&lt;/math&gt;||\prec<br /> |-<br /> | &lt;math&gt;\succ&lt;/math&gt;||\succ||&lt;math&gt;\vdash&lt;/math&gt;||\vdash||&lt;math&gt;\dashv&lt;/math&gt;||\dashv<br /> |-<br /> | &lt;math&gt;\preceq&lt;/math&gt;||\preceq||&lt;math&gt;\succeq&lt;/math&gt;||\succeq||&lt;math&gt;\models&lt;/math&gt;||\models<br /> |-<br /> | &lt;math&gt;\perp&lt;/math&gt;||\perp||&lt;math&gt;\parallel&lt;/math&gt;||\parallel||<br /> |-<br /> | &lt;math&gt;\mid&lt;/math&gt;||\mid||&lt;math&gt;\bumpeq&lt;/math&gt;||\bumpeq||<br /> |}<br /> Negations of many of these relations can be formed by just putting \not before the symbol, or by slipping an n between the \ and the word. Here are a couple examples, plus many other negations; it works for many of the many others as well.<br /> <br /> {| class=&quot;latextable&quot;}<br /> !Symbol!!Command!!Symbol!!Command!!Symbol!!Command<br /> |-<br /> |&lt;math&gt;\nmid&lt;/math&gt;||\nmid||&lt;math&gt;\nleq&lt;/math&gt;||\nleq||&lt;math&gt;\ngeq&lt;/math&gt;||\ngeq<br /> |-<br /> | &lt;math&gt;\nsim&lt;/math&gt;||\nsim||&lt;math&gt;\ncong&lt;/math&gt;||\ncong||&lt;math&gt;\nparallel&lt;/math&gt;||\nparallel<br /> |-<br /> | &lt;math&gt;\not&lt;&lt;/math&gt;||\not&lt;||&lt;math&gt;\not&gt;&lt;/math&gt;||\not&gt;||&lt;math&gt;\not=&lt;/math&gt;||\not= or \neq<br /> |-<br /> | &lt;math&gt;\not\le&lt;/math&gt;||\not\le||&lt;math&gt;\not\ge&lt;/math&gt;||\not\ge||&lt;math&gt;\not\sim&lt;/math&gt;||\not\sim<br /> |-<br /> |&lt;math&gt;\not \approx&lt;/math&gt;||\not\approx||&lt;math&gt;\not\cong&lt;/math&gt;||\not\cong||&lt;math&gt;\not\equiv&lt;/math&gt;||\not\equiv<br /> |-<br /> | &lt;math&gt;\not\parallel&lt;/math&gt;||\not\parallel||&lt;math&gt;\nless&lt;/math&gt;||\nless||&lt;math&gt;\ngtr&lt;/math&gt;||\ngtr<br /> |-<br /> | &lt;math&gt;\lneq&lt;/math&gt;||\lneq||&lt;math&gt;\gneq&lt;/math&gt;||\gneq||&lt;math&gt;\lnsim&lt;/math&gt;||\lnsim<br /> |-<br /> | &lt;math&gt;\lneqq&lt;/math&gt;||\lneqq||&lt;math&gt;\gneqq&lt;/math&gt;||\gneqq<br /> |}<br /> <br /> To use other relations not listed here, such as =, &gt;, and &lt;, in LaTeX, you must use the symbols on your keyboard, they are not available in &lt;math&gt;\LaTeX&lt;/math&gt;.<br /> <br /> ==Greek Letters==<br /> {| class=&quot;latextable&quot;<br /> |+ Lowercase Letters<br /> !Symbol!!Command!!Symbol!!Command!!Symbol!!Command!!Symbol!!Command<br /> |-<br /> |&lt;math&gt;\alpha&lt;/math&gt;||\alpha||&lt;math&gt;\beta&lt;/math&gt;||\beta||&lt;math&gt;\gamma&lt;/math&gt;||\gamma||&lt;math&gt;\delta&lt;/math&gt;||\delta<br /> |-<br /> | &lt;math&gt;\epsilon&lt;/math&gt;||\epsilon||&lt;math&gt;\varepsilon&lt;/math&gt;||\varepsilon||&lt;math&gt;\zeta&lt;/math&gt;||\zeta||&lt;math&gt;\eta&lt;/math&gt;||\eta<br /> |-<br /> | &lt;math&gt;\theta&lt;/math&gt;||\theta||&lt;math&gt;\vartheta&lt;/math&gt;||\vartheta||&lt;math&gt;\iota&lt;/math&gt;||\iota||&lt;math&gt;\kappa&lt;/math&gt;||\kappa<br /> |-<br /> | &lt;math&gt;\lambda&lt;/math&gt;||\lambda||&lt;math&gt;\mu&lt;/math&gt;||\mu||&lt;math&gt;\nu&lt;/math&gt;||\nu||&lt;math&gt;\xi&lt;/math&gt;||\xi<br /> |-<br /> |&lt;math&gt;\pi&lt;/math&gt;||\pi||&lt;math&gt;\varpi&lt;/math&gt;||\varpi||&lt;math&gt;\rho&lt;/math&gt;||\rho||&lt;math&gt;\varrho&lt;/math&gt;||\varrho<br /> |-<br /> | &lt;math&gt;\sigma&lt;/math&gt;||\sigma||&lt;math&gt;\varsigma&lt;/math&gt;||\varsigma||&lt;math&gt;\tau&lt;/math&gt;||\tau||&lt;math&gt;\upsilon&lt;/math&gt;||\upsilon<br /> |-<br /> | &lt;math&gt;\phi&lt;/math&gt;||\phi||&lt;math&gt;\varphi&lt;/math&gt;||\varphi||&lt;math&gt;\chi&lt;/math&gt;||\chi||&lt;math&gt;\psi&lt;/math&gt;||\psi<br /> |-<br /> | &lt;math&gt;\omega&lt;/math&gt;||\omega<br /> |}<br /> <br /> <br /> {| class=&quot;latextable&quot;<br /> |+ Capital Letters<br /> !Symbol!!Command!!Symbol!!Command!!Symbol!!Command!!Symbol!!Command<br /> |-<br /> |&lt;math&gt;\Gamma&lt;/math&gt;||\Gamma||&lt;math&gt;\Delta&lt;/math&gt;||\Delta||&lt;math&gt;\Theta&lt;/math&gt;||\Theta||&lt;math&gt;\Lambda&lt;/math&gt;||\Lambda<br /> |-<br /> | &lt;math&gt;\Xi&lt;/math&gt;||\Xi||&lt;math&gt;\Pi&lt;/math&gt;||\Pi||&lt;math&gt;\Sigma&lt;/math&gt;||\Sigma||&lt;math&gt;\Upsilon&lt;/math&gt;||\Upsilon<br /> |-<br /> | &lt;math&gt;\Phi&lt;/math&gt;||\Phi||&lt;math&gt;\Psi&lt;/math&gt;||\Psi||&lt;math&gt;\Omega&lt;/math&gt;||\Omega<br /> |}<br /> <br /> <br /> == Arrows ==<br /> <br /> {| class=&quot;latextable&quot;<br /> !Symbol !! Command!!Symbol !! Command<br /> |-<br /> |&lt;math&gt;\gets&lt;/math&gt;||\gets||&lt;math&gt;\to&lt;/math&gt;||\to<br /> |-<br /> |&lt;math&gt;\leftarrow&lt;/math&gt;||\leftarrow||&lt;math&gt;\Leftarrow&lt;/math&gt;||\Leftarrow<br /> |-<br /> |&lt;math&gt;\rightarrow&lt;/math&gt;||\rightarrow||&lt;math&gt;\Rightarrow&lt;/math&gt;||\Rightarrow<br /> |-<br /> |&lt;math&gt;\leftrightarrow&lt;/math&gt;||\leftrightarrow||&lt;math&gt;\Leftrightarrow&lt;/math&gt;||\Leftrightarrow<br /> |-<br /> |&lt;math&gt;\mapsto&lt;/math&gt;||\mapsto||&lt;math&gt;\hookleftarrow&lt;/math&gt;||\hookleftarrow<br /> |-<br /> |&lt;math&gt;\leftharpoonup&lt;/math&gt;||\leftharpoonup||&lt;math&gt;\leftharpoondown&lt;/math&gt;||\leftharpoondown<br /> |-<br /> |&lt;math&gt;\rightleftharpoons&lt;/math&gt;||\rightleftharpoons||&lt;math&gt;\longleftarrow&lt;/math&gt;||\longleftarrow<br /> |-<br /> |&lt;math&gt;\Longleftarrow&lt;/math&gt;||\Longleftarrow||&lt;math&gt;\longrightarrow&lt;/math&gt;||\longrightarrow<br /> |-<br /> |&lt;math&gt;\Longrightarrow&lt;/math&gt;||\Longrightarrow||&lt;math&gt;\longleftrightarrow&lt;/math&gt;||\longleftrightarrow<br /> |-<br /> |&lt;math&gt;\Longleftrightarrow&lt;/math&gt;||\Longleftrightarrow||&lt;math&gt;\longmapsto&lt;/math&gt;||\longmapsto<br /> |-<br /> |&lt;math&gt;\hookrightarrow&lt;/math&gt;||\hookrightarrow||&lt;math&gt;\rightharpoonup&lt;/math&gt;||\rightharpoonup<br /> |-<br /> |&lt;math&gt;\rightharpoondown&lt;/math&gt;||\rightharpoondown||&lt;math&gt;\leadsto&lt;/math&gt;||\leadsto<br /> |-<br /> |&lt;math&gt;\uparrow&lt;/math&gt;||\uparrow||&lt;math&gt;\Uparrow&lt;/math&gt;||\Uparrow<br /> |-<br /> |&lt;math&gt;\downarrow&lt;/math&gt;||\downarrow||&lt;math&gt;\Downarrow&lt;/math&gt;||\Downarrow<br /> |-<br /> |&lt;math&gt;\updownarrow&lt;/math&gt;||\updownarrow||&lt;math&gt;\Updownarrow&lt;/math&gt;||\Updownarrow<br /> |-<br /> |&lt;math&gt;\nearrow&lt;/math&gt;||\nearrow||&lt;math&gt;\searrow&lt;/math&gt;||\searrow<br /> |-<br /> |&lt;math&gt;\swarrow&lt;/math&gt;||\swarrow||&lt;math&gt;\nwarrow&lt;/math&gt;||\nwarrow<br /> |}<br /> (For those of you who hate typing long strings of letters, \iff and \implies can be used in place of \Longleftrightarrow and \Longrightarrow respectively.)<br /> <br /> ==Dots==<br /> {| class=&quot;latextable&quot;<br /> !Symbol!!Command!!Symbol!!Command!!<br /> |- <br /> |&lt;math&gt;\cdot&lt;/math&gt;||\cdot|| |&lt;math&gt;\vdots&lt;/math&gt;||\vdots|| <br /> |- <br /> |&lt;math&gt;\dots&lt;/math&gt;||\dots|| |&lt;math&gt;\ddots&lt;/math&gt;||\ddots||<br /> |-<br /> |&lt;math&gt;\cdots&lt;/math&gt;||\cdots|| |&lt;math&gt;\iddots&lt;/math&gt;||\iddots||<br /> <br /> |}<br /> <br /> ==Accents==<br /> {| class=&quot;latextable&quot;<br /> !Symbol !! Command!!Symbol !! Command!!Symbol !! Command<br /> |-<br /> |&lt;math&gt;\hat{x}&lt;/math&gt;||\hat{x}||&lt;math&gt;\check{x}&lt;/math&gt;||\check{x}||&lt;math&gt;\dot{x}&lt;/math&gt;||\dot{x}<br /> |-<br /> |&lt;math&gt;\breve{x}&lt;/math&gt;||\breve{x}||&lt;math&gt;\acute{x}&lt;/math&gt;||\acute{x}||&lt;math&gt;\ddot{x}&lt;/math&gt;||\ddot{x}<br /> |-<br /> |&lt;math&gt;\grave{x}&lt;/math&gt;||\grave{x}||&lt;math&gt;\tilde{x}&lt;/math&gt;||\tilde{x}||&lt;math&gt;\mathring{x}&lt;/math&gt;||\mathring{x}<br /> |-<br /> |&lt;math&gt;\bar{x}&lt;/math&gt;||\bar{x}||&lt;math&gt;\vec{x}&lt;/math&gt;||\vec{x}<br /> |}<br /> When applying accents to i and j, you can use \imath and \jmath to keep the dots from interfering with the accents:<br /> {| class=&quot;latextable&quot;<br /> !Symbol !! Command!!Symbol !! Command<br /> |-<br /> |&lt;math&gt;\vec{\jmath}&lt;/math&gt;||\vec{\jmath}||&lt;math&gt;\tilde{\imath}&lt;/math&gt;||\tilde{\imath}<br /> |}<br /> \tilde and \hat have wide versions that allow you to accent an expression:<br /> {| class=&quot;latextable&quot;<br /> !Symbol !! Command!!Symbol !! Command<br /> |-<br /> |&lt;math&gt;\widehat{7+x}&lt;/math&gt;||\widehat{7+x}||&lt;math&gt;\widetilde{abc}&lt;/math&gt;||\widetilde{abc}<br /> |}<br /> <br /> ==Others==<br /> {| class=&quot;latextable&quot;<br /> !Symbol!!Command!!Symbol!!Command!!Symbol!!Command <br /> |-<br /> |&lt;math&gt;\infty&lt;/math&gt;||\infty||&lt;math&gt;\triangle&lt;/math&gt;||\triangle||&lt;math&gt;\angle&lt;/math&gt;||\angle<br /> |-<br /> |&lt;math&gt;\aleph&lt;/math&gt;||\aleph||&lt;math&gt;\hbar&lt;/math&gt;||\hbar||&lt;math&gt;\imath&lt;/math&gt;||\imath<br /> |-<br /> |&lt;math&gt;\jmath&lt;/math&gt;||\jmath||&lt;math&gt;\ell&lt;/math&gt;||\ell||&lt;math&gt;\wp&lt;/math&gt;||\wp<br /> |-<br /> |&lt;math&gt;\Re&lt;/math&gt;||\Re||&lt;math&gt;\Im&lt;/math&gt;||\Im||&lt;math&gt;\mho&lt;/math&gt;||\mho<br /> |-<br /> |&lt;math&gt;\prime&lt;/math&gt;||\prime||&lt;math&gt;\emptyset&lt;/math&gt;||\emptyset||&lt;math&gt;\nabla&lt;/math&gt;||\nabla<br /> |-<br /> |&lt;math&gt;\surd&lt;/math&gt;||\surd||&lt;math&gt;\partial&lt;/math&gt;||\partial||&lt;math&gt;\top&lt;/math&gt;||\top<br /> |-<br /> |&lt;math&gt;\bot&lt;/math&gt;||\bot||&lt;math&gt;\vdash&lt;/math&gt;||\vdash||&lt;math&gt;\dashv&lt;/math&gt;||\dashv<br /> |-<br /> |&lt;math&gt;\forall&lt;/math&gt;||\forall||&lt;math&gt;\exists&lt;/math&gt;||\exists||&lt;math&gt;\neg&lt;/math&gt;||\neg<br /> |-<br /> |&lt;math&gt;\flat&lt;/math&gt;||\flat||&lt;math&gt;\natural&lt;/math&gt;||\natural||&lt;math&gt;\sharp&lt;/math&gt;||\sharp<br /> |-<br /> |&lt;math&gt;\backslash&lt;/math&gt;||\backslash||&lt;math&gt;\Box&lt;/math&gt;||\Box||&lt;math&gt;\Diamond&lt;/math&gt;||\Diamond<br /> |-<br /> |&lt;math&gt;\clubsuit&lt;/math&gt;||\clubsuit||&lt;math&gt;\diamondsuit&lt;/math&gt;||\diamondsuit||&lt;math&gt;\heartsuit&lt;/math&gt;||\heartsuit<br /> |-<br /> |[[Image:Spadesuit.gif]]||\spadesuit||&lt;math&gt;\Join&lt;/math&gt;||\Join||&lt;math&gt;\blacksquare&lt;/math&gt;||\blacksquare<br /> |-<br /> |&lt;math&gt;\S&lt;/math&gt;||\S||&lt;math&gt;\P&lt;/math&gt;||\P||&lt;math&gt;\copyright&lt;/math&gt;||\copyright<br /> |-<br /> |&lt;math&gt;\pounds&lt;/math&gt;||\pounds||&lt;math&gt;\overarc{ABC}&lt;/math&gt;||\overarc{ABC}||&lt;math&gt;\underarc{XYZ}&lt;/math&gt;||\underarc{XYZ}<br /> |-<br /> |&lt;math&gt;\bigstar&lt;/math&gt;||\bigstar||&lt;math&gt;\in&lt;/math&gt;||\in||&lt;math&gt;\cup&lt;/math&gt;||\cup<br /> |-<br /> |&lt;math&gt;\square&lt;/math&gt;||\square||<br /> |-<br /> |&lt;math&gt;\smiley&lt;/math&gt;||\smiley||<br /> |-<br /> |&lt;math&gt;\mathbb{R}&lt;/math&gt;||\mathbb{R} (represents all real numbers)||<br /> |-<br /> |&lt;math&gt;\checkmark&lt;/math&gt;||\checkmark||<br /> |-<br /> |&lt;math&gt;\cancer&lt;/math&gt;||\cancer||<br /> |}<br /> <br /> ==Command Symbols==<br /> Some symbols are used in commands so they need to be treated in a special way.