https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Pieslinger&feedformat=atom AoPS Wiki - User contributions [en] 2021-04-20T20:16:09Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=1998_AHSME_Problems/Problem_24&diff=79940 1998 AHSME Problems/Problem 24 2016-08-10T14:14:22Z <p>Pieslinger: /* Solution */</p> <hr /> <div>== Problem ==<br /> Call a &lt;math&gt;7&lt;/math&gt;-digit telephone number &lt;math&gt;d_1d_2d_3-d_4d_5d_6d_7&lt;/math&gt; ''memorable'' if the prefix [[sequence]] &lt;math&gt;d_1d_2d_3&lt;/math&gt; is exactly the same as either of the sequences &lt;math&gt;d_4d_5d_6&lt;/math&gt; or &lt;math&gt;d_5d_6d_7&lt;/math&gt; (possibly both). Assuming that each &lt;math&gt;d_i&lt;/math&gt; can be any of the ten decimal digits &lt;math&gt;0,1,2, \ldots, 9&lt;/math&gt;, the number of different memorable telephone numbers is<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 19,810<br /> \qquad\mathrm{(B)}\ 19,910<br /> \qquad\mathrm{(C)}\ 19,990<br /> \qquad\mathrm{(D)}\ 20,000<br /> \qquad\mathrm{(E)}\ 20,100&lt;/math&gt;<br /> == Solution A ==<br /> <br /> In this problem, we only need to consider the digits &lt;math&gt;\overline{d_4d_5d_6d_7}&lt;/math&gt;. Each possibility of &lt;math&gt;\overline{d_4d_5d_6d_7}&lt;/math&gt; gives &lt;math&gt;2&lt;/math&gt; possibilities for &lt;math&gt;\overline{d_1d_2d_3}&lt;/math&gt;, which are &lt;math&gt;\overline{d_1d_2d_3}=\overline{d_4d_5d_6}&lt;/math&gt; and &lt;math&gt;\overline{d_1d_2d_3}=\overline{d_5d_6d_7}&lt;/math&gt; with the exception of the case of &lt;math&gt;d_4=d_5=d_6=d_7&lt;/math&gt;, which only gives one sequence. After accounting for overcounting, the answer is &lt;math&gt;(10 \times 10 \times 10 \times 10) \times 2 - 10=19990 \Rightarrow \boxed{\mathrm{(C)}}&lt;/math&gt;<br /> <br /> == Solution B ==<br /> Let &lt;math&gt;A&lt;/math&gt; represent the set of telephone numbers with &lt;math&gt;\overline{d_1d_2d_3} = \overline{d_4d_5d_6}&lt;/math&gt; (of which there are &lt;math&gt;1000&lt;/math&gt; possibilities for &lt;math&gt;\overline{d_1d_2d_3}&lt;/math&gt; and &lt;math&gt;10&lt;/math&gt; for &lt;math&gt;d_7&lt;/math&gt;), and &lt;math&gt;B&lt;/math&gt; those such that &lt;math&gt;\overline{d_1d_2d_3} = \overline{d_5d_6d_7}&lt;/math&gt;. Then &lt;math&gt;A \cap B&lt;/math&gt; (the telephone numbers that belong to both &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt;) is the set of telephone numbers such that &lt;math&gt;d_1 = d_2 = d_3 = d_4 = d_5 = d_6 = d_7&lt;/math&gt;, of which there are &lt;math&gt;10&lt;/math&gt; possibilities. By the [[Principle of Inclusion-Exclusion]],<br /> <br /> &lt;cmath&gt;|A \cup B| = |A| + |B| - |A \cap B| = 1000 \times 10 + 1000 \times 10 - 10 = 19990 \Rightarrow \mathrm{(C)}&lt;/cmath&gt;<br /> <br /> == See also ==<br /> {{AHSME box|year=1998|num-b=23|num-a=25}}<br /> <br /> [[Category:Introductory Combinatorics Problems]]<br /> {{MAA Notice}}</div> Pieslinger https://artofproblemsolving.com/wiki/index.php?title=1950_AHSME_Problems/Problem_23&diff=67079 1950 AHSME Problems/Problem 23 2015-01-28T00:33:10Z <p>Pieslinger: </p> <hr /> <div>==Problem==<br /> A man buys a house for &amp;#036;10,000 and rents it. He puts &lt;math&gt;12\frac{1}{2}\%&lt;/math&gt; of each month's rent aside for repairs and upkeep; pays &amp;#036;325 a year taxes and realizes &lt;math&gt;5\frac{1}{2}\%&lt;/math&gt; on his investment. The monthly rent (in dollars) is:<br /> <br /> &lt;math&gt; \textbf{(A)} \ \ 64.82\qquad\textbf{(B)} \ \ 83.33\qquad\textbf{(C)} \ \ 72.08\qquad\textbf{(D)} \ \ 45.83\qquad\textbf{(E)} \ \ 177.08 &lt;/math&gt;<br /> <br /> == Solution ==<br /> &lt;math&gt;12\frac{1}{2}\%&lt;/math&gt; is the same as &lt;math&gt;\frac{1}{8}&lt;/math&gt;, so the man sets one eighth of each month's rent aside, so he only gains &lt;math&gt;\frac{7}{8}&lt;/math&gt; of his rent. He also pays &amp;#036;325 each year, and he realizes &lt;math&gt;5.5\%&lt;/math&gt;, or &amp;#036;550, on his investment. Therefore he must have collected a total of &amp;#036;325 +&amp;#036;550 = &amp;#036;875 in rent. This was for the whole year, so he collected &lt;math&gt;\frac{875}{12}&lt;/math&gt; dollars each month as rent. This is only &lt;math&gt;\frac{7}{8}&lt;/math&gt; of the monthly rent, so the monthly rent in dollars is &lt;math&gt;\frac{875}{12}\cdot \frac{8}{7}=\boxed{\textbf{(B)}\ \ 83.33}&lt;/math&gt;<br /> <br /> == See Also ==<br /> {{AHSME 50p box|year=1950|num-b=22|num-a=24}}<br /> <br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Pieslinger https://artofproblemsolving.com/wiki/index.php?title=1950_AHSME_Problems/Problem_18&diff=67078 1950 AHSME Problems/Problem 18 2015-01-28T00:27:44Z <p>Pieslinger: </p> <hr /> <div>== Problem==<br /> <br /> Of the following <br /> <br /> (1) &lt;math&gt; a(x-y)=ax-ay &lt;/math&gt;<br /> <br /> (2) &lt;math&gt; a^{x-y}=a^x-a^y &lt;/math&gt;<br /> <br /> (3) &lt;math&gt; \log (x-y)=\log x-\log y &lt;/math&gt;<br /> <br /> (4) &lt;math&gt; \frac{\log x}{\log y}=\log{x}-\log{y} &lt;/math&gt;<br /> <br /> (5) &lt;math&gt; a(xy)=ax\times ay &lt;/math&gt;<br /> <br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ \text{Only 1 and 4 are true}\qquad\\ \textbf{(B)}\ \text{Only 1 and 5 are true}\qquad\\ \textbf{(C)}\ \text{Only 1 and 3 are true}\qquad\\ \textbf{(D)}\ \text{Only 1 and 2 are true}\qquad\\ \textbf{(E)}\ \text{Only 1 is true} &lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> The distributive property doesn't apply to logarithms or in the ways illustrated, and only applies to addition and subtraction. Also, &lt;math&gt;a^{x-y} = \frac{a^x}{a^y}&lt;/math&gt;, so &lt;math&gt;\textbf{(E)} \text{ Only 1 is true}&lt;/math&gt;<br /> ==See Also==<br /> <br /> {{AHSME 50p box|year=1950|num-b=17|num-a=19}}<br /> <br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Pieslinger https://artofproblemsolving.com/wiki/index.php?title=1950_AHSME_Problems/Problem_7&diff=67077 1950 AHSME Problems/Problem 7 2015-01-28T00:23:56Z <p>Pieslinger: </p> <hr /> <div>== Problem==<br /> <br /> If the digit &lt;math&gt;1&lt;/math&gt; is placed after a two digit number whose tens' digit is &lt;math&gt;t&lt;/math&gt;, and units' digit is &lt;math&gt;u&lt;/math&gt;, the new number is:<br /> <br /> &lt;math&gt; \textbf{(A)}\ 10t+u+1\qquad\textbf{(B)}\ 100t+10u+1\qquad\textbf{(C)}\ 1000t+10u+1\qquad\textbf{(D)}\ t+u+1\qquad\\ \textbf{(E)}\ \text{None of these answers} &lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> By placing the digit &lt;math&gt;1&lt;/math&gt; after a two digit number, you are changing the units place to &lt;math&gt;1&lt;/math&gt; and moving everything else up a place. Therefore the answer is &lt;math&gt;\boxed{\textbf{(B)}\ 100t+10u+1}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> <br /> {{AHSME 50p box|year=1950|num-b=6|num-a=8}}<br /> <br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Pieslinger https://artofproblemsolving.com/wiki/index.php?title=1986_AIME_Problems/Problem_8&diff=58880 1986 AIME Problems/Problem 8 2014-01-25T18:10:32Z <p>Pieslinger: /* Solution 1 */</p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt;S&lt;/math&gt; be the sum of the base &lt;math&gt;10&lt;/math&gt; [[logarithm]]s of all the [[proper divisor]]s (all [[divisor]]s of a number excluding itself) of &lt;math&gt;1000000&lt;/math&gt;. What is the integer nearest to &lt;math&gt;S&lt;/math&gt;?<br /> <br /> __TOC__<br /> == Solution ==<br /> === Solution 1 ===<br /> The [[prime factorization]] of &lt;math&gt;1000000 = 2^65^6&lt;/math&gt;, so there are &lt;math&gt;(6 + 1)(6 + 1) = 49&lt;/math&gt; divisors, of which &lt;math&gt;48&lt;/math&gt; are proper. The sum of multiple logarithms of the same base is equal to the logarithm of the products of the numbers. <br /> <br /> Writing out the first few terms, we see that the answer is equal to &lt;cmath&gt;\log 1 + \log 2 + \log 4 + \ldots + \log 1000000 = \log (2^05^0)(2^15^0)(2^25^0)\cdots (2^65^6).&lt;/cmath&gt; Each power of &lt;math&gt;2&lt;/math&gt; appears &lt;math&gt;7&lt;/math&gt; times; and the same goes for &lt;math&gt;5&lt;/math&gt;. So the overall power of &lt;math&gt;2&lt;/math&gt; and &lt;math&gt;5&lt;/math&gt; is &lt;math&gt;7(1+2+3+4+5+6) = 7 \cdot 21 = 147&lt;/math&gt;. However, since the question asks for proper divisors, we exclude &lt;math&gt;2^65^6&lt;/math&gt;, so each power is actually &lt;math&gt;141&lt;/math&gt; times. The answer is thus &lt;math&gt;S = \log 2^{141}5^{141} = \log 10^{141} = \boxed{141}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> Since the prime factorization of &lt;math&gt;10^6&lt;/math&gt; is &lt;math&gt;2^6 \cdot 5^6&lt;/math&gt;, the number of factors in &lt;math&gt;10^6&lt;/math&gt; is &lt;math&gt;7 \cdot 7=49&lt;/math&gt;. You can pair them up into groups of two so each group multiplies to &lt;math&gt;10^6&lt;/math&gt;. Note that &lt;math&gt;\log