<br /> {| class=&quot;latextable&quot;<br /> !Symbol!!Command!!Symbol!!Command!!Symbol!!Command!!Symbol!!Command<br /> |-<br /> |&lt;math&gt;\textdollar&lt;/math&gt;||\textdollar or \\$||&lt;math&gt;\&amp;&lt;/math&gt;||\&amp;||&lt;math&gt;\%&lt;/math&gt;||\%||&lt;math&gt;\#&lt;/math&gt;||\#<br /> |-<br /> |&lt;math&gt;\_&lt;/math&gt;||\_||&lt;math&gt;\{&lt;/math&gt;||\{||&lt;math&gt;\}&lt;/math&gt;||\}||&lt;math&gt;\backslash&lt;/math&gt;||\backslash<br /> |}<br /> <br /> (Warning: Using &lt;nowiki&gt;$&lt;/nowiki&gt; for &lt;math&gt;\textdollar&lt;/math&gt; will result in &lt;math&gt;\&lt;/math&gt;. This is a bug as far as we know. Depending on the version of &lt;math&gt;\LaTeX&lt;/math&gt; this is not always a problem.)<br /> <br /> ==European Language Symbols==<br /> {| class=&quot;latextable&quot;<br /> !Symbol!!Command!!Symbol!!Command!!Symbol!!Command!!Symbol!!Command<br /> |-<br /> |&lt;math&gt;{\oe}&lt;/math&gt;||{\oe}||&lt;math&gt;{\ae}&lt;/math&gt;||{\ae}||&lt;math&gt;{\o}&lt;/math&gt;||{\o}<br /> |-<br /> |&lt;math&gt;{\OE}&lt;/math&gt;||{\OE}||&lt;math&gt;{\AE}&lt;/math&gt;||{\AE}||&lt;math&gt;{\AA}&lt;/math&gt;||{\AA}||&lt;math&gt;{\O}&lt;/math&gt;||{\O}<br /> |-<br /> |&lt;math&gt;{\l}&lt;/math&gt;||{\l}||&lt;math&gt;{\ss}&lt;/math&gt;||{\ss}||&lt;math&gt;\text{!}&lt;/math&gt;||!`<br /> |-<br /> |&lt;math&gt;{\L}&lt;/math&gt;||{\L}||&lt;math&gt;{\SS}&lt;/math&gt;||{\SS}||<br /> |}<br /> <br /> ==Bracketing Symbols==<br /> In mathematics, sometimes we need to enclose expressions in brackets or braces or parentheses. Some of these work just as you'd imagine in LaTeX; type ( and ) for parentheses, [ and ] for brackets, and | and | for absolute value. However, other symbols have special commands:<br /> {| class=&quot;latextable&quot;<br /> !Symbol !! Command!!Symbol !! Command!!Symbol !! Command<br /> |-<br /> |&lt;math&gt;\{&lt;/math&gt;||\{||&lt;math&gt;\}&lt;/math&gt;||\}||&lt;math&gt;\|&lt;/math&gt;||&lt;nowiki&gt;\|&lt;/nowiki&gt;<br /> |-<br /> | &lt;math&gt;\backslash&lt;/math&gt;||\backslash||&lt;math&gt;\lfloor&lt;/math&gt;||\lfloor||&lt;math&gt;\rfloor&lt;/math&gt;||\rfloor<br /> |-<br /> | &lt;math&gt;\lceil&lt;/math&gt;||\lceil||&lt;math&gt;\rceil&lt;/math&gt;||\rceil||&lt;math&gt;\langle&lt;/math&gt;||\langle<br /> |-<br /> | &lt;math&gt;\rangle&lt;/math&gt;||\rangle<br /> |}<br /> You might notice that if you use any of these to typeset an expression that is vertically large, like<br /> <br /> :&lt;tt&gt;(\frac{a}{x} )^2&lt;/tt&gt;<br /> <br /> the parentheses don't come out the right size:<br /> <br /> :&lt;math&gt;(\frac{a}{x})^2&lt;/math&gt;<br /> <br /> If we put \left and \right before the relevant parentheses, we get a prettier expression:<br /> <br /> :&lt;tt&gt;\left(\frac{a}{x} \right)^2&lt;/tt&gt;<br /> <br /> gives<br /> <br /> :&lt;math&gt;\left(\frac{a}{x} \right)^2&lt;/math&gt;<br /> <br /> For systems of equations or piecewise functions, use the cases environment:<br /> <br /> &lt;tt&gt;f(x) = \begin{cases} x^2 &amp; x \ge 0 \\ x &amp; x &lt; 0 \end{cases}&lt;/tt&gt;<br /> <br /> which gives<br /> <br /> &lt;math&gt;f(x) = \begin{cases} x^2 &amp; x \ge 0 \\ x &amp; x &lt; 0 \end{cases}&lt;/math&gt;<br /> <br /> In addition to the &lt;tt&gt;\left&lt;/tt&gt; and &lt;tt&gt;\right&lt;/tt&gt; commands, when doing floor or ceiling functions with fractions, using <br /> <br /> &lt;tt&gt;\left\lceil\frac{x}{y}\right\rceil&lt;/tt&gt; <br /> <br /> and &lt;tt&gt;\left\lfloor\frac{x}{y}\right\rfloor &lt;/tt&gt;<br /> <br /> give both &lt;math&gt;\left\lceil\frac{x}{y}\right\rceil\text{ and }\left\lfloor\frac{x}{y}\right\rfloor\text{, respectively.}&lt;/math&gt;<br /> <br /> <br /> And, if you type this<br /> <br /> &lt;tt&gt;\underbrace{a_0+a_1+a_2+\cdots+a_n}_{x}&lt;/tt&gt;<br /> <br /> Gives<br /> <br /> &lt;math&gt;\underbrace{a_0+a_1+a_2+\cdots+a_n}_{x}&lt;/math&gt;<br /> <br /> Or<br /> <br /> &lt;tt&gt;\overbrace{a_0+a_1+a_2+\cdots+a_n}^{x}&lt;/tt&gt;<br /> <br /> Gives<br /> <br /> &lt;math&gt;\overbrace{a_0+a_1+a_2+\cdots+a_n}^{x}&lt;/math&gt;<br /> <br /> <br /> \left and \right can also be used to resize the following symbols:<br /> {| class=&quot;latextable&quot;<br /> !Symbol !! Command!!Symbol !! Command!!Symbol !! Command<br /> |-<br /> |&lt;math&gt;\uparrow&lt;/math&gt;||\uparrow||&lt;math&gt;\downarrow&lt;/math&gt;||\downarrow||&lt;math&gt;\updownarrow&lt;/math&gt;||\updownarrow<br /> |-<br /> | &lt;math&gt;\Uparrow&lt;/math&gt;||\Uparrow||&lt;math&gt;\Downarrow&lt;/math&gt;||\Downarrow||&lt;math&gt;\Updownarrow&lt;/math&gt;||\Updownarrow<br /> |}<br /> <br /> ==Multi-Size Symbols==<br /> Some symbols render differently in inline math mode and in display mode. Display mode occurs when you use &lt;nowiki&gt;$...$&lt;/nowiki&gt; or &lt;nowiki&gt;$$...$$&lt;/nowiki&gt;, or environments like \begin{equation}...\end{equation}, \begin{align}...\end{align}. Read more in the [[LaTeX:Commands|commands]] section of the guide about how symbols which take arguments above and below the symbols, such as a summation symbol, behave in the two modes.<br /> <br /> In each of the following, the two images show the symbol in display mode, then in inline mode.<br /> <br /> {| class=&quot;latextable&quot;<br /> !Symbol !! Command!!Symbol !! Command!!Symbol !! Command<br /> |-<br /> |&lt;math&gt;\sum \textstyle\sum&lt;/math&gt;||\sum||&lt;math&gt;\int \textstyle\int&lt;/math&gt;||\int||&lt;math&gt;\oint \textstyle\oint&lt;/math&gt;||\oint<br /> |-<br /> |&lt;math&gt;\prod \textstyle\prod&lt;/math&gt;||\prod||&lt;math&gt;\coprod \textstyle\coprod&lt;/math&gt;||\coprod||&lt;math&gt;\bigcap \textstyle\bigcap&lt;/math&gt;||\bigcap<br /> |-<br /> |&lt;math&gt;\bigcup \textstyle\bigcup&lt;/math&gt;||\bigcup||&lt;math&gt;\bigsqcup \textstyle\bigsqcup&lt;/math&gt;||\bigsqcup||&lt;math&gt;\bigvee \textstyle\bigvee&lt;/math&gt;||\bigvee<br /> |-<br /> |&lt;math&gt;\bigwedge \textstyle\bigwedge&lt;/math&gt;||\bigwedge||&lt;math&gt;\bigodot \textstyle\bigodot&lt;/math&gt;||\bigodot||&lt;math&gt;\bigotimes \textstyle\bigotimes&lt;/math&gt;||\bigotimes<br /> |-<br /> |&lt;math&gt;\bigoplus \textstyle\bigoplus&lt;/math&gt;||\bigoplus||&lt;math&gt;\biguplus \textstyle\biguplus&lt;/math&gt;||\biguplus<br /> |}<br /> <br /> ==See Also==<br /> *[[LaTeX:Commands | Next: Commands]]</div> Scrabbler94 https://artofproblemsolving.com/wiki/index.php?title=Euclidean_algorithm&diff=121170 Euclidean algorithm 2020-04-20T04:49:24Z <p>Scrabbler94: /* Introductory */</p> <hr /> <div>The '''Euclidean algorithm''' (also known as the '''Euclidean division algorithm''' or '''Euclid's algorithm''') is an algorithm that finds the [[greatest common divisor]] (GCD) of two elements of a [[Euclidean domain]], the most common of which is the [[nonnegative]] [[integer]]s &lt;math&gt;\mathbb{Z}{\geq 0}&lt;/math&gt;, without [[factoring]] them.<br /> <br /> ==Main idea and Informal Description==<br /> The basic idea is to repeatedly use the fact that &lt;math&gt;\gcd({a,b}) \equiv \gcd({b,a - b})&lt;/math&gt;<br /> <br /> If we have two non-negative integers &lt;math&gt;a,b&lt;/math&gt; with &lt;math&gt;a|b&lt;/math&gt; and &lt;math&gt;b\ne0&lt;/math&gt;, then the greatest common divisor is &lt;math&gt;{a}&lt;/math&gt;. If &lt;math&gt;a\ge b&gt;0&lt;/math&gt;, then the set of common divisors of &lt;math&gt;{a}&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; is the same as the set of common divisors of &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;r&lt;/math&gt; where &lt;math&gt;r&lt;/math&gt; is the [[remainder]] of division of &lt;math&gt;{a}&lt;/math&gt; by &lt;math&gt;b&lt;/math&gt;. Indeed, we have &lt;math&gt;a=mb+r&lt;/math&gt; with some integer &lt;math&gt;m&lt;/math&gt;, so, if &lt;math&gt;{d}&lt;/math&gt; divides both &lt;math&gt;{a}&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;, it must divide both &lt;math&gt;{a}&lt;/math&gt; and &lt;math&gt;mb&lt;/math&gt; and, thereby, their difference &lt;math&gt;r&lt;/math&gt;. Similarly, if &lt;math&gt;{d}&lt;/math&gt; divides both &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;r&lt;/math&gt;, it should divide &lt;math&gt;{a}&lt;/math&gt; as well. Thus, the greatest common divisors of &lt;math&gt;{a}&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; and of &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;r&lt;/math&gt; coincide: &lt;math&gt;GCD(a,b)=GCD(b,r)&lt;/math&gt;. But the pair &lt;math&gt;(b,r)&lt;/math&gt; consists of smaller numbers than the pair &lt;math&gt;(a,b)&lt;/math&gt;! So, we reduced our task to a simpler one. And we can do this reduction again and again until the smaller number becomes &lt;math&gt;0&lt;/math&gt;.<br /> <br /> == General Form ==<br /> Start with any two elements &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; of a [[Euclidean Domain]]<br /> <br /> * If &lt;math&gt;b=0&lt;/math&gt;, then &lt;math&gt;\gcd(a,b)=a&lt;/math&gt;.<br /> * Otherwise take the remainder when &lt;math&gt;{a}&lt;/math&gt; is divided by &lt;math&gt;a \pmod{b}&lt;/math&gt;, and find &lt;math&gt;\gcd(a,a \bmod {b})&lt;/math&gt;.<br /> * Repeat this until the remainder is 0.&lt;br&gt;<br /> <br /> &lt;math&gt;a \pmod{b} \equiv r_1&lt;/math&gt;&lt;br&gt;<br /> &lt;math&gt;b \pmod {r_1} \equiv r_2&lt;/math&gt;&lt;br&gt;<br /> &lt;math&gt; \vdots&lt;/math&gt; &lt;br&gt;<br /> &lt;math&gt;r_{n-1} \pmod r_n \equiv 0&lt;/math&gt;&lt;br&gt;<br /> Then &lt;math&gt;\gcd({a,b}) = r_n&lt;/math&gt;&lt;br&gt;<br /> <br /> Usually the Euclidean algorithm is written down just as a chain of divisions with remainder:<br /> <br /> for &lt;math&gt;r_{k+1} &lt; r_k &lt; r_{k-1}&lt;/math&gt;&lt;br&gt;<br /> &lt;math&gt;a = b \cdot q_1+r_1&lt;/math&gt;&lt;br&gt;<br /> &lt;math&gt;b = r_1 \cdot q_2 + r_2&lt;/math&gt;&lt;br&gt;<br /> &lt;math&gt;r_1 = r_2 \cdot q_3 + r_3&lt;/math&gt;&lt;br&gt;<br /> &lt;math&gt;\vdots&lt;/math&gt;&lt;br&gt;<br /> &lt;math&gt;r_{n-1} = r_n \cdot q_{n+1} +0&lt;/math&gt;&lt;br&gt;<br /> and so &lt;math&gt;\gcd({a,b}) = r_n&lt;/math&gt;&lt;br&gt;<br /> <br /> == Example ==<br /> To see how it works, just take an example. Say &lt;math&gt;a = 93, b=42&lt;/math&gt;. &lt;br/&gt;<br /> We have &lt;math&gt;93 \equiv 9 \pmod{42}&lt;/math&gt;, so &lt;math&gt;\gcd(93,42) = \gcd(42,9)&lt;/math&gt;. &lt;br/&gt; <br /> Similarly, &lt;math&gt;42 \equiv 6 \pmod{9}&lt;/math&gt;, so &lt;math&gt;\gcd(42,9) = \gcd(9,6)&lt;/math&gt;. &lt;br/&gt;<br /> Continuing, &lt;math&gt;9 \equiv 3 \pmod{6}&lt;/math&gt;, so &lt;math&gt;\gcd(9,6) = \gcd(6,3)&lt;/math&gt;. &lt;br/&gt;<br /> Then &lt;math&gt;6 \equiv 0 \pmod{3}&lt;/math&gt;, so &lt;math&gt;\gcd(6,3) = \gcd(3,0) = 3&lt;/math&gt;. &lt;br/&gt;<br /> Thus &lt;math&gt;\gcd(93,42) = 3&lt;/math&gt;.