n+\log{(10^6/n)}=\log{n}+\log{10^6}-\log{n}=6&lt;/math&gt;. Thus, the sum of the logs of the divisors is half the number of divisors of &lt;math&gt;10^6 \cdot 6 -6&lt;/math&gt; (since they are asking only for proper divisors), and the answer is &lt;math&gt;(49/2)\cdot 6-6=141&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=1986|num-b=7|num-a=9}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Pieslinger https://artofproblemsolving.com/wiki/index.php?title=1986_AIME_Problems/Problem_8&diff=58879 1986 AIME Problems/Problem 8 2014-01-25T18:09:07Z <p>Pieslinger: /* Solution 1 */</p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt;S&lt;/math&gt; be the sum of the base &lt;math&gt;10&lt;/math&gt; [[logarithm]]s of all the [[proper divisor]]s (all [[divisor]]s of a number excluding itself) of &lt;math&gt;1000000&lt;/math&gt;. What is the integer nearest to &lt;math&gt;S&lt;/math&gt;?<br /> <br /> __TOC__<br /> == Solution ==<br /> === Solution 1 ===<br /> The [[prime factorization]] of &lt;math&gt;1000000 = 2^65^6&lt;/math&gt;, so there are &lt;math&gt;(6 + 1)(6 + 1) = 49&lt;/math&gt; divisors, of which &lt;math&gt;48&lt;/math&gt; are proper. The sum of multiple logarithms of the same base is equal to the logarithm of the products of the numbers. <br /> <br /> Writing out the first few terms, we see that the answer is equal to &lt;cmath&gt;\log 1 + \log 2 + \log 4 + \ldots + \log 1000000 = \log (2^05^0)(2^15^0)(2^25^0)\cdots(2^05^1)(2^15^1)\cdots (2^55^6)(2^65^6).&lt;/cmath&gt; Each power of &lt;math&gt;2&lt;/math&gt; appears &lt;math&gt;7&lt;/math&gt; times; and the same goes for &lt;math&gt;5&lt;/math&gt;. So the overall power of &lt;math&gt;2&lt;/math&gt; and &lt;math&gt;5&lt;/math&gt; is &lt;math&gt;7(1+2+3+4+5+6) = 7 \cdot 21 = 147&lt;/math&gt;. However, since the question asks for proper divisors, we exclude &lt;math&gt;2^65^6&lt;/math&gt;, so each power is actually &lt;math&gt;141&lt;/math&gt; times. The answer is thus &lt;math&gt;S = \log 2^{141}5^{141} = \log 10^{141} = \boxed{141}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> Since the prime factorization of &lt;math&gt;10^6&lt;/math&gt; is &lt;math&gt;2^6 \cdot 5^6&lt;/math&gt;, the number of factors in &lt;math&gt;10^6&lt;/math&gt; is &lt;math&gt;7 \cdot 7=49&lt;/math&gt;. You can pair them up into groups of two so each group multiplies to &lt;math&gt;10^6&lt;/math&gt;. Note that &lt;math&gt;\log n+\log{(10^6/n)}=\log{n}+\log{10^6}-\log{n}=6&lt;/math&gt;. Thus, the sum of the logs of the divisors is half the number of divisors of &lt;math&gt;10^6 \cdot 6 -6&lt;/math&gt; (since they are asking only for proper divisors), and the answer is &lt;math&gt;(49/2)\cdot 6-6=141&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=1986|num-b=7|num-a=9}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Pieslinger https://artofproblemsolving.com/wiki/index.php?title=2004_AMC_10B_Problems/Problem_11&diff=44887 2004 AMC 10B Problems/Problem 11 2012-02-19T18:07:22Z <p>Pieslinger: </p> <hr /> <div>== Problem ==<br /> <br /> Two eight-sided dice each have faces numbered 1 through 8. When the dice are rolled, each face has an equal probability of appearing on the top. What is the probability that the product of the two top numbers is greater than their sum?<br /> <br /> &lt;math&gt; \mathrm{(A) \ } \frac{1}{2} \qquad \mathrm{(B) \ } \frac{47}{64} \qquad \mathrm{(C) \ } \frac{3}{4} \qquad \mathrm{(D) \ } \frac{55}{64} \qquad \mathrm{(E) \ } \frac{7}{8} &lt;/math&gt;<br /> <br /> ==Solutions==<br /> ===Solution 1===<br /> We have &lt;math&gt;1\times n = n &lt; 1 + n&lt;/math&gt;, hence if at least one of the numbers is &lt;math&gt;1&lt;/math&gt;, the sum is larger. There &lt;math&gt;15&lt;/math&gt; such possibilities.<br /> <br /> We have &lt;math&gt;2\times 2 = 2+2&lt;/math&gt;.<br /> <br /> For &lt;math&gt;n&gt;2&lt;/math&gt; we already have &lt;math&gt;2\times n = n + n &gt; 2 + n&lt;/math&gt;, hence all other cases are good.<br /> <br /> Out of the &lt;math&gt;8\times 8&lt;/math&gt; possible cases, we found that in &lt;math&gt;15+1=16&lt;/math&gt; the sum is greater than or equal to the product, hence in &lt;math&gt;64-16=48&lt;/math&gt; cases the sum is smaller, satisfying the condition. Therefore the answer is &lt;math&gt;\frac{48}{64} = \boxed{\frac34}&lt;/math&gt;.