<br /> <br /> * &lt;math&gt;93 = 2 \cdot 42 + 9 \qquad (1)&lt;/math&gt;<br /> * &lt;math&gt;42 = 4 \cdot 9 + 6 \qquad (2)&lt;/math&gt;<br /> * &lt;math&gt;9 = 1 \cdot 6 + 3 \qquad (3)&lt;/math&gt;<br /> * &lt;math&gt;6 = 2 \cdot 3 + 0 \qquad (4)&lt;/math&gt;<br /> <br /> == Extended Euclidean Algorithm ==<br /> An added bonus of the Euclidean algorithm is the &quot;linear representation&quot; of the greatest common divisor. This allows us to write &lt;math&gt;\gcd(a,b)=ax+by&lt;/math&gt;, where &lt;math&gt;x,y&lt;/math&gt; are some elements from the same [[Euclidean Domain]] as &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; that can be determined using the algorithm. We can work backwards from whichever step is the most convenient.<br /> <br /> Continuing the previous example, our goal is to find &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; such that &lt;math&gt;93a + 42b = \gcd(93,42) = 3.&lt;/math&gt; We can work backwards from equation &lt;math&gt;(3)&lt;/math&gt; since &lt;math&gt;3&lt;/math&gt; appears there:<br /> <br /> &lt;math&gt;3 = 9 - 1 \cdot 6&lt;/math&gt;<br /> <br /> We currently have &lt;math&gt;3&lt;/math&gt; as a linear combination of &lt;math&gt;6&lt;/math&gt; and &lt;math&gt;9&lt;/math&gt;. Our goal is to replace &lt;math&gt;6&lt;/math&gt; and &lt;math&gt;9&lt;/math&gt; so that we have a linear combination of &lt;math&gt;42&lt;/math&gt; and &lt;math&gt;93&lt;/math&gt; only. We start by rearranging &lt;math&gt;(2)&lt;/math&gt; to &lt;math&gt;6 = 42 - 4 \cdot 9&lt;/math&gt; so we can substitute &lt;math&gt;6&lt;/math&gt; to express &lt;math&gt;3&lt;/math&gt; as a linear combination of &lt;math&gt;9&lt;/math&gt; and &lt;math&gt;42&lt;/math&gt;:<br /> <br /> &lt;math&gt;3 = 9 - 1 \cdot (42 - 4 \cdot 9)&lt;/math&gt; &lt;br&gt;<br /> &lt;math&gt;3 = -1 \cdot 42 + 5 \cdot 9.&lt;/math&gt; &lt;br&gt;<br /> <br /> Continuing, we rearrange &lt;math&gt;(1)&lt;/math&gt; to substitute &lt;math&gt;9 = 93 - 2 \cdot 42&lt;/math&gt;:<br /> <br /> &lt;math&gt;3 = -1 \cdot 42 + 5 \cdot (93 - 2 \cdot 42)&lt;/math&gt; &lt;br&gt;<br /> &lt;math&gt;3 = -11 \cdot 42 + 5 \cdot 93. \qquad (5)&lt;/math&gt; &lt;br&gt;<br /> <br /> We have found one linear combination. To find others, since &lt;math&gt;42 \cdot 93 - 93 \cdot 42 = 0&lt;/math&gt;, dividing both sides by &lt;math&gt;\gcd(93,42) = 3&lt;/math&gt; gives &lt;math&gt;14 \cdot 93 - 31 \cdot 42 = 0&lt;/math&gt;. We can add &lt;math&gt;k&lt;/math&gt; times this equation to &lt;math&gt;(5)&lt;/math&gt;, so we can write &lt;math&gt;3&lt;/math&gt; as a linear combination of &lt;math&gt;93&lt;/math&gt; and &lt;math&gt;42&lt;/math&gt;<br /> <br /> &lt;math&gt;3 = (14k + 5) \cdot 93 + (-31k - 11) \cdot 42&lt;/math&gt;<br /> <br /> for any integer &lt;math&gt;k&lt;/math&gt;.<br /> <br /> == Problems ==<br /> ===Introductory===<br /> <br /> ===Intermediate===<br /> * [[1985 AIME Problems/Problem 13]]<br /> * [[1959 IMO Problems/Problem 1]] (Note: this problem is widely regarded as the easiest problem ever asked in the IMO)<br /> <br /> ===Olympiad===<br /> <br /> ==See Also==<br /> *[[Divisibility]]<br /> <br /> [[Category:Algorithms]]<br /> [[Category:Number theory]]</div> Scrabbler94 https://artofproblemsolving.com/wiki/index.php?title=AMC_historical_results&diff=118695 AMC historical results 2020-03-03T06:19:11Z <p>Scrabbler94: average scores are not yet available</p> <hr /> <div>&lt;!-- Post AMC statistics and lists of high scorers here so that the AMC page doesn't get cluttered. --&gt;<br /> This is the '''AMC historical results''' page. This page should include results for the [[AIME]] as well. For [[USAMO]] results, see [[USAMO historical results]].<br /> ==2020==<br /> ===AMC 10A===<br /> *Average score: N/A<br /> *HR: 94.5<br /> *AIME floor: 103.5<br /> *DHR: 124.5<br /> <br /> ===AMC 10B===<br /> *Average score: N/A<br /> *HR: 99<br /> *AIME floor: 102 <br /> *DHR: 120<br /> <br /> ===AMC 12A===<br /> *Average score: 42.99<br /> *AIME floor: 87<br /> *DHR: 123<br /> <br /> ===AMC 12B===<br /> *Average score: 42.68<br /> *AIME floor: 87<br /> *DHR: 120<br /> <br /> ==2019==<br /> ===AMC 10A===<br /> *Average score: 51.65<br /> *HR: 96<br /> *AIME floor: 103.5<br /> *DHR: 123<br /> <br /> ===AMC 10B===<br /> *Average score: 58.42<br /> *HR: 102<br /> *AIME floor: 108<br /> *DHR: 121.5<br /> <br /> ===AMC 12A===<br /> *Average score: 49.22<br /> *AIME floor: 84<br /> *DHR: 121.5<br /> <br /> ===AMC 12B===<br /> *Average score: 56.74<br /> *AIME floor: 94.5 <br /> *DHR: 123<br /> <br /> ===AIME I===<br /> *Average score: 5.87<br /> *Median score: 6<br /> *USAMO cutoff: 220 (AMC 12A), 230.5 (AMC 12B)<br /> *USAJMO cutoff: 209.5 (AMC 10A), 216 (AMC 10B)<br /> <br /> ===AIME II===<br /> *Average score: 6.47<br /> *Median score: 6<br /> *USAMO cutoff: 230.5 (AMC 12A), 236 (AMC 12B)<br /> *USAJMO cutoff: 216.5 (AMC 10A), 220.5 (AMC 10B)<br /> ===AMC 8===<br /> *DHR: 23<br /> *HR: 19<br /> <br /> ==2018==<br /> ===AMC 10A===<br /> *Average score: 53.84<br /> *HR: 100.5<br /> *AIME floor: 111<br /> *DHR: 127.5<br /> <br /> ===AMC 10B===<br /> *Average score: 57.81<br /> *HR: 97.5<br /> *AIME floor: 108<br /> *DHR: 123<br /> <br /> ===AMC 12A===<br /> *Average score: 56.36<br /> *AIME floor: 93<br /> *DHR: 120<br /> <br /> ===AMC 12B===<br /> *Average score: 57.85<br /> *AIME floor: 99<br /> *DHR: 126<br /> <br /> ===AIME I===<br /> *Average score: 5.09<br /> *Median score: 5<br /> *USAMO cutoff: 215 (AMC 12A), 235 (AMC 12B)<br /> *USAJMO cutoff: 222 (AMC 10A), 212 (AMC 10B)<br /> <br /> ===AIME II===<br /> *Average score: 5.48<br /> *Median score: 5<br /> *USAMO cutoff: 216 (AMC 12A), 230.5 (AMC 12B)<br /> *USAJMO cutoff: 222 (AMC 10A), 212 (AMC 10B)<br /> <br /> ===AMC 8===<br /> *Average score: 8.51<br /> *Honor roll: 15<br /> *DHR: 19<br /> <br /> ==2017==<br /> ===AMC 10A===<br /> *Average score: 59.33<br /> *AIME floor: 112.5<br /> *DHR: 127.5<br /> <br /> ===AMC 10B===<br /> *Average score: 66.56<br /> *AIME floor: 120<br /> *DHR: 136.5<br /> <br /> ===AMC 12A===<br /> *Average score: 60.32<br /> *AIME floor: 96<br /> *DHR: 115.5<br /> <br /> ===AMC 12B===<br /> *Average score: 58.35<br /> *AIME floor: 100.5<br /> *DHR: 129<br /> <br /> ===AIME I===<br /> *Average score:5.70<br /> *Median score: 6<br /> *USAMO cutoff: 225 (AMC 12A), 235 (AMC 12B)<br /> *USAJMO cutoff: 224.5 (AMC 10A), 233 (AMC 10B)<br /> <br /> ===AIME II===<br /> *Average score: 5.64<br /> *Median score: 6<br /> *USAMO cutoff: 221 (AMC 12A), 230.5 (AMC 12B)<br /> *USAJMO cutoff: 219 (AMC 10A), 225 (AMC 10B)<br /> ===AMC 8===<br /> *Honor roll: 17<br /> *DHR: 20<br /> <br /> ==2016==<br /> ===AMC 10A===<br /> *Average score: 65.3<br /> *AIME floor: 110<br /> *DHR: 120<br /> <br /> ===AMC 10B===<br /> *Average score: 65.4<br /> *AIME floor: 110<br /> *DHR: 124.5<br /> <br /> ===AMC 12A===<br /> *Average score: 59.06<br /> *AIME floor: 93<br /> *DHR: 111<br /> <br /> ===AMC 12B===<br /> *Average score: 67.96<br /> *AIME floor: 100.5<br /> *DHR: 127.5<br /> <br /> ===AIME I===<br /> *Average score: 5.83<br /> *Median score: 6<br /> *USAMO cutoff: 220<br /> *USAJMO cutoff: 210.5<br /> <br /> ===AIME II===<br /> *Average score: 4.33<br /> *Median score: 4<br /> *USAMO cutoff: 205<br /> *USAJMO cutoff: 200<br /> <br /> ===AMC 8===<br /> *Honor roll: 18<br /> *DHR: 22<br /> <br /> ==2015==<br /> ===AMC 10A===<br /> *Average score: 73.39<br /> *AIME floor: 106.5<br /> *DHR: 115.5<br /> <br /> ===AMC 10B===<br /> *Average score: 76.10<br /> *AIME floor: 120<br /> *DHR: 132<br /> <br /> ===AMC 12A===<br /> *Average score: 69.90<br /> *AIME floor: 99<br /> *DHR: 117<br /> <br /> ===AMC 12B===<br /> *Average score: 66.92<br /> *AIME floor: 100.5<br /> *DHR: 126<br /> <br /> ===AIME I===<br /> *Average score: 5.29<br /> *Median score: 5<br /> *USAMO cutoff: 219.0<br /> *USAJMO cutoff: 213.0<br /> <br /> ===AIME II===<br /> *Average score: 6.63<br /> *Median score: 6<br /> *USAMO cutoff: 229.0<br /> *USAJMO cutoff: 223.5<br /> <br /> ===AMC 8===<br /> *Honor roll: 16<br /> *DHR: 21<br /> <br /> ==2014==<br /> ===AMC 10A===<br /> *Average score: 63.83<br /> *AIME floor: 120<br /> *DHR: 132<br /> <br /> ===AMC 10B===<br /> *Average score: 71.44<br /> *AIME floor: 120<br /> *DHR: 132<br /> <br /> ===AMC 12A===<br /> *Average score: 64.01<br /> *AIME floor: 93<br /> *DHR: 109.5<br /> <br /> ===AMC 12B===<br /> *Average score: 68.11<br /> *AIME floor: 100.5<br /> *DHR: 121.5<br /> <br /> ===AIME I===<br /> *Average score: 4.88<br /> *Median score: 5<br /> *USAMO cutoff: 211.5<br /> *USAJMO cutoff: 211<br /> <br /> ===AIME II===<br /> *Average score: 5.49<br /> *Median score: 5<br /> *USAMO cutoff: 211.5<br /> *USAJMO cutoff: 211<br /> <br /> ===AMC 8===<br /> *Honor roll: 19<br /> *DHR: 23<br /> <br /> ==2013==<br /> ===AMC 10A===<br /> *Average score: 72.50<br /> *AIME floor: 108<br /> *DHR: 117<br /> <br /> ===AMC 10B===<br /> *Average score: 72.62<br /> *AIME floor: 120<br /> *DHR: 129<br /> <br /> ===AMC 12A===<br /> *Average score: 65.06<br /> *AIME floor: 88.5<br /> *DHR: 106.5<br /> <br /> ===AMC 12B===<br /> *Average score: 64.21<br /> *AIME floor: 93<br /> *DHR: 108<br /> <br /> ===AIME I===<br /> *Average score: 4.69<br /> *Median score: 4<br /> *USAMO cutoff: 209<br /> *USAJMO cutoff: 210.5<br /> <br /> ===AIME II===<br /> *Average score: 6.56<br /> *Median score: 6<br /> *USAMO cutoff: 209<br /> *USAJMO cutoff: 210.5<br /> <br /> ===AMC 8===<br /> *Honor roll: 18<br /> *DHR: 22<br /> <br /> ==2012==<br /> ===AMC 10A===<br /> *Average score: 72.51<br /> *AIME floor: 115.5<br /> *DHR: 121.5<br /> <br /> ===AMC 10B===<br /> *Average score: 76.59<br /> *AIME floor: 120<br /> *DHR: 133.5<br /> <br /> ===AMC 12A===<br /> *Average score: 64.62<br /> *AIME floor: 94.5<br /> *DHR: 109.5<br /> <br /> ===AMC 12B===<br /> *Average score: 70.08<br /> *AIME floor: 99<br /> *DHR: 114<br /> <br /> ===AIME I===<br /> *Average score: 5.13<br /> *Median score: <br /> *USAMO cutoff: 204.5<br /> *USAJMO cutoff: 204<br /> <br /> ===AIME II===<br /> *Average score: 4.94<br /> *Median score: <br /> *USAMO cutoff: 204.5<br /> *USAJMO cutoff: 204<br /> <br /> ===AMC 8===<br /> *Honor roll: 18<br /> *DHR: 22<br /> <br /> ==2011==<br /> ===AMC 10A===<br /> *Average score: 64.24<br /> *AIME floor: 117<br /> *DHR: 129<br /> <br /> ===AMC 10B===<br /> *Average score: 71.78<br /> *AIME floor: 117<br /> *DHR: 133.5<br /> <br /> ===AMC 12A===<br /> *Average score: 65.38<br /> *AIME floor: 93<br /> *DHR: 112.5<br /> <br /> ===AMC 12B===<br /> *Average score: 64.71<br /> *AIME floor: 97.5<br /> *DHR: 121.5<br /> <br /> ===AIME I===<br /> *Average score: 2.23<br /> *Median score: <br /> *USAMO cutoff: 188<br /> *USAJMO cutoff: 179<br /> <br /> ===AIME II===<br /> *Average score: 5.47<br /> *Median score: <br /> *USAMO cutoff: 