<br /> <br /> ===Solution 2===<br /> Let the two rolls be &lt;math&gt;m&lt;/math&gt;, and &lt;math&gt;n&lt;/math&gt;. <br /> <br /> From the restriction: <br /> &lt;math&gt;mn &gt; m + n&lt;/math&gt;<br /> <br /> &lt;math&gt;mn - m - n &gt; 0&lt;/math&gt;<br /> <br /> &lt;math&gt;mn - m - n + 1 &gt; 1&lt;/math&gt;<br /> <br /> &lt;math&gt;(m-1)(n-1) &gt; 1&lt;/math&gt;<br /> <br /> Since &lt;math&gt;m-1&lt;/math&gt; and &lt;math&gt;n-1&lt;/math&gt; are non-negative integers between &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;8&lt;/math&gt;, either &lt;math&gt;(m-1)(n-1) = 0&lt;/math&gt;, &lt;math&gt;(m-1)(n-1) = 1&lt;/math&gt;, or &lt;math&gt;(m-1)(n-1) &gt; 1&lt;/math&gt;<br /> <br /> &lt;math&gt;(m-1)(n-1) = 0&lt;/math&gt; if and only if &lt;math&gt;m=1&lt;/math&gt; or &lt;math&gt;n=1&lt;/math&gt;. <br /> <br /> There are &lt;math&gt;8&lt;/math&gt; ordered pairs &lt;math&gt;(m,n)&lt;/math&gt; with &lt;math&gt;m=1&lt;/math&gt;, &lt;math&gt;8&lt;/math&gt; ordered pairs with &lt;math&gt;n=1&lt;/math&gt;, and &lt;math&gt;1&lt;/math&gt; ordered pair with &lt;math&gt;m=1&lt;/math&gt; and &lt;math&gt;n=1&lt;/math&gt;. So, there are &lt;math&gt;8+8-1 = 15&lt;/math&gt; ordered pairs &lt;math&gt;(m,n)&lt;/math&gt; such that &lt;math&gt;(m-1)(n-1) = 0&lt;/math&gt;. <br /> <br /> &lt;math&gt;(m-1)(n-1) = 1&lt;/math&gt; if and only if &lt;math&gt;m-1=1&lt;/math&gt; and &lt;math&gt;n-1=1&lt;/math&gt; or equivalently &lt;math&gt;m=2&lt;/math&gt; and &lt;math&gt;n=2&lt;/math&gt;. This gives &lt;math&gt;1&lt;/math&gt; ordered pair &lt;math&gt;(m,n) = (2,2)&lt;/math&gt;. <br /> <br /> So, there are a total of &lt;math&gt;15+1=16&lt;/math&gt; ordered pairs &lt;math&gt;(m,n)&lt;/math&gt; with &lt;math&gt;(m-1)(n-1) &lt; 1&lt;/math&gt;. <br /> <br /> Since there are a total of &lt;math&gt;8\cdot8 = 64&lt;/math&gt; ordered pairs &lt;math&gt;(m,n)&lt;/math&gt;, there are &lt;math&gt;64-16 = 48&lt;/math&gt; ordered pairs &lt;math&gt;(m,n)&lt;/math&gt; with &lt;math&gt;(m-1)(n-1) &gt; 1&lt;/math&gt;. <br /> <br /> Thus, the desired probability is &lt;math&gt;\frac{48}{64} = \frac{3}{4} \Rightarrow C&lt;/math&gt;.<br /> <br /> == See also ==<br /> <br /> {{AMC10 box|year=2004|ab=B|num-b=10|num-a=12}}</div> Pieslinger https://artofproblemsolving.com/wiki/index.php?title=2004_AMC_10B_Problems/Problem_11&diff=44886 2004 AMC 10B Problems/Problem 11 2012-02-19T18:07:00Z <p>Pieslinger: </p> <hr /> <div>== Problem ==<br /> <br /> Two eight-sided dice each have faces numbered 1 through 8. When the dice are rolled, each face has an equal probability of appearing on the top. What is the probability that the product of the two top numbers is greater than their sum?<br /> <br /> &lt;math&gt; \mathrm{(A) \ } \frac{1}{2} \qquad \mathrm{(B) \ } \frac{47}{64} \qquad \mathrm{(C) \ } \frac{3}{4} \qquad \mathrm{(D) \ } \frac{55}{64} \qquad \mathrm{(E) \ } \frac{7}{8} &lt;/math&gt;<br /> <br /> ==Solutions==<br /> ===Solution 1===<br /> We have &lt;math&gt;1\times n = n &lt; 1 + n&lt;/math&gt;, hence if at least one of the numbers is &lt;math&gt;1&lt;/math&gt;, the sum is larger. There &lt;math&gt;15&lt;/math&gt; such possibilities.<br /> <br /> We have &lt;math&gt;2\times 2 = 2+2&lt;/math&gt;.<br /> <br /> For &lt;math&gt;n&gt;2&lt;/math&gt; we already have &lt;math&gt;2\times n = n + n &gt; 2 + n&lt;/math&gt;, hence all other cases are good.<br /> <br /> Out of the &lt;math&gt;8\times 8&lt;/math&gt; possible cases, we found that in &lt;math&gt;15+1=16&lt;/math&gt; the sum is greater than or equal to the product, hence in &lt;math&gt;64-16=48&lt;/math&gt; the sum is smaller, satisfying the condition. Therefore the answer is &lt;math&gt;\frac{48}{64} = \boxed{\frac34}&lt;/math&gt;.<br /> <br /> ===Solution 2===<br /> Let the two rolls be &lt;math&gt;m&lt;/math&gt;, and &lt;math&gt;n&lt;/math&gt;. <br /> <br /> From the restriction: <br /> &lt;math&gt;mn &gt; m + n&lt;/math&gt;<br /> <br /> &lt;math&gt;mn - m - n &gt; 0&lt;/math&gt;<br /> <br /> &lt;math&gt;mn - m - n + 1 &gt; 1&lt;/math&gt;<br /> <br /> &lt;math&gt;(m-1)(n-1) &gt; 1&lt;/math&gt;<br /> <br /> Since &lt;math&gt;m-1&lt;/math&gt; and &lt;math&gt;n-1&lt;/math&gt; are non-negative integers between &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;8&lt;/math&gt;, either &lt;math&gt;(m-1)(n-1) = 0&lt;/math&gt;, &lt;math&gt;(m-1)(n-1) = 1&lt;/math&gt;, or &lt;math&gt;(m-1)(n-1) &gt; 1&lt;/math&gt;<br /> <br /> &lt;math&gt;(m-1)(n-1) = 0&lt;/math&gt; if and only if &lt;math&gt;m=1&lt;/math&gt; or &lt;math&gt;n=1&lt;/math&gt;. <br /> <br /> There are &lt;math&gt;8&lt;/math&gt; ordered pairs &lt;math&gt;(m,n)&lt;/math&gt; with &lt;math&gt;m=1&lt;/math&gt;, &lt;math&gt;8&lt;/math&gt; ordered pairs with &lt;math&gt;n=1&lt;/math&gt;, and &lt;math&gt;1&lt;/math&gt; ordered pair with &lt;math&gt;m=1&lt;/math&gt; and &lt;math&gt;n=1&lt;/math&gt;. So, there are &lt;math&gt;8+8-1 = 15&lt;/math&gt; ordered pairs &lt;math&gt;(m,n)&lt;/math&gt; such that &lt;math&gt;(m-1)(n-1) = 0&lt;/math&gt;. <br /> <br /> &lt;math&gt;(m-1)(n-1) = 1&lt;/math&gt; if and only if &lt;math&gt;m-1=1&lt;/math&gt; and &lt;math&gt;n-1=1&lt;/math&gt; or equivalently &lt;math&gt;m=2&lt;/math&gt; and &lt;math&gt;n=2&lt;/math&gt;. This gives &lt;math&gt;1&lt;/math&gt; ordered pair &lt;math&gt;(m,n) = (2,2)&lt;/math&gt;. <br /> <br /> So, there are a total of &lt;math&gt;15+1=16&lt;/math&gt; ordered pairs &lt;math&gt;(m,n)&lt;/math&gt; with &lt;math&gt;(m-1)(n-1) &lt; 1&lt;/math&gt;. <br /> <br /> Since there are a total of &lt;math&gt;8\cdot8 = 64&lt;/math&gt; ordered pairs &lt;math&gt;(m,n)&lt;/math&gt;, there are &lt;math&gt;64-16 = 48&lt;/math&gt; ordered pairs &lt;math&gt;(m,n)&lt;/math&gt; with &lt;math&gt;(m-1)(n-1) &gt; 1&lt;/math&gt;. <br /> <br /> Thus, the desired probability is &lt;math&gt;\frac{48}{64} = \frac{3}{4} \Rightarrow C&lt;/math&gt;.<br /> <br /> == See also ==<br /> <br /> {{AMC10 box|year=2004|ab=B|num-b=10|num-a=12}}</div> Pieslinger https://artofproblemsolving.com/wiki/index.php?title=2004_AMC_12B_Problems/Problem_6&diff=44885 2004 AMC 12B Problems/Problem 6 2012-02-19T17:54:02Z <p>Pieslinger: </p> <hr /> <div>{{duplicate|[[2004 AMC 12B Problems|2004 AMC 12B #6]] and [[2004 AMC 10B Problems|2004 AMC 10B #8]]}}<br /> <br /> == Problem ==<br /> Minneapolis-St. Paul International Airport is 8 miles southwest of downtown St. Paul and 10 miles southeast of downtown Minneapolis. Which of the following is closest to the number of miles between downtown St. Paul and downtown Minneapolis? <br /> <br /> &lt;math&gt;(\mathrm {A}) 13\qquad (\mathrm {B}) 14 \qquad (\mathrm {C}) 15 \qquad (\mathrm {D}) 16 \qquad (\mathrm {E}) 17&lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> The directions &quot;southwest&quot; and &quot;southeast&quot; are orthogonal. Thus the described situation is a right triangle with legs 8 miles and 10 miles long. The hypotenuse length is &lt;math&gt;\sqrt{8^2 + 10^2} \sim 12.8&lt;/math&gt;, and thus the answer is &lt;math&gt;\mathrm{(A)}&lt;/math&gt;.<br /> <br /> Without a calculator one can note that &lt;math&gt;8^2 + 10^2 = 164 &lt; 169 = 13^2\rightarrow\boxed{A}&lt;/math&gt;.<br /> <br /> == See Also ==<br /> <br /> {{AMC12 box|year=2004|ab=B|num-b=5|num-a=7}}<br /> {{AMC10 box|year=2004|ab=B|num-b=7|num-a=9}}</div> Pieslinger https://artofproblemsolving.com/wiki/index.php?title=2004_AMC_10B_Problems/Problem_3&diff=44884 2004 AMC 10B Problems/Problem 3 2012-02-19T17:51:28Z <p>Pieslinger: </p> <hr /> <div>==Problem==<br /> At each basketball practice last week, Jenny made twice as many free throws as she made at the previous practice. At her fifth practice she made 48 free throws. How many free throws did she make at the first practice?<br /> <br /> &lt;math&gt; \mathrm{(A) \ } 3 \qquad \mathrm{(B) \ } 6 \qquad \mathrm{(C) \ } 9 \qquad \mathrm{(D) \ } 12\qquad \mathrm{(E) \ } 15 &lt;/math&gt;<br /> ==Solution==<br /> <br /> At the fourth practice she made &lt;math&gt;48/2=24&lt;/math&gt; throws, at the third one it was &lt;math&gt;24/2=12&lt;/math&gt;, then we get &lt;math&gt;12/2=6&lt;/math&gt; throws for the second practice, and finally &lt;math&gt;6/2=3\Rightarrow\boxed{A}&lt;/math&gt; throws at the first one.<br /> <br /> == See also ==<br /> <br /> {{AMC10 box|year=2004|ab=B|num-b=2|num-a=4}}</div> Pieslinger https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_10A_Problems/Problem_13&diff=44470 2006 AMC 10A Problems/Problem 13 2012-02-05T19:28:38Z <p>Pieslinger: /* Solution */</p> <hr /> <div>== Problem ==<br /> A player pays $5 to play a game. A die is rolled. If the number on the die is [[odd integer | odd]], the game is lost. If the number on the die is [[even integer | even]], the die is rolled again. In this case the player wins if the second number matches the first and loses otherwise. How much should the player win if the game is fair? (In a fair game the [[probability]] of winning times the amount won is what the player should pay.) <br /> <br /> &lt;math&gt;\mathrm{(A) \ }$12\qquad\mathrm{(B) \ } $30\qquad\mathrm{(C) \ }$50\qquad\mathrm{(D) \ } $60\qquad\mathrm{(E) \ }$100\qquad&lt;/math&gt;<br /> == Solution ==<br /> <br /> The probability of rolling an even number on the first turn is &lt;math&gt;\frac{1}{2}&lt;/math&gt; and the probability of rolling the same number on the next turn is &lt;math&gt;\frac{1}{6}&lt;/math&gt;The probability of winning is &lt;math&gt;\frac{1}{12}&lt;/math&gt;. If the game is to be fair, the amount paid, 5 dollars, must be &lt;math&gt;\frac{1}{12}&lt;/math&gt; the amount of prize money, so the answer is<br /> '''D.''' 60<br /> <br /> == See also ==<br /> {{AMC10 box|year=2006|ab=A|num-b=12|num-a=14}}<br /> <br /> [[Category:Introductory Combinatorics Problems]]</div> Pieslinger https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_10A_Problems/Problem_13&diff=44469 2006 AMC 10A Problems/Problem 13 2012-02-05T19:26:27Z <p>Pieslinger: /* Solution */</p> <hr /> <div>== Problem ==<br /> A player pays $5 to play a game. A die is rolled. If the number on the die is [[odd integer | odd]], the game is lost. If the number on the die is [[even integer | even]], the die is rolled again. In this case the player wins if the second number matches the first and loses otherwise. How much should the player win if the game is fair? (In a fair game the [[probability]] of winning times the amount won is what the player should pay.) <br /> <br /> &lt;math&gt;\mathrm{(A) \ }$12\qquad\mathrm{(B) \ } $30\qquad\mathrm{(C) \ }$50\qquad\mathrm{(D) \ } $60\qquad\mathrm{(E) \ }$100\qquad&lt;/math&gt;<br /> == Solution ==<br /> The probability of rolling an even number on the first turn is &lt;math&gt;\frac{1}{2}&lt;/math&gt; and the probability of rolling the same number on the next turn is &lt;math&gt;\frac{1}{6}&lt;/math&gt;, so the probability of winning is &lt;math&gt;\frac{1}{12}&lt;/math&gt;. If the game is to be fair, the amount paid, 5 dollars, must be &lt;math&gt;\frac{1}{12}&lt;/math&gt; the amount of prize money, so the answer is '''D.''' 60<br /> <br /> == See also ==<br /> {{AMC10 box|year=2006|ab=A|num-b=12|num-a=14}}<br /> <br /> [[Category:Introductory Combinatorics Problems]]</div> Pieslinger https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_10B_Problems/Problem_23&diff=44465 2003 AMC 10B Problems/Problem 23 2012-02-05T18:14:37Z <p>Pieslinger: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> <br /> A regular octagon &lt;math&gt; ABCDEFGH &lt;/math&gt; has an area of one square unit. What is the area of the rectangle &lt;math&gt; ABEF &lt;/math&gt;?<br /> <br /> &lt;asy&gt; unitsize(1cm); defaultpen(linewidth(.8pt)+fontsize(8pt)); pair C=dir(22.5), B=dir(67.5), A=dir(112.5), H=dir(157.5), G=dir(202.5), F=dir(247.5), E=dir(292.5), D=dir(337.5); draw(A--B--C--D--E--F--G--H--cycle); label(&quot;$A$&quot;,A,NNW); label(&quot;$B$&quot;,B,NNE); label(&quot;$C$&quot;,C,ENE); label(&quot;$D$&quot;,D,ESE); label(&quot;$E$&quot;,E,SSE); label(&quot;$F$&quot;,F,SSW); label(&quot;$G$&quot;,G,WSW); label(&quot;$H$&quot;,H,WNW);&lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ 1-\frac{\sqrt2}{2}\qquad\textbf{(B)}\ \frac{\sqrt2}{4}\qquad\textbf{(C)}\ \sqrt2-1\qquad\textbf{(D)}\ \frac{1}2\qquad\textbf{(E)}\ \frac{1+\sqrt2}{4} &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> Here is an easy way to look at this, where &lt;math&gt;p&lt;/math&gt; is the perimeter, and &lt;math&gt;a&lt;/math&gt; is the [[apothem]]:<br /> <br /> Area of Octagon: &lt;math&gt; \frac{ap}{2}=1 &lt;/math&gt;.<br /> <br /> Area of Rectangle: &lt;math&gt; \frac{p}{8}\times 2a=\frac{ap}{4} &lt;/math&gt;.<br /> <br /> You can see from this that the octagon's area is twice as large as the rectangle's area is &lt;math&gt;\boxed{\textbf{(D)}\ \frac{1}{2}}&lt;/math&gt;.<br /> ==Solution 2==<br /> <br /> Here is a less complicated way than that of the user above. If you draw lines connecting opposite vertices and draw the rectangle ABEF, you can see that two of the triangles share the same base and height with half the rectangle. Therefore, the rectangle's area is the same as 4 of these triangles, and is &lt;math&gt;\boxed{\textbf{(D)}\ \frac{1}{2}}&lt;/math&gt; the area of the octagon<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2003|ab=B|num-b=22|num-a=24}}</div> Pieslinger https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_10B_Problems/Problem_23&diff=44464 2003 AMC 10B Problems/Problem 23 2012-02-05T18:13:06Z <p>Pieslinger: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> A regular octagon &lt;math&gt; ABCDEFGH &lt;/math&gt; has an area of one square unit. What is the area of the rectangle &lt;math&gt; ABEF &lt;/math&gt;?