215.5<br /> *USAJMO cutoff: 196.5<br /> <br /> ===AMC 8===<br /> *Honor roll: 17<br /> *DHR: 22<br /> <br /> ==2010==<br /> ===AMC 10A===<br /> *Average score: 68.11<br /> *AIME floor: 115.5<br /> <br /> ===AMC 10B===<br /> *Average score: 68.57<br /> *AIME floor: 118.5<br /> <br /> ===AMC 12A===<br /> *Average score: 61.02<br /> *AIME floor: 88.5<br /> <br /> ===AMC 12B===<br /> *Average score: 59.58<br /> *AIME floor: 88.5<br /> <br /> ===AIME I===<br /> *Average score: 5.90<br /> *Median score: <br /> *USAMO cutoff: 208.5 (204.5 for non juniors and seniors)<br /> *USAJMO cutoff: 188.5<br /> <br /> ===AIME II===<br /> *Average score: 3.39<br /> *Median score: <br /> *USAMO cutoff: 208.5 (204.5 for non juniors and seniors)<br /> *USAJMO cutoff: 188.5<br /> <br /> ===AMC 8===<br /> *Honor roll: 17<br /> *DHR: 22<br /> <br /> ==2009==<br /> ===AMC 10A===<br /> *Average score: 67.41<br /> *AIME floor: 120<br /> <br /> ===AMC 10B===<br /> *Average score: 74.73<br /> *AIME floor: 120<br /> <br /> ===AMC 12A===<br /> *Average score: 66.37<br /> *AIME floor: 97.5<br /> <br /> ===AMC 12B===<br /> *Average score: 71.88<br /> *AIME floor: 100 (Top 5% (1.00))<br /> <br /> ===AIME I===<br /> *Average score: 4.17<br /> *Median score: 4<br /> *USAMO floor: <br /> <br /> ===AIME II===<br /> *Average score: 3.27<br /> *Median score: 3<br /> *USAMO floor:<br /> <br /> ===AMC 8===<br /> *Honor roll: 17<br /> *DHR: 20<br /> <br /> ==2008==<br /> ===AMC 10A===<br /> *Average score: 60.25<br /> *AIME floor:117<br /> <br /> ===AMC 10B===<br /> *Average score: <br /> *AIME floor:120<br /> <br /> ===AMC 12A===<br /> *Average score: 65.6<br /> *AIME floor: 97.5<br /> <br /> ===AMC 12B===<br /> *Average score: 68.9<br /> *AIME floor: 97.5<br /> <br /> ===AIME I===<br /> *Average score: <br /> *Median score: <br /> *USAMO floor: <br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AMC 8===<br /> *Honor roll: 19<br /> *DHR: 22<br /> <br /> ==2007==<br /> <br /> ===AMC 10A===<br /> *Average score: 67.9<br /> *AIME floor: 117<br /> <br /> ===AMC 10B=== &lt;dab&gt;<br /> *Average score: 61.5<br /> *AIME floor: 115.5<br /> <br /> ===AMC 12A===<br /> *Average score: 66.8<br /> *AIME floor: 97.5<br /> <br /> ===AMC 12B===<br /> *Average score: 73.1<br /> *AIME floor: 100<br /> <br /> ===AIME I===<br /> *Average score: 5<br /> *Median score: 3<br /> *USAMO floor: 6<br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AMC 8===<br /> *Honor roll: 17<br /> *DHR: 21<br /> <br /> ==2006==<br /> ===AMC 10A===<br /> *Average score: 79.0<br /> *AIME floor: 120<br /> <br /> ===AMC 10B===<br /> *Average score: 68.5<br /> *AIME floor: 120<br /> <br /> ===AMC 12A===<br /> *Average score: 85.7<br /> *AIME floor: 100<br /> <br /> ===AMC 12B===<br /> *Average score: 85.5<br /> *AIME floor: 100<br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AIME II===<br /> *Average score: <br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AMC 8===<br /> *Honor roll: 17<br /> *DHR: 21<br /> <br /> ==2005==<br /> ===AMC 10A===<br /> *Average score: 74.0<br /> *AIME floor: 120<br /> <br /> ===AMC 10B===<br /> *Average score: 79.0<br /> *AIME floor: 120<br /> <br /> ===AMC 12A===<br /> *Average score: 78.7<br /> *AIME floor: 100<br /> <br /> ===AMC 12B===<br /> *Average score: 83.4<br /> *AIME floor: 100<br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AMC 8===<br /> *Honor roll: 16<br /> *DHR: 20<br /> <br /> ==2004==<br /> ===AMC 10A===<br /> *Average score: 69.1<br /> *AIME floor: 110<br /> <br /> ===AMC 10B===<br /> *Average score: 80.4<br /> *AIME floor: 115<br /> <br /> ===AMC 12A===<br /> *Average score: 73.9<br /> *AIME floor: 100<br /> <br /> ===AMC 12B===<br /> *Average score: 84.5<br /> *AIME floor: 100<br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AMC 8===<br /> *Honor roll: 17<br /> *DHR: 21<br /> <br /> ==2003==<br /> ===AMC 10A===<br /> *Average score: 74.4<br /> *AIME floor: 119<br /> <br /> ===AMC 10B===<br /> *Average score: 79.6<br /> *AIME floor: 121<br /> <br /> ===AMC 12A===<br /> *Average score: 77.8<br /> *AIME floor: 100<br /> <br /> ===AMC 12B===<br /> *Average score: 76.6<br /> *AIME floor: 100<br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AMC 8===<br /> *Honor roll: 18<br /> *DHR: 22<br /> <br /> ==2002==<br /> ===AMC 10A===<br /> *Average score: 68.5<br /> *AIME floor: 115<br /> <br /> ===AMC 10B===<br /> *Average score: 74.9<br /> *AIME floor: 118<br /> <br /> ===AMC 12A===<br /> *Average score: 72.7<br /> *AIME floor: 100<br /> <br /> ===AMC 12B===<br /> *Average score: 80.8<br /> *AIME floor: 100<br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2001==<br /> ===AMC 10===<br /> *Average score: 67.8<br /> *AIME floor: 116<br /> <br /> ===AMC 12===<br /> *Average score: 56.6<br /> *AIME floor: 84<br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2000==<br /> ===AMC 10===<br /> *Average score: &lt;math&gt;64.2&lt;/math&gt;<br /> *AIME floor: &lt;math&gt;110&lt;/math&gt;<br /> <br /> ===AMC 12===<br /> *Average score: &lt;math&gt;64.9&lt;/math&gt;<br /> *AIME floor: &lt;math&gt;92&lt;/math&gt;<br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==1999==<br /> ===AHSME===<br /> *Average score: &lt;math&gt;68.8&lt;/math&gt;<br /> *AIME floor:<br /> <br /> ===AIME===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:</div> Scrabbler94 https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10B_Problems/Problem_5&diff=117629 2020 AMC 10B Problems/Problem 5 2020-02-10T19:38:07Z <p>Scrabbler94: Undo revision 117585 by Fidgetboss 4000 (talk)</p> <hr /> <div>==Problem==<br /> <br /> How many distinguishable arrangements are there of 1 brown tile, 1 purple tile, 2 green tiles, and 3 yellow tiles in a row from left to right? (Tiles of the same color are indistinguishable.)<br /> <br /> &lt;math&gt;\textbf{(A)}\ 210 \qquad\textbf{(B)}\ 420 \qquad\textbf{(C)}\ 630 \qquad\textbf{(D)}\ 840 \qquad\textbf{(E)}\ 1050&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Let's first find how many possibilities there would be if they were all distinguishable, then divide out the ones we overcounted.<br /> <br /> There are &lt;math&gt;7!&lt;/math&gt; ways to order &lt;math&gt;7&lt;/math&gt; objects. However, since there's &lt;math&gt;3!=6&lt;/math&gt; ways to switch the yellow tiles around without changing anything (since they're indistinguishable) and &lt;math&gt;2!=2&lt;/math&gt; ways to order the green tiles, we have to divide out these possibilities.<br /> <br /> &lt;math&gt;\frac{7!}{6\cdot2}=\boxed{\textbf{(B) }420}&lt;/math&gt; ~quacker88<br /> <br /> ==Video Solution==<br /> https://youtu.be/Gkm5rU5MlOU<br /> <br /> ~IceMatrix<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2020|ab=B|num-b=4|num-a=6}}<br /> {{MAA Notice}}</div> Scrabbler94 https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10B_Problems/Problem_5&diff=117628 2020 AMC 10B Problems/Problem 5 2020-02-10T19:37:49Z <p>Scrabbler94: Undo revision 117586 by Fidgetboss 4000 (talk)</p> <hr /> <div>==Problem==<br /> <br /> How many distinguishable arrangements are there of 1 brown tile, 1 purple tile, 2 green tiles, and 3 yellow tiles in a row from left to right? (Tiles of the same color are indistinguishable.)<br /> <br /> &lt;math&gt;\textbf{(A)}\ 210 \qquad\textbf{(B)}\ 420 \qquad\textbf{(C)}\ 630 \qquad\textbf{(D)}\ 840 \qquad\textbf{(E)}\ 1050&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Let's first find how many possibilities there would be if they were all distinguishable, then divide out the ones we overcounted.<br /> <br /> There are &lt;math&gt;7!&lt;/math&gt; ways to order &lt;math&gt;7&lt;/math&gt; objects. However, since there's &lt;math&gt;3!=6&lt;/math&gt; ways to switch the yellow tiles around without changing anything (since they're indistinguishable) and &lt;math&gt;2!=2&lt;/math&gt; ways to order the green tiles, we have to divide out these possibilities.<br /> <br /> &lt;math&gt;\frac{7!}{6\cdot2}=\boxed{\textbf{(B) }420}&lt;/math&gt; ~quacker88<br /> <br /> ==Solution 2 (2 second solve)==<br /> <br /> Be a memer and guess &lt;math&gt;\boxed{\textbf{(B) }420}&lt;/math&gt;, which happens to be the correct answer.<br /> -fidgetboss_4000 <br /> <br /> ==Video Solution==<br /> https://youtu.be/Gkm5rU5MlOU<br /> <br /> ~IceMatrix<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2020|ab=B|num-b=4|num-a=6}}<br /> {{MAA Notice}}</div> Scrabbler94 https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems&diff=116523 2020 AMC 10A Problems 2020-02-02T02:42:31Z <p>Scrabbler94: /* Problem 12 */ sp. isosceles</p> <hr /> <div>{{AMC10 Problems|year=2020|ab=A}}<br /> <br /> ==Problem 1==<br /> <br /> What value of &lt;math&gt;x&lt;/math&gt; satisfies<br /> <br /> &lt;cmath&gt;x- \frac{3}{4} = \frac{5}{12} - \frac{1}{3}?&lt;/cmath&gt;<br /> &lt;math&gt;\textbf{(A)}\ -\frac{2}{3}\qquad\textbf{(B)}\ \frac{7}{36}\qquad\textbf{(C)}\ \frac{7}{12}\qquad\textbf{(D)}\ \frac{2}{3}\qquad\textbf{(E)}\ \frac{5}{6}&lt;/math&gt;<br /> <br /> [[2020 AMC 10A Problems/Problem 1|Solution]]<br /> <br /> ==Problem 2==<br /> <br /> The numbers &lt;math&gt;3, 5, 7, a,&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; have an average (arithmetic mean) of &lt;math&gt;15&lt;/math&gt;. What is the average of &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 0 \qquad\textbf{(B) } 15 \qquad\textbf{(C) } 30 \qquad\textbf{(D) } 45 \qquad\textbf{(E) } 60&lt;/math&gt;<br /> <br /> [[2020 AMC 10A Problems/Problem 2|Solution]]<br /> <br /> ==Problem 3==<br /> <br /> Assuming &lt;math&gt;a\neq3&lt;/math&gt;, &lt;math&gt;b\neq4&lt;/math&gt;, and &lt;math&gt;c\neq5&lt;/math&gt;, what is the value in simplest form of the following expression?<br /> &lt;cmath&gt;\frac{a-3}{5-c} \cdot \frac{b-4}{3-a} \cdot \frac{c-5}{4-b}&lt;/cmath&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } -1 \qquad \textbf{(B) } 1 \qquad \textbf{(C) } \frac{abc}{60} \qquad \textbf{(D) } \frac{1}{abc} - \frac{1}{60} \qquad \textbf{(E) } \frac{1}{60} - \frac{1}{abc}&lt;/math&gt;<br /> <br /> [[2020 AMC 10A Problems/Problem 3|Solution]]<br /> <br /> ==Problem 4==<br /> <br /> A driver travels for &lt;math&gt;2&lt;/math&gt; hours at &lt;math&gt;60&lt;/math&gt; miles per hour, during which her car gets &lt;math&gt;30&lt;/math&gt; miles per gallon of gasoline. She is paid &lt;math&gt;\0.50&lt;/math&gt; per mile, and her only expense is gasoline at &lt;math&gt;\$2.00&lt;/math&gt; per gallon. What is her net rate of pay, in dollars per hour, after this expense?<br /> <br /> &lt;math&gt;\textbf{(A) }20 \qquad\textbf{(B) }22 \qquad\textbf{(C) }24 \qquad\textbf{(D) } 25\qquad\textbf{(E) } 26&lt;/math&gt;<br /> <br /> [[2020 AMC 10A Problems/Problem 4|Solution]]<br /> <br /> ==Problem 5==<br /> <br /> What is the sum of all real numbers &lt;math&gt;x&lt;/math&gt; for which &lt;math&gt;|x^2-12x+34|=2?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 12 \qquad \textbf{(B) } 15 \qquad \textbf{(C) } 18 \qquad \textbf{(D) } 21 \qquad \textbf{(E) } 25&lt;/math&gt;<br /> <br /> [[2020 AMC 10A Problems/Problem 5|Solution]]<br /> <br /> ==Problem 6==<br /> <br /> How many &lt;math&gt;4&lt;/math&gt;-digit positive integers (that is, integers between &lt;math&gt;1000&lt;/math&gt; and &lt;math&gt;9999&lt;/math&gt;, inclusive) having only even digits are divisible by &lt;math&gt;5?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 80 \qquad \textbf{(B) } 100 \qquad \textbf{(C) } 125 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 500&lt;/math&gt;<br /> <br /> [[2020 AMC 10A Problems/Problem 6|Solution]]<br /> <br /> ==Problem 7==<br /> <br /> The &lt;math&gt;25&lt;/math&gt; integers from &lt;math&gt;-10&lt;/math&gt; to &lt;math&gt;14,&lt;/math&gt; inclusive, can be arranged to form a &lt;math&gt;5&lt;/math&gt;-by-&lt;math&gt;5&lt;/math&gt; square in which the sum of the numbers in each row, the sum of the numbers in each column, and the sum of the numbers along each of the main diagonals are all the same. What is the value of this common sum?<br /> <br /> &lt;math&gt;\textbf{(A) }2 \qquad\textbf{(B) } 5\qquad\textbf{(C) } 10\qquad\textbf{(D) } 25\qquad\textbf{(E) } 50&lt;/math&gt;<br /> <br /> [[2020 AMC 10A Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8==<br /> <br /> What is the value of<br /> <br /> &lt;math&gt;1+2+3-4+5+6+7-8+\cdots+197+198+199-200?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 9,800 \qquad \textbf{(B) } 9,900 \qquad \textbf{(C) } 10,000 \qquad \textbf{(D) } 10,100 \qquad \textbf{(E) } 10,200&lt;/math&gt;<br /> <br /> [[2020 AMC 10A Problems/Problem 8|Solution]]<br /> <br /> ==Problem 9==<br /> <br /> A single bench section at a school event can hold either &lt;math&gt;7&lt;/math&gt; adults or &lt;math&gt;11&lt;/math&gt; children. When &lt;math&gt;N&lt;/math&gt; bench sections are connected end to end, an equal number of adults and children seated together will occupy all the bench space. What is the least possible positive integer value of &lt;math&gt;N?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 9 \qquad \textbf{(B) } 18 \qquad \textbf{(C) } 27 \qquad \textbf{(D) } 36 \qquad \textbf{(E) } 77&lt;/math&gt;<br /> <br /> [[2020 AMC 10A Problems/Problem 9|Solution]]<br /> <br /> ==Problem 10==<br /> <br /> Seven cubes, whose volumes are &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;8&lt;/math&gt;, &lt;math&gt;27&lt;/math&gt;, &lt;math&gt;64&lt;/math&gt;, &lt;math&gt;125&lt;/math&gt;, &lt;math&gt;216&lt;/math&gt;, and &lt;math&gt;343&lt;/math&gt; cubic units, are stacked vertically to form a tower in which the volumes of the cubes decrease from bottom to top. Except for the bottom cube, the bottom face of each cube lies completely on top of the cube below it. What is the total surface area of the tower (including the bottom) in square units?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 644\qquad\textbf{(B)}\ 658\qquad\textbf{(C)}\ 664\qquad\textbf{(D)}\ 720\qquad\textbf{(E)}\ 749&lt;/math&gt;<br /> <br /> [[2020 AMC 10A Problems/Problem 10|Solution]]<br /> <br /> ==Problem 11==<br /> <br /> What is the median of the following list of &lt;math&gt;4040&lt;/math&gt; numbers&lt;math&gt;?&lt;/math&gt;<br /> <br /> &lt;cmath&gt;1, 2, 3, ..., 2020, 1^2, 2^2, 3^2, ..., 2020^2&lt;/cmath&gt;<br /> &lt;math&gt;\textbf{(A)}\ 1974.5\qquad\textbf{(B)}\ 1975.5\qquad\textbf{(C)}\ 1976.5\qquad\textbf{(D)}\ 1977.5\qquad\textbf{(E)}\ 1978.5&lt;/math&gt;<br /> <br /> [[2020 AMC 10A Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> <br /> Triangle &lt;math&gt;AMC&lt;/math&gt; is isosceles with &lt;math&gt;AM = AC&lt;/math&gt;. Medians &lt;math&gt;\overline{MV}&lt;/math&gt; and &lt;math&gt;\overline{CU}&lt;/math&gt; are perpendicular to each other, and &lt;math&gt;MV=CU=12&lt;/math&gt;. What is the area of &lt;math&gt;\triangle AMC?&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> draw((-4,0)--(4,0)--(0,12)--cycle);<br /> draw((-2,6)--(4,0));<br /> draw((2,6)--(-4,0));<br /> label(&quot;M&quot;, (-4,0), W);<br /> label(&quot;C&quot;, (4,0), E);<br /> label(&quot;A&quot;, (0, 12), N);<br /> label(&quot;V&quot;, (2, 6), NE);<br /> label(&quot;U&quot;, (-2, 6), NW);<br /> label(&quot;P&quot;, (0, 3.6), S);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 48 \qquad \textbf{(B) } 72 \qquad \textbf{(C) } 96 \qquad \textbf{(D) } 144 \qquad \textbf{(E) } 192&lt;/math&gt;<br /> <br /> [[2020 AMC 10A Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> <br /> A frog sitting at the point &lt;math&gt;(1, 2)&lt;/math&gt; begins a sequence of jumps, where each jump is parallel to one of the coordinate axes and has length &lt;math&gt;1&lt;/math&gt;, and the direction of each jump (up, down, right, or left) is chosen independently at random. The sequence ends when the frog reaches a side of the square with vertices &lt;math&gt;(0,0), (0,4), (4,4),&lt;/math&gt; and &lt;math&gt;(4,0)&lt;/math&gt;. What is the probability that the sequence of jumps ends on a vertical side of the square&lt;math&gt;?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac12\qquad\textbf{(B)}\ \frac 58\qquad\textbf{(C)}\ \frac 23\qquad\textbf{(D)}\ \frac34\qquad\textbf{(E)}\ \frac 78&lt;/math&gt;<br /> <br /> [[2020 AMC 10A Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> <br /> Real numbers &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; satisfy &lt;math&gt;x + y = 4&lt;/math&gt; and &lt;math&gt;x \cdot y = -2&lt;/math&gt;. What is the value of&lt;cmath&gt;x + \frac{x^3}{y^2} + \frac{y^3}{x^2} + y?&lt;/cmath&gt;&lt;math&gt;\textbf{(A)}\ 360\qquad\textbf{(B)}\ 400\qquad\textbf{(C)}\ 420\qquad\textbf{(D)}\ 440\qquad\textbf{(E)}\ 480&lt;/math&gt;<br /> <br /> [[2020 AMC 10A Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> <br /> A positive integer divisor of &lt;math&gt;12!&lt;/math&gt; is chosen at random. The probability that the divisor chosen is a perfect square can be expressed as &lt;math&gt;\frac{m}{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. What is &lt;math&gt;m+n&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 3\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 23&lt;/math&gt;<br /> <br /> [[2020 AMC 10A Problems/Problem 15|Solution]]<br /> <br /> ==Problem 16==<br /> <br /> A point is chosen at random within the square in the coordinate plane whose vertices are &lt;math&gt;(0, 0), (2020, 0), (2020, 2020),&lt;/math&gt; and &lt;math&gt;(0, 2020)&lt;/math&gt;. The probability that the point is within &lt;math&gt;d&lt;/math&gt; units of a lattice point is &lt;math&gt;\tfrac{1}{2}&lt;/math&gt;. (A point &lt;math&gt;(x, y)&lt;/math&gt; is a lattice point if &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; are both integers.) What is &lt;math&gt;d&lt;/math&gt; to the nearest tenth&lt;math&gt;?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 0.3 \qquad \textbf{(B) } 0.4 \qquad \textbf{(C) } 0.5 \qquad \textbf{(D) } 0.6 \qquad \textbf{(E) } 0.7&lt;/math&gt;<br /> <br /> [[2020 AMC 10A Problems/Problem 16|Solution]]<br /> <br /> ==Problem 17==<br /> <br /> Define&lt;cmath&gt;P(x) =(x-1^2)(x-2^2)\cdots(x-100^2).&lt;/cmath&gt;How many integers &lt;math&gt;n&lt;/math&gt; are there such that &lt;math&gt;P(n)\leq 0&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 4900 \qquad \textbf{(B) } 4950\qquad \textbf{(C) } 5000\qquad \textbf{(D) } 5050 \qquad \textbf{(E) } 5100&lt;/math&gt;<br /> <br /> [[2020 AMC 10A Problems/Problem 17|Solution]]<br /> <br /> ==Problem 18==<br /> <br /> Let &lt;math&gt;(a,b,c,d)&lt;/math&gt; be an ordered quadruple of not necessarily distinct integers, each one of them in the set &lt;math&gt;{0,1,2,3}.&lt;/math&gt; For how many such quadruples is it true that &lt;math&gt;a\cdot d-b\cdot c&lt;/math&gt; is odd? (For example, &lt;math&gt;(0,3,1,1)&lt;/math&gt; is one such quadruple, because &lt;math&gt;0\cdot 1-3\cdot 1 = -3&lt;/math&gt; is odd.)<br /> <br /> &lt;math&gt;\textbf{(A) } 48 \qquad \textbf{(B) } 64 \qquad \textbf{(C) } 96 \qquad \textbf{(D) } 128 \qquad \textbf{(E) } 192&lt;/math&gt;<br /> <br /> [[2020 AMC 10A Problems/Problem 18|Solution]]<br /> <br /> ==Problem 19==<br /> <br /> As shown in the figure below, a regular dodecahedron (the polyhedron consisting of &lt;math&gt;12&lt;/math&gt; congruent regular pentagonal faces) floats in space with two horizontal faces. Note that there is a ring of five slanted faces adjacent to the top face, and a ring of five slanted faces adjacent to the bottom face. How many ways are there to move from the top face to the bottom face via a sequence of adjacent faces so that each face is visited at most once and moves are not permitted from the bottom ring to the top ring?<br /> <br /> &lt;asy&gt;<br /> import graph;<br /> unitsize(5cm);<br /> pair A = (0.082, 0.378);<br /> pair B = (0.091, 0.649);<br /> pair C = (0.249, 0.899);<br /> pair D = (0.479, 0.939);<br /> pair E = (0.758, 0.893);<br /> pair F = (0.862, 0.658);<br /> pair G = (0.924, 0.403);<br /> pair H = (0.747, 0.194);<br /> pair I = (0.526, 0.075);<br /> pair J = (0.251, 0.170);<br /> pair K = (0.568, 0.234);<br /> pair L = (0.262, 0.449);<br /> pair M = (0.373, 0.813);<br /> pair N = (0.731, 0.813);<br /> pair O = (0.851, 0.461);<br /> path[] f;<br /> f = A--B--C--M--L--cycle;<br /> f = C--D--E--N--M--cycle;<br /> f = E--F--G--O--N--cycle;<br /> f = G--H--I--K--O--cycle;<br /> f = I--J--A--L--K--cycle;<br /> f = K--L--M--N--O--cycle;<br /> draw(f);<br /> axialshade(f, white, M, gray(0.5), (C+2*D)/3);<br /> draw(f);<br /> filldraw(f, gray);<br /> filldraw(f, gray);<br /> axialshade(f, white, L, gray(0.7), J);<br /> draw(f);<br /> draw(f);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 125 \qquad \textbf{(B) } 250 \qquad \textbf{(C) } 405 \qquad \textbf{(D) } 640 \qquad \textbf{(E) } 810&lt;/math&gt;<br /> <br /> [[2020 AMC 10A Problems/Problem 19|Solution]]<br /> <br /> ==Problem 20==<br /> <br /> Quadrilateral &lt;math&gt;ABCD&lt;/math&gt; satisfies &lt;math&gt;\angle ABC = \angle ACD = 90^{\circ}, AC=20,&lt;/math&gt; and &lt;math&gt;CD=30.&lt;/math&gt; Diagonals &lt;math&gt;\overline{AC}&lt;/math&gt; and &lt;math&gt;\overline{BD}&lt;/math&gt; intersect at point &lt;math&gt;E,&lt;/math&gt; and &lt;math&gt;AE=5.&lt;/math&gt; What is the area of quadrilateral &lt;math&gt;ABCD?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 330 \qquad \textbf{(B) } 340 \qquad \textbf{(C) } 350 \qquad \textbf{(D) } 360 \qquad \textbf{(E) } 370&lt;/math&gt;<br /> <br /> [[2020 AMC 10A Problems/Problem 20|Solution]]<br /> <br /> ==Problem 21==<br /> <br /> There exists a unique strictly increasing sequence of nonnegative integers &lt;math&gt;a_1 &lt; a_2 &lt; … &lt; a_k&lt;/math&gt; such that&lt;cmath&gt;\frac{2^{289}+1}{2^{17}+1} = 2^{a_1} + 2^{a_2} + … + 2^{a_k}.&lt;/cmath&gt;What is &lt;math&gt;k?