<br /> <br /> &lt;asy&gt; unitsize(1cm); defaultpen(linewidth(.8pt)+fontsize(8pt)); pair C=dir(22.5), B=dir(67.5), A=dir(112.5), H=dir(157.5), G=dir(202.5), F=dir(247.5), E=dir(292.5), D=dir(337.5); draw(A--B--C--D--E--F--G--H--cycle); label(&quot;$A$&quot;,A,NNW); label(&quot;$B$&quot;,B,NNE); label(&quot;$C$&quot;,C,ENE); label(&quot;$D$&quot;,D,ESE); label(&quot;$E$&quot;,E,SSE); label(&quot;$F$&quot;,F,SSW); label(&quot;$G$&quot;,G,WSW); label(&quot;$H$&quot;,H,WNW);&lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ 1-\frac{\sqrt2}{2}\qquad\textbf{(B)}\ \frac{\sqrt2}{4}\qquad\textbf{(C)}\ \sqrt2-1\qquad\textbf{(D)}\ \frac{1}2\qquad\textbf{(E)}\ \frac{1+\sqrt2}{4} &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> Here is an easy way to look at this, where &lt;math&gt;p&lt;/math&gt; is the perimeter, and &lt;math&gt;a&lt;/math&gt; is the [[apothem]]:<br /> <br /> Area of Octagon: &lt;math&gt; \frac{ap}{2}=1 &lt;/math&gt;.<br /> <br /> Area of Rectangle: &lt;math&gt; \frac{p}{8}\times 2a=\frac{ap}{4} &lt;/math&gt;.<br /> <br /> You can see from this that the octagon's area is twice as large as the rectangle's area is &lt;math&gt;\boxed{\textbf{(D)}\ \frac{1}{2}}&lt;/math&gt;.<br /> ==Solution 2==<br /> <br /> If you draw lines connecting opposite vertices and draw the rectangle ABEF, you can see that two of the triangles share the same base and height with half the rectangle. Therefore, the rectangle's area is the same as 4 of these triangles, and is &lt;math&gt;\boxed{\textbf{(D)}\ \frac{1}{2}}&lt;/math&gt; the area of the octagon<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2003|ab=B|num-b=22|num-a=24}}</div> Pieslinger https://artofproblemsolving.com/wiki/index.php?title=1990_AJHSME_Problems/Problem_19&diff=42311 1990 AJHSME Problems/Problem 19 2011-09-15T00:54:12Z <p>Pieslinger: </p> <hr /> <div>==Problem==<br /> <br /> There are &lt;math&gt;120&lt;/math&gt; seats in a row. What is the fewest number of seats that must be occupied so the next person to be seated must sit next to someone?<br /> <br /> &lt;math&gt;\text{(A)}\ 30 \qquad \text{(B)}\ 40 \qquad \text{(C)}\ 41 \qquad \text{(D)}\ 60 \qquad \text{(E)}\ 119&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> p is a person seated, o is an empty seat<br /> <br /> The pattern of seating that results in the fewest occupied seats is opoopoopoo...po<br /> we can group the seats in 3s<br /> opo opo opo ...opo<br /> <br /> there are a total of &lt;math&gt;\boxed{40}&lt;/math&gt; groups<br /> ==See Also==<br /> <br /> {{AJHSME box|year=1990|num-b=18|num-a=20}}<br /> [[Category:Introductory Combinatorics Problems]]</div> Pieslinger https://artofproblemsolving.com/wiki/index.php?title=AMC_12&diff=39080 AMC 12 2011-06-04T00:49:59Z <p>Pieslinger: </p> <hr /> <div>The '''American Mathematics Contest 12''' ('''AMC 12''') is the first exam in the series of exams used to challenge bright students, grades 12 and below, on the path toward choosing the team that represents the United States at the [[International Mathematics Olympiad]] (IMO).<br /> <br /> High scoring AMC 12 students are invited to take the more challenging [[American Invitational Mathematics Examination]] (AIME).<br /> <br /> The AMC 12 is administered by the [[American Mathematics Competitions]] (AMC). [[Art of Problem Solving]] (AoPS) is a proud sponsor of the AMC!<br /> <br /> The AMC 12 used to be the [[American High School Mathematics Examination]] from 1951 to 1999.<br /> <br /> <br /> == Format ==<br /> <br /> The AMC 12 is a 25 question, 75 minute multiple choice test. Problems generally increase in difficulty as the exam progresses. Ever since 2008, calculators have been banned from use during the test.<br /> <br /> The AMC 12 is scored in a way that penalizes guesses. Correct answers are worth 6 points, incorrect questions are worth 0 points, and unanswered questions are worth 1.5 points, to give a total score out of 150 points. From 2002 to 2006, the number of points for an unanswered question was 2.5 points and before 2002 it was 2 points. Students that score over 100 points or in the top 5% of the AMC 12 contest are invited to take the [[AIME]].