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 117 \qquad \textbf{(B) } 136 \qquad \textbf{(C) } 137 \qquad \textbf{(D) } 273 \qquad \textbf{(E) } 306&lt;/math&gt;<br /> <br /> [[2020 AMC 10A Problems/Problem 21|Solution]]<br /> <br /> ==Problem 22==<br /> <br /> For how many positive integers &lt;math&gt;n \le 1000&lt;/math&gt; is&lt;cmath&gt;\left\lfloor \dfrac{998}{n} \right\rfloor+\left\lfloor \dfrac{999}{n} \right\rfloor+\left\lfloor \dfrac{1000}{n}\right \rfloor&lt;/cmath&gt;not divisible by &lt;math&gt;3&lt;/math&gt;? (Recall that &lt;math&gt;\lfloor x \rfloor&lt;/math&gt; is the greatest integer less than or equal to &lt;math&gt;x&lt;/math&gt;.)<br /> <br /> &lt;math&gt;\textbf{(A) } 22 \qquad\textbf{(B) } 23 \qquad\textbf{(C) } 24 \qquad\textbf{(D) } 25 \qquad\textbf{(E) } 26&lt;/math&gt;<br /> <br /> [[2020 AMC 10A Problems/Problem 22|Solution]]<br /> <br /> ==Problem 23==<br /> <br /> Let &lt;math&gt;T&lt;/math&gt; be the triangle in the coordinate plane with vertices &lt;math&gt;(0,0), (4,0),&lt;/math&gt; and &lt;math&gt;(0,3).&lt;/math&gt; Consider the following five isometries (rigid transformations) of the plane: rotations of &lt;math&gt;90^{\circ}, 180^{\circ},&lt;/math&gt; and &lt;math&gt;270^{\circ}&lt;/math&gt; counterclockwise around the origin, reflection across the &lt;math&gt;x&lt;/math&gt;-axis, and reflection across the &lt;math&gt;y&lt;/math&gt;-axis. How many of the &lt;math&gt;125&lt;/math&gt; sequences of three of these transformations (not necessarily distinct) will return &lt;math&gt;T&lt;/math&gt; to its original position? (For example, a &lt;math&gt;180^{\circ}&lt;/math&gt; rotation, followed by a reflection across the &lt;math&gt;x&lt;/math&gt;-axis, followed by a reflection across the &lt;math&gt;y&lt;/math&gt;-axis will return &lt;math&gt;T&lt;/math&gt; to its original position, but a &lt;math&gt;90^{\circ}&lt;/math&gt; rotation, followed by a reflection across the &lt;math&gt;x&lt;/math&gt;-axis, followed by another reflection across the &lt;math&gt;x&lt;/math&gt;-axis will not return &lt;math&gt;T&lt;/math&gt; to its original position.)<br /> <br /> &lt;math&gt;\textbf{(A) } 12 \qquad \textbf{(B) } 15 \qquad \textbf{(C) } 17 \qquad \textbf{(D) } 20 \qquad \textbf{(E) } 25&lt;/math&gt;<br /> <br /> [[2020 AMC 10A Problems/Problem 23|Solution]]<br /> <br /> ==Problem 24==<br /> <br /> Let &lt;math&gt;n&lt;/math&gt; be the least positive integer greater than &lt;math&gt;1000&lt;/math&gt; for which&lt;cmath&gt;\gcd(63, n+120) =21\quad \text{and} \quad \gcd(n+63, 120)=60.&lt;/cmath&gt;What is the sum of the digits of &lt;math&gt;n&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 12 \qquad\textbf{(B) } 15 \qquad\textbf{(C) } 18 \qquad\textbf{(D) } 21\qquad\textbf{(E) } 24&lt;/math&gt;<br /> <br /> [[2020 AMC 10A Problems/Problem 24|Solution]]<br /> <br /> ==Problem 25==<br /> <br /> Jason rolls three fair standard six-sided dice. Then he looks at the rolls and chooses a subset of the dice (possibly empty, possibly all three dice) to reroll. After rerolling, he wins if and only if the sum of the numbers face up on the three dice is exactly &lt;math&gt;7&lt;/math&gt;. Jason always plays to optimize his chances of winning. What is the probability that he chooses to reroll exactly two of the dice?<br /> <br /> &lt;math&gt;\textbf{(A) } \frac{7}{36} \qquad\textbf{(B) } \frac{5}{24} \qquad\textbf{(C) } \frac{2}{9} \qquad\textbf{(D) } \frac{17}{72} \qquad\textbf{(E) } \frac{1}{4}&lt;/math&gt;<br /> <br /> [[2020 AMC 10A Problems/Problem 25|Solution]]<br /> <br /> ==See also==<br /> {{AMC10 box|year=2020|ab=A|before=[[2019 AMC 10B Problems]]|after=[[2020 AMC 10B Problems]]}}<br /> {{MAA Notice}}</div> Scrabbler94 https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_10A_Problems&diff=115479 2014 AMC 10A Problems 2020-01-25T00:30:54Z <p>Scrabbler94: revert this page's vandalism</p> <hr /> <div>==Problem 1==<br /> <br /> What is &lt;math&gt;10 \cdot \left(\tfrac{1}{2} + \tfrac{1}{5} + \tfrac{1}{10}\right)^{-1}?&lt;/math&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ 3\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ \frac{25}{2}\qquad\textbf{(D)}\ \frac{170}{3}\qquad\textbf{(E)}\ 170&lt;/math&gt;<br /> <br /> [[2014 AMC 10A Problems/Problem 1|Solution]]<br /> <br /> ==Problem 2==<br /> <br /> Roy's cat eats &lt;math&gt;\frac{1}{3}&lt;/math&gt; of a can of cat food every morning and &lt;math&gt;\frac{1}{4}&lt;/math&gt; of a can of cat food every evening. Before feeding his cat on Monday morning, Roy opened a box containing &lt;math&gt;6&lt;/math&gt; cans of cat food. On what day of the week did the cat finish eating all the cat food in the box?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \text{Tuesday}\qquad\textbf{(B)}\ \text{Wednesday}\qquad\textbf{(C)}\ \text{Thursday}\qquad\textbf{(D)}\ \text{Friday}\qquad\textbf{(E)}\ \text{Saturday}&lt;/math&gt;<br /> <br /> [[2014 AMC 10A Problems/Problem 2|Solution]]<br /> ==Problem 3==<br /> <br /> Bridget bakes 48 loaves of bread for her bakery. She sells half of them in the morning for &lt;math&gt;\textdollar 2.50&lt;/math&gt; each. In the afternoon she sells two thirds of what she has left, and because they are not fresh, she charges only half price. In the late afternoon she sells the remaining loaves at a dollar each. Each loaf costs &lt;math&gt;\textdollar 0.75&lt;/math&gt; for her to make. In dollars, what is her profit for the day?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 24\qquad\textbf{(B)}\ 36\qquad\textbf{(C)}\ 44\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 52&lt;/math&gt;<br /> <br /> [[2014 AMC 10A Problems/Problem 3|Solution]]<br /> ==Problem 4==<br /> <br /> Walking down Jane Street, Ralph passed four houses in a row, each painted a different color. He passed the orange house before the red house, and he passed the blue house before the yellow house. The blue house was not next to the yellow house. How many orderings of the colored houses are possible?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ 6&lt;/math&gt;<br /> <br /> [[2014 AMC 10A Problems/Problem 4|Solution]]<br /> ==Problem 5==<br /> <br /> On an algebra quiz, &lt;math&gt;10\%&lt;/math&gt; of the students scored &lt;math&gt;70&lt;/math&gt; points, &lt;math&gt;35\%&lt;/math&gt; scored &lt;math&gt;80&lt;/math&gt; points, &lt;math&gt;30\%&lt;/math&gt; scored &lt;math&gt;90&lt;/math&gt; points, and the rest scored &lt;math&gt;100&lt;/math&gt; points. What is the difference between the mean and median score of the students' scores on this quiz?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5&lt;/math&gt;<br /> <br /> [[2014 AMC 10A Problems/Problem 5|Solution]]<br /> <br /> ==Problem 6==<br /> <br /> Suppose that &lt;math&gt;a&lt;/math&gt; cows give &lt;math&gt;b&lt;/math&gt; gallons of milk in &lt;math&gt;c&lt;/math&gt; days. At this rate, how many gallons of milk will &lt;math&gt;d&lt;/math&gt; cows give in &lt;math&gt;e&lt;/math&gt; days?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{bde}{ac}\qquad\textbf{(B)}\ \frac{ac}{bde}\qquad\textbf{(C)}\ \frac{abde}{c}\qquad\textbf{(D)}\ \frac{bcde}{a}\qquad\textbf{(E)}\ \frac{abc}{de}&lt;/math&gt;<br /> <br /> [[2014 AMC 10A Problems/Problem 6|Solution]]<br /> <br /> ==Problem 7==<br /> <br /> Nonzero real numbers &lt;math&gt;x&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt;, &lt;math&gt;a&lt;/math&gt;, and &lt;math&gt;b&lt;/math&gt; satisfy &lt;math&gt;x &lt; a&lt;/math&gt; and &lt;math&gt;y &lt; b&lt;/math&gt;. How many of the following inequalities must be true?<br /> <br /> &lt;math&gt;\textbf{(I)}\ x+y &lt; a+b\qquad&lt;/math&gt;<br /> <br /> &lt;math&gt;<br /> \textbf{(II)}\ x-y &lt; a-b\qquad&lt;/math&gt;<br /> <br /> &lt;math&gt;<br /> \textbf{(III)}\ xy &lt; ab\qquad&lt;/math&gt;<br /> <br /> &lt;math&gt;<br /> \textbf{(IV)}\ \frac{x}{y} &lt; \frac{a}{b}&lt;/math&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4&lt;/math&gt;<br /> <br /> [[2014 AMC 10A Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8==<br /> <br /> Which of the following numbers is a perfect square?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \dfrac{14!15!}2\qquad\textbf{(B)}\ \dfrac{15!16!}2\qquad\textbf{(C)}\ \dfrac{16!17!}2\qquad\textbf{(D)}\ \dfrac{17!18!}2\qquad\textbf{(E)}\ \dfrac{18!19!}2 &lt;/math&gt;<br /> <br /> [[2014 AMC 10A Problems/Problem 8|Solution]]<br /> <br /> ==Problem 9==<br /> <br /> The two legs of a right triangle, which are altitudes, have lengths &lt;math&gt;2\sqrt3&lt;/math&gt; and &lt;math&gt;6&lt;/math&gt;. How long is the third altitude of the triangle?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5 &lt;/math&gt;<br /> <br /> [[2014 AMC 10A Problems/Problem 9|Solution]]<br /> ==Problem 10==<br /> <br /> Five positive consecutive integers starting with &lt;math&gt;a&lt;/math&gt; have average &lt;math&gt;b&lt;/math&gt;. What is the average of &lt;math&gt;5&lt;/math&gt; consecutive integers that start with &lt;math&gt;b&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ a+3\qquad\textbf{(B)}\ a+4\qquad\textbf{(C)}\ a+5\qquad\textbf{(D)}\ a+6\qquad\textbf{(E)}\ a+7&lt;/math&gt;<br /> <br /> [[2014 AMC 10A Problems/Problem 10|Solution]]<br /> <br /> ==Problem 11==<br /> <br /> A customer who intends to purchase an appliance has three coupons, only one of which may be used:<br /> <br /> Coupon 1: &lt;math&gt;10\%&lt;/math&gt; off the listed price if the listed price is at least &lt;math&gt;\textdollar50&lt;/math&gt;<br /> <br /> Coupon 2: &lt;math&gt;\textdollar 20&lt;/math&gt; off the listed price if the listed price is at least &lt;math&gt;\textdollar100&lt;/math&gt;<br /> <br /> Coupon 3: &lt;math&gt;18\%&lt;/math&gt; off the amount by which the listed price exceeds &lt;math&gt;\textdollar100&lt;/math&gt;<br /> <br /> For which of the following listed prices will coupon &lt;math&gt;1&lt;/math&gt; offer a greater price reduction than either coupon &lt;math&gt;2&lt;/math&gt; or coupon &lt;math&gt;3&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }\textdollar179.95\qquad<br /> \textbf{(B) }\textdollar199.95\qquad<br /> \textbf{(C) }\textdollar219.95\qquad<br /> \textbf{(D) }\textdollar239.95\qquad<br /> \textbf{(E) }\textdollar259.95\qquad&lt;/math&gt;<br /> <br /> [[2014 AMC 10A Problems/Problem 11|Solution]]<br /> ==Problem 12==<br /> <br /> A regular hexagon has side length 6. Congruent arcs with radius 3 are drawn with the center at each of the vertices, creating circular sectors as shown. The region inside the hexagon but outside the sectors is shaded as shown. What is the area of the shaded region?<br /> <br /> &lt;asy&gt;<br /> size(125);<br /> defaultpen(linewidth(0.8));<br /> path hexagon=(2*dir(0))--(2*dir(60))--(2*dir(120))--(2*dir(180))--(2*dir(240))--(2*dir(300))--cycle;<br /> fill(hexagon,lightgrey);<br /> for(int i=0;i&lt;=5;i=i+1)<br /> {<br /> path arc=2*dir(60*i)--arc(2*dir(60*i),1,120+60*i,240+60*i)--cycle;<br /> unfill(arc);<br /> draw(arc);<br /> }<br /> draw(hexagon,linewidth(1.8));&lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ 27\sqrt{3}-9\pi\qquad\textbf{(B)}\ 27\sqrt{3}-6\pi\qquad\textbf{(C)}\ 54\sqrt{3}-18\pi\qquad\textbf{(D)}\ 54\sqrt{3}-12\pi\qquad\textbf{(E)}\ 108\sqrt{3}-9\pi &lt;/math&gt;<br /> <br /> [[2014 AMC 10A Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> <br /> Equilateral &lt;math&gt;\triangle ABC&lt;/math&gt; has side length &lt;math&gt;1&lt;/math&gt;, and squares &lt;math&gt;ABDE&lt;/math&gt;, &lt;math&gt;BCHI&lt;/math&gt;, &lt;math&gt;CAFG&lt;/math&gt; lie outside the triangle. What is the area of hexagon &lt;math&gt;DEFGHI&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> import graph;<br /> size(6cm);<br /> pen dps = linewidth(0.7) + fontsize(8); defaultpen(dps);<br /> pair B = (0,0);<br /> pair C = (1,0);<br /> pair A = rotate(60,B)*C;<br /> <br /> pair E = rotate(270,A)*B;<br /> pair D = rotate(270,E)*A;<br /> <br /> pair F = rotate(90,A)*C;<br /> pair G = rotate(90,F)*A;<br /> <br /> pair I = rotate(270,B)*C;<br /> pair H = rotate(270,I)*B;<br /> <br /> draw(A--B--C--cycle);<br /> draw(A--E--D--B);<br /> draw(A--F--G--C);<br /> draw(B--I--H--C);<br /> <br /> draw(E--F);<br /> draw(D--I);<br /> draw(I--H);<br /> draw(H--G);<br /> <br /> label(&quot;$A$&quot;,A,N);<br /> label(&quot;$B$&quot;,B,SW);<br /> label(&quot;$C$&quot;,C,SE);<br /> label(&quot;$D$&quot;,D,W);<br /> label(&quot;$E$&quot;,E,W);<br /> label(&quot;$F$&quot;,F,E);<br /> label(&quot;$G$&quot;,G,E);<br /> label(&quot;$H$&quot;,H,SE);<br /> label(&quot;$I$&quot;,I,SW);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ \dfrac{12+3\sqrt3}4\qquad\textbf{(B)}\ \dfrac92\qquad\textbf{(C)}\ 3+\sqrt3\qquad\textbf{(D)}\ \dfrac{6+3\sqrt3}2\qquad\textbf{(E)}\ 6 &lt;/math&gt;<br /> <br /> [[2014 AMC 10A Problems/Problem 13|Solution]]<br /> ==Problem 14==<br /> <br /> The &lt;math&gt;y&lt;/math&gt;-intercepts, &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt;, of two perpendicular lines intersecting at the point &lt;math&gt;A(6,8)&lt;/math&gt; have a sum of zero. What is the area of &lt;math&gt;\triangle APQ&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 45\qquad\textbf{(B)}\ 48\qquad\textbf{(C)}\ 54\qquad\textbf{(D)}\ 60\qquad\textbf{(E)}\ 72 &lt;/math&gt;<br /> <br /> [[2014 AMC 10A Problems/Problem 14|Solution]]<br /> ==Problem 15==<br /> <br /> David drives from his home to the airport to catch a flight. He drives &lt;math&gt;35&lt;/math&gt; miles in the first hour, but realizes that he will be &lt;math&gt;1&lt;/math&gt; hour late if he continues at this speed. He increases his speed by &lt;math&gt;15&lt;/math&gt; miles per hour for the rest of the way to the airport and arrives &lt;math&gt;30&lt;/math&gt; minutes early. How many miles is the airport from his home?<br /> <br /> &lt;math&gt;\textbf{(A) }140\qquad<br /> \textbf{(B) }175\qquad<br /> \textbf{(C) }210\qquad<br /> \textbf{(D) }245\qquad<br /> \textbf{(E) }280\qquad&lt;/math&gt;<br /> <br /> [[2014 AMC 10A Problems/Problem 15|Solution]]<br /> ==Problem 16==<br /> <br /> In rectangle &lt;math&gt;ABCD&lt;/math&gt;, &lt;math&gt;AB=1&lt;/math&gt;, &lt;math&gt;BC=2&lt;/math&gt;, and points &lt;math&gt;E&lt;/math&gt;, &lt;math&gt;F&lt;/math&gt;, and &lt;math&gt;G&lt;/math&gt; are midpoints of &lt;math&gt;\overline{BC}&lt;/math&gt;, &lt;math&gt;\overline{CD}&lt;/math&gt;, and &lt;math&gt;\overline{AD}&lt;/math&gt;, respectively. Point &lt;math&gt;H&lt;/math&gt; is the midpoint of &lt;math&gt;\overline{GE}&lt;/math&gt;. What is the area of the shaded region?<br /> <br /> &lt;asy&gt;<br /> import graph;<br /> size(9cm);<br /> pen dps = fontsize(10); defaultpen(dps);<br /> pair D = (0,0);<br /> pair F = (1/2,0);<br /> pair C = (1,0);<br /> pair G = (0,1);<br /> pair E = (1,1);<br /> pair A = (0,2);<br /> pair B = (1,2);<br /> pair H = (1/2,1);<br /> <br /> // do not look<br /> pair X = (1/3,2/3);<br /> pair Y = (2/3,2/3);<br /> <br /> draw(A--B--C--D--cycle);<br /> draw(G--E);<br /> draw(A--F--B);<br /> draw(D--H--C);<br /> filldraw(H--X--F--Y--cycle,grey);<br /> <br /> label(&quot;$A$&quot;,A,NW);<br /> label(&quot;$B$&quot;,B,NE);<br /> label(&quot;$C$&quot;,C,SE);<br /> label(&quot;$D$&quot;,D,SW);<br /> label(&quot;$E$&quot;,E,E);<br /> label(&quot;$F$&quot;,F,S);<br /> label(&quot;$G$&quot;,G,W);<br /> label(&quot;$H$&quot;,H,N);<br /> <br /> label(&quot;$\frac12$&quot;,(0.25,0),S);<br /> label(&quot;$\frac12$&quot;,(0.75,0),S);<br /> label(&quot;$1$&quot;,(1,0.5),E);<br /> label(&quot;$1$&quot;,(1,1.5),E);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ \dfrac1{12}\qquad\textbf{(B)}\ \dfrac{\sqrt3}{18}\qquad\textbf{(C)}\ \dfrac{\sqrt2}{12}\qquad\textbf{(D)}\ \dfrac{\sqrt3}{12}\qquad\textbf{(E)}\ \dfrac16 &lt;/math&gt;<br /> <br /> [[2014 AMC 10A Problems/Problem 16|Solution]]<br /> <br /> ==Problem 17==<br /> <br /> Three fair six-sided dice are rolled. What is the probability that the values shown on two of the dice sum to the value shown on the remaining die?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \dfrac16\qquad\textbf{(B)}\ \dfrac{13}{72}\qquad\textbf{(C)}\ \dfrac7{36}\qquad\textbf{(D)}\ \dfrac5{24}\qquad\textbf{(E)}\ \dfrac29 &lt;/math&gt;<br /> <br /> [[2014 AMC 10A Problems/Problem 17|Solution]]<br /> ==Problem 18==<br /> <br /> A square in the coordinate plane has vertices whose &lt;math&gt;y&lt;/math&gt;-coordinates are &lt;math&gt;0&lt;/math&gt;, &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;4&lt;/math&gt;, and &lt;math&gt;5&lt;/math&gt;. What is the area of the square?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 16\qquad\textbf{(B)}\ 17\qquad\textbf{(C)}\ 25\qquad\textbf{(D)}\ 26\qquad\textbf{(E)}\ 27 &lt;/math&gt;<br /> <br /> [[2014 AMC 10A Problems/Problem 18|Solution]]<br /> <br /> ==Problem 19==<br /> <br /> Four cubes with edge lengths &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;, and &lt;math&gt;4&lt;/math&gt; are stacked as shown. What is the length of the portion of &lt;math&gt;\overline{XY}&lt;/math&gt; contained in the cube with edge length &lt;math&gt;3&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \dfrac{3\sqrt{33}}5\qquad\textbf{(B)}\ 2\sqrt3\qquad\textbf{(C)}\ \dfrac{2\sqrt{33}}3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 3\sqrt2 &lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> dotfactor = 3;<br /> size(10cm);<br /> dot((0, 10));<br /> label(&quot;$X$&quot;, (0,10),W,fontsize(8pt));<br /> dot((6,2));<br /> label(&quot;$Y$&quot;, (6,2),E,fontsize(8pt));<br /> draw((0, 0)--(0, 10)--(1, 10)--(1, 9)--(2, 9)--(2, 7)--(3, 7)--(3,4)--(4, 4)--(4, 0)--cycle);<br /> draw((0,9)--(1, 9)--(1.5, 9.5)--(1.5, 10.5)--(0.5, 10.5)--(0, 10));<br /> draw((1, 10)--(1.5,10.5));<br /> draw((1.5, 10)--(3,10)--(3,8)--(2,7)--(0,7));<br /> draw((2,9)--(3,10));<br /> draw((3,8.5)--(4.5,8.5)--(4.5,5.5)--(3,4)--(0,4));<br /> draw((3,7)--(4.5,8.5));<br /> draw((4.5,6)--(6,6)--(6,2)--(4,0));<br /> draw((4,4)--(6,6));<br /> label(&quot;$1$&quot;, (1,9.5), W,fontsize(8pt));<br /> label(&quot;$2$&quot;, (2,8), W,fontsize(8pt));<br /> label(&quot;$3$&quot;, (3,5.5), W,fontsize(8pt));<br /> label(&quot;$4\$&quot;, (4,2), W,fontsize(8pt));<br /> &lt;/asy&gt;<br /> [[2014 AMC 10A Problems/Problem 19|Solution]]<br /> ==Problem 20==<br /> <br /> The product &lt;math&gt;(8)(888\dots8)&lt;/math&gt;, where the second factor has &lt;math&gt;k&lt;/math&gt; digits, is an integer whose digits have a sum of &lt;math&gt;1000&lt;/math&gt;. What is &lt;math&gt;k&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 901\qquad\textbf{(B)}\ 911\qquad\textbf{(C)}\ 919\qquad\textbf{(D)}\ 991\qquad\textbf{(E)}\ 999 &lt;/math&gt;<br /> <br /> [[2014 AMC 10A Problems/Problem 20|Solution]]<br /> <br /> ==Problem 21==<br /> Positive integers &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are such that the graphs of &lt;math&gt;y=ax+5&lt;/math&gt; and &lt;math&gt;y=3x+b&lt;/math&gt; intersect the &lt;math&gt;x&lt;/math&gt;-axis at the same point. What is the sum of all possible &lt;math&gt;x&lt;/math&gt;-coordinates of these points of intersection?<br /> <br /> &lt;math&gt; \textbf{(A)}\ {-20}\qquad\textbf{(B)}\ {-18}\qquad\textbf{(C)}\ {-15}\qquad\textbf{(D)}\ {-12}\qquad\textbf{(E)}\ {-8} &lt;/math&gt;<br /> <br /> [[2014 AMC 10A Problems/Problem 21|Solution]]<br /> <br /> ==Problem 22==<br /> <br /> In rectangle &lt;math&gt;ABCD&lt;/math&gt;, &lt;math&gt;AB=20&lt;/math&gt; and &lt;math&gt;BC=10&lt;/math&gt;. Let &lt;math&gt;E&lt;/math&gt; be a point on &lt;math&gt;\overline{CD}&lt;/math&gt; such that &lt;math&gt;\angle CBE=15^\circ&lt;/math&gt;. What is &lt;math&gt;AE&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \dfrac{20\sqrt3}3\qquad\textbf{(B)}\ 10\sqrt3\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 11\sqrt3\qquad\textbf{(E)}\ 20 &lt;/math&gt;<br /> <br /> [[2014 AMC 10A Problems/Problem 22|Solution]]<br /> ==Problem 23==<br /> <br /> A rectangular piece of paper whose length is &lt;math&gt;\sqrt3&lt;/math&gt; times the width has area &lt;math&gt;A&lt;/math&gt;. The paper is divided into three equal sections along the opposite lengths, and then a dotted line is drawn from the first divider to the second divider on the opposite side as shown. The paper is then folded flat along this dotted line to create a new shape with area &lt;math&gt;B&lt;/math&gt;. What is the ratio &lt;math&gt;B:A&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> import graph;<br /> size(6cm);<br /> <br /> real L = 0.05;<br /> <br /> pair A = (0,0);<br /> pair B = (sqrt(3),0);<br /> pair C = (sqrt(3),1);<br /> pair D = (0,1);<br /> <br /> pair X1 = (sqrt(3)/3,0);<br /> pair X2= (2*sqrt(3)/3,0);<br /> pair Y1 = (2*sqrt(3)/3,1);<br /> pair Y2 = (sqrt(3)/3,1);<br /> <br /> dot(X1);<br /> dot(Y1);<br /> <br /> draw(A--B--C--D--cycle, linewidth(2));<br /> draw(X1--Y1,dashed);<br /> <br /> draw(X2--(2*sqrt(3)/3,L));<br /> draw(Y2--(sqrt(3)/3,1-L));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ 1:2\qquad\textbf{(B)}\ 3:5\qquad\textbf{(C)}\ 2:3\qquad\textbf{(D)}\ 3:4\qquad\textbf{(E)}\ 4:5 &lt;/math&gt;<br /> <br /> [[2014 AMC 10A Problems/Problem 23|Solution]]<br /> <br /> ==Problem 24==<br /> <br /> A sequence of natural numbers is constructed by listing the first &lt;math&gt;4&lt;/math&gt;, then skipping one, listing the next &lt;math&gt;5&lt;/math&gt;, skipping &lt;math&gt;2&lt;/math&gt;, listing &lt;math&gt;6&lt;/math&gt;, skipping &lt;math&gt;3&lt;/math&gt;, and, on the &lt;math&gt;n&lt;/math&gt;th iteration, listing &lt;math&gt;n+3&lt;/math&gt; and skipping &lt;math&gt;n&lt;/math&gt;. The sequence begins &lt;math&gt;1,2,3,4,6,7,8,9,10,13&lt;/math&gt;. What is the &lt;math&gt;500,000&lt;/math&gt;th number in the sequence?