[http://www.unl.edu/amc/e-exams/e7-aime/adminaime.html]<br /> <br /> == Curriculum ==<br /> The AMC 12 tests [[mathematical problem solving]] with [[arithmetic]], [[algebra]], [[counting]], [[geometry]], [[number theory]], and [[probability]] and other secondary school math topics. Problems are designed to be solvable by students without any background in calculus.<br /> <br /> == Resources ==<br /> === Links ===<br /> * [http://www.unl.edu/amc/ AMC homepage], their [http://www.unl.edu/amc/e-exams/e6-amc12/amc12.shtml AMC 12 page], and [http://www.unl.edu/amc/mathclub/index.html practice problems]<br /> * The [[AoPS]] [http://www.artofproblemsolving.com/Resources/AoPS_R_Contests_AMC12.php AMC 12 guide].<br /> * [http://www.artofproblemsolving.com/Forum/index.php?f=133 AMC Forum] for discussion of the AMC and problems from AMC exams.<br /> * The [http://www.artofproblemsolving.com/Forum/resources.php AoPS Contest Archive] includes problems and solutions from [http://www.artofproblemsolving.com/Forum/resources.php?c=182 past AMC exams].<br /> * [[AMC 12 Problems and Solutions]]<br /> <br /> === Recommended reading ===<br /> * [http://www.artofproblemsolving.com/Store/contests.php?contest=amc Problem and solution books for past AMC exams].<br /> <br /> * [http://www.artofproblemsolving.com/Store/viewitem.php?item=intro:algebra Introduction to Algebra] by [[Richard Rusczyk]]. <br /> * [http://www.artofproblemsolving.com/Store/viewitem.php?item=intro:counting Introduction to Counting &amp; Probability] by Dr. [[David Patrick]]. <br /> * [http://www.artofproblemsolving.com/Store/viewitem.php?item=intro:geometry Introduction to Geometry] by [[Richard Rusczyk]]. <br /> * [http://www.artofproblemsolving.com/Store/viewitem.php?item=intro:nt Introduction to Number Theory] by [[Mathew Crawford]].<br /> * [http://www.artofproblemsolving.com/Store/viewitem.php?item=interm:algebra Intermediate Algebra] by [[Richard Rusczyk]] and [[Mathew Crawford]].<br /> * [http://www.artofproblemsolving.com/Store/viewitem.php?item=interm:counting Intermediate Counting &amp; Probability] by Dr. [[David Patrick]]. <br /> * [http://www.artofproblemsolving.com/Store/viewitem.php?item=precalc Precalculus] by [[Richard Rusczyk]].<br /> * [http://www.artofproblemsolving.com/Store/viewitem.php?item=ps:aops1 The Art of Problem Solving Volume 1] by [[Sandor Lehoczky]] and [[Richard Rusczyk]].<br /> * [http://www.artofproblemsolving.com/Store/viewitem.php?item=ps:aops2 The Art of Problem Solving Volume 2] by [[Sandor Lehoczky]] and [[Richard Rusczyk]].<br /> <br /> === AMC Preparation Classes ===<br /> * [[AoPS]] hosts an [http://www.artofproblemsolving.com/School/index.php online school] teaching introductory and intermediate classes in topics covered by the AMC 12 as well as an AMC 12 preparation class.<br /> * [[AoPS]] holds many free [[Math Jams]], some of which are devoted to discussing problems on the AMC 10 and AMC 12. [http://www.artofproblemsolving.com/School/mathjams.php Math Jam Schedule]<br /> * [[EPGY]] offers an AMC 12 preparation class.<br /> <br /> == See also ==<br /> * [[Mathematics competitions]]<br /> * [[ARML]]<br /> * [[Mathematics summer programs]]<br /> <br /> <br /> <br /> [[Category:Mathematics competitions]]</div> Pieslinger https://artofproblemsolving.com/wiki/index.php?title=AMC_12&diff=39079 AMC 12 2011-06-04T00:49:10Z <p>Pieslinger: Undo revision 39078 by Pieslinger (talk)</p> <hr /> <div>The '''American Mathematics Contest 12''' ('''AMC 12''') is the first exam in the series of exams used to challenge bright students, grades 12 and below, on the path toward choosing the team that represents the United States at the [[International Mathematics Olympiad]] (IMO).<br /> <br /> High scoring AMC 12 students are invited to take the more challenging [[American Invitational Mathematics Examination]] (AIME).<br /> <br /> The AMC 12 is administered by the [[American Mathematics Competitions]] (AMC). [[Art of Problem Solving]] (AoPS) is a proud sponsor of the AMC!<br /> <br /> The AMC 12 used to be the [[American High School Mathematics Examination]] from 1951 to 1999.<br /> <br /> <br /> == Format ==<br /> <br /> The AMC 12 is a 25 question, 75 minute multiple choice test. Problems generally increase in difficulty as the exam progresses. Ever since 2008, calculators have been banned.<br /> <br /> The AMC 12 is scored in a way that penalizes guesses. Correct answers are worth 6 points, incorrect questions are worth 0 points, and unanswered questions are worth 1.5 points, to give a total score out of 