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 996,506\qquad\textbf{(B)}\ 996,507\qquad\textbf{(C)}\ 996,508\qquad\textbf{(D)}\ 996,509\qquad\textbf{(E)}\ 996,510 &lt;/math&gt;<br /> <br /> [[2014 AMC 10A Problems/Problem 24|Solution]]<br /> <br /> ==Problem 25==<br /> <br /> The number &lt;math&gt;5^{867}&lt;/math&gt; is between &lt;math&gt;2^{2013}&lt;/math&gt; and &lt;math&gt;2^{2014}&lt;/math&gt;. How many pairs of integers &lt;math&gt;(m,n)&lt;/math&gt; are there such that &lt;math&gt;1\leq m\leq 2012&lt;/math&gt; and &lt;cmath&gt;5^n&lt;2^m&lt;2^{m+2}&lt;5^{n+1}?&lt;/cmath&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }278\qquad\textbf{(B) }279\qquad\textbf{(C) }280\qquad\textbf{(D) }281\qquad\textbf{(E) }282\qquad&lt;/math&gt;<br /> <br /> [[2014 AMC 10A Problems/Problem 25|Solution]]<br /> ==See also==<br /> {{AMC10 box|year=2014|ab=A|before=[[2013 AMC 10B Problems]]|after=[[2014 AMC 10B Problems]]}}<br /> * [[AMC 10]]<br /> * [[AMC 10 Problems and Solutions]]<br /> * [[2014 AMC 10A]]<br /> * [[Mathematics competition resources]]<br /> {{MAA Notice}}</div> Scrabbler94 https://artofproblemsolving.com/wiki/index.php?title=2004_AMC_10B_Problems/Problem_21&diff=115361 2004 AMC 10B Problems/Problem 21 2020-01-24T04:04:23Z <p>Scrabbler94: /* Solution */ solution 2 isn't really correct as the positions of the terms which appear in both sequences change after shifting.</p> <hr /> <div>==Problem==<br /> <br /> Let &lt;math&gt;1&lt;/math&gt;; &lt;math&gt;4&lt;/math&gt;; &lt;math&gt;\ldots&lt;/math&gt; and &lt;math&gt;9&lt;/math&gt;; &lt;math&gt;16&lt;/math&gt;; &lt;math&gt;\ldots&lt;/math&gt; be two arithmetic progressions. The set &lt;math&gt;S&lt;/math&gt; is the union of the first &lt;math&gt;2004&lt;/math&gt; terms of each sequence. How many distinct numbers are in &lt;math&gt;S&lt;/math&gt;?<br /> <br /> &lt;math&gt; \mathrm{(A) \ } 3722 \qquad \mathrm{(B) \ } 3732 \qquad \mathrm{(C) \ } 3914 \qquad \mathrm{(D) \ } 3924 \qquad \mathrm{(E) \ } 4007 &lt;/math&gt;<br /> ==Solution==<br /> The two sets of terms are &lt;math&gt;A=\{ 3k+1 : 0\leq k &lt; 2004 \}&lt;/math&gt; and &lt;math&gt;B=\{ 7l+9 : 0\leq l&lt;2004\}&lt;/math&gt;. <br /> <br /> Now &lt;math&gt;S=A\cup B&lt;/math&gt;. We can compute &lt;math&gt;|S|=|A\cup B|=|A|+|B|-|A\cap B|=4008-|A\cap B|&lt;/math&gt;. We will now find &lt;math&gt;|A\cap B|&lt;/math&gt;. <br /> <br /> Consider the numbers in &lt;math&gt;B&lt;/math&gt;. We want to find out how many of them lie in &lt;math&gt;A&lt;/math&gt;. In other words, we need to find out the number of valid values of &lt;math&gt;l&lt;/math&gt; for which &lt;math&gt;7l+9\in A&lt;/math&gt;.<br /> <br /> The fact &quot;&lt;math&gt;7l+9\in A&lt;/math&gt;&quot; can be rewritten as &quot;&lt;math&gt;1\leq 7l+9 \leq 3\cdot 2003 + 1&lt;/math&gt;, and &lt;math&gt;7l+9\equiv 1\pmod 3&lt;/math&gt;&quot;.<br /> <br /> The first condition gives &lt;math&gt;0\leq l\leq 857&lt;/math&gt;, the second one gives &lt;math&gt;l\equiv 1\pmod 3&lt;/math&gt;.<br /> <br /> Thus the good values of &lt;math&gt;l&lt;/math&gt; are &lt;math&gt;\{1,4,7,\dots,856\}&lt;/math&gt;, and their count is &lt;math&gt;858/3 = 286&lt;/math&gt;.<br /> <br /> Therefore &lt;math&gt;|A\cap B|=286&lt;/math&gt;, and thus &lt;math&gt;|S|=4008-|A\cap B|=\boxed{3722}&lt;/math&gt;.<br /> <br /> == See also ==<br /> <br /> {{AMC10 box|year=2004|ab=B|num-b=20|num-a=22}}<br /> {{MAA Notice}}</div> Scrabbler94 https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10B_Problems/Problem_21&diff=112736 2009 AMC 10B Problems/Problem 21 2019-12-09T18:59:05Z <p>Scrabbler94: Solution 3 is incorrect</p> <hr /> <div>== Problem ==<br /> What is the remainder when &lt;math&gt;3^0 + 3^1 + 3^2 + \cdots + 3^{2009}&lt;/math&gt; is divided by 8?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 0\qquad<br /> \mathrm{(B)}\ 1\qquad<br /> \mathrm{(C)}\ 2\qquad<br /> \mathrm{(D)}\ 4\qquad<br /> \mathrm{(E)}\ 6&lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> === Solution 1 ===<br /> <br /> The sum of any four consecutive powers of 3 is divisible by &lt;math&gt;3^0 + 3^1 + 3^2 +3^3 = 40&lt;/math&gt; and hence is divisible by 8. Therefore<br /> : &lt;math&gt;(3^2 + 3^3 + 3^4 + 3^5) + \cdots + (3^{2006} + 3^{2007} + 3^{2008} + 3^{2009})&lt;/math&gt;<br /> is divisible by 8. So the required remainder is &lt;math&gt;3^0 + 3^1 = \boxed {4}&lt;/math&gt;. The answer is &lt;math&gt;\mathrm{(D)}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> We have &lt;math&gt;3^2 = 9 \equiv 1 \pmod 8&lt;/math&gt;. Hence for any &lt;math&gt;k&lt;/math&gt; we have &lt;math&gt;3^{2k}\equiv 1^k = 1 \pmod 8&lt;/math&gt;, and then &lt;math&gt;3^{2k+1} = 3\cdot 3^{2k} \equiv 3\cdot 1 = 3 \pmod 8&lt;/math&gt;.<br /> <br /> Therefore our sum gives the same remainder modulo &lt;math&gt;8&lt;/math&gt; as &lt;math&gt;1 + 3 + 1 + 3 + 1 + \cdots + 1 + 3&lt;/math&gt;. There are &lt;math&gt;2010&lt;/math&gt; terms in the sum, hence there are &lt;math&gt;2010/2 = 1005&lt;/math&gt; pairs &lt;math&gt;1+3&lt;/math&gt;, and thus the sum is &lt;math&gt;1005 \cdot 4 = 4020 \equiv 20 \equiv \boxed{4} \pmod 8&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AMC10 box|year=2009|ab=B|num-b=20|num-a=22}}<br /> {{MAA Notice}}</div> Scrabbler94 https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_12B_Problems/Problem_11&diff=112360 2005 AMC 12B Problems/Problem 11 2019-11-29T18:43:31Z <p>Scrabbler94: Solution 3 is incorrect. Cannot simply apply expected value arguments like that.</p> <hr /> <div>{{duplicate|[[2005 AMC 12B Problems|2005 AMC 12B #11]] and [[2005 AMC 10B Problems|2005 AMC 10B #15]]}}<br /> == Problem ==<br /> An envelope contains eight bills: &lt;math&gt;2&lt;/math&gt; ones, &lt;math&gt;2&lt;/math&gt; fives, &lt;math&gt;2&lt;/math&gt; tens, and &lt;math&gt;2&lt;/math&gt; twenties. Two bills are drawn at random without replacement. What is the probability that their sum is &amp;#36;&lt;math&gt;20&lt;/math&gt; or more?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ {{{\frac{1}{4}}}} \qquad \mathrm{(B)}\ {{{\frac{2}{5}}}} \qquad \mathrm{(C)}\ {{{\frac{3}{7}}}} \qquad \mathrm{(D)}\ {{{\frac{1}{2}}}} \qquad \mathrm{(E)}\ {{{\frac{2}{3}}}}&lt;/math&gt;<br /> <br /> == Solution 1==<br /> The only way to get a total of &amp;#36;&lt;math&gt;20&lt;/math&gt; or more is if you pick a twenty and another bill, or if you pick both tens. There are a total of &lt;math&gt;\dbinom{8}{2}=\dfrac{8\times7}{2\times1}=28&lt;/math&gt; ways to choose &lt;math&gt;2&lt;/math&gt; bills out of &lt;math&gt;8&lt;/math&gt;. There are &lt;math&gt;12&lt;/math&gt; ways to choose a twenty and some other non-twenty bill. There is &lt;math&gt;1&lt;/math&gt; way to choose both twenties, and also &lt;math&gt;1&lt;/math&gt; way to choose both tens. Adding these up, we find that there are a total of &lt;math&gt;14&lt;/math&gt; ways to attain a sum of &lt;math&gt;20&lt;/math&gt; or greater, so there is a total probability of &lt;math&gt;\dfrac{14}{28}=\boxed{\mathrm{(D)}\ \dfrac{1}{2}}&lt;/math&gt;.<br /> <br /> == Solution 2==<br /> Another way to do this problem is to use complementary counting, i.e. how many ways that the sum is less than 20. Now, you do not have to consider the 2 twenties, so you have 6 bills left. &lt;math&gt;\dbinom{6}{2} = \dfrac{6\times5}{2\times1} = 15&lt;/math&gt; ways. However, you counted the case when you have 2 tens, so you need to subtract 1, and you get 14. Finding the ways to get 20 or higher, you subtract 14 from 28 and get 14. So the answer is &lt;math&gt;\dfrac{14}{28} = \boxed{\mathrm{(D)}\ \dfrac{1}{2}}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AMC10 box|year=2005|ab=B|num-b=14|num-a=16}}<br /> {{AMC12 box|year=2005|ab=B|num-b=10|num-a=12}}<br /> {{MAA Notice}}</div> Scrabbler94 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_13&diff=112224 2019 AMC 8 Problems/Problem 13 2019-11-27T06:23:11Z <p>Scrabbler94: solutions 1&amp;2 identical; for completeness, solution should check that 110 actually works</p> <hr /> <div>==Problem 13==<br /> A ''palindrome'' is a number that has the same value when read from left to right or from right to left. (For example 12321 is a palindrome.) Let &lt;math&gt;N&lt;/math&gt; be the least three-digit integer which is not a palindrome but which is the sum of three distinct two-digit palindromes. What is the sum of the digits of &lt;math&gt;N&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }2\qquad\textbf{(B) }3\qquad\textbf{(C) }4\qquad\textbf{(D) }5\qquad\textbf{(E) }6&lt;/math&gt;<br /> <br /> ==Solution==<br /> Note that the only positive 2-digit palindromes are multiples of 11, namely &lt;math&gt;11, 22, \ldots, 99&lt;/math&gt;. Since &lt;math&gt;N&lt;/math&gt; is the sum of 2-digit palindromes, &lt;math&gt;N&lt;/math&gt; is necessarily a multiple of 11. The smallest 3-digit multiple of 11 which is not a palindrome is 110, so &lt;math&gt;N=110&lt;/math&gt; is a candidate solution. We must check that 110 can be written as the sum of three distinct 2-digit palindromes; this suffices as &lt;math&gt;110=77+22+11&lt;/math&gt;. Then &lt;math&gt;N = 110&lt;/math&gt;, and the sum of the digits of &lt;math&gt;N&lt;/math&gt; is &lt;math&gt;1+1+0 = \boxed{\textbf{(A) }2}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2019|num-b=12|num-a=14}}<br /> <br /> {{MAA Notice}}</div> Scrabbler94 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_25&diff=112183 2019 AMC 8 Problems/Problem 25 2019-11-25T22:52:10Z <p>Scrabbler94: /* Solution 1 */ improve clarity, rigor of solution 1</p> <hr /> <div>==Problem 25==<br /> Alice has &lt;math&gt;24&lt;/math&gt; apples. In how many ways can she share them with Becky and Chris so that each of the three people has at least two apples?<br /> <br /> ==Solution 1==<br /> We use [[stars and bars]]. The problem asks for the number of integer solutions &lt;math&gt;(a,b,c)&lt;/math&gt; such that &lt;math&gt;a+b+c = 24&lt;/math&gt; and &lt;math&gt;a,b,c \ge 2&lt;/math&gt;. We can subtract 2 from &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, &lt;math&gt;c&lt;/math&gt;, so that we equivalently seek the number of non-negative integer solutions to &lt;math&gt;a' + b' + c' = 18&lt;/math&gt;. By stars and bars (using 18 stars and 2 bars), the number of solutions is &lt;math&gt;\binom{18+2}{2} = \binom{20}{2} = \boxed{\textbf{(C) }190}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> Let's say you assume that Alice has 2 apples. There are 19 ways to split the rest of the apples with Becky and Chris. If Alice has 3 apples, there are 18 ways to split the rest of the apples with Becky and Chris. If Alice has 4 apples, there are 17 ways to split the rest. So the total number of ways to split 24 apples between the three friends is equal to 19 + 18 + 17...…… + 1 = 20 (19/2) = &lt;math&gt;\boxed{\textbf{(C)}\ 190}&lt;/math&gt;<br /> <br /> ~heeeeeeheeeeeee<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2019|num-b=24|after=Last Problem}}<br /> <br /> {{MAA Notice}}</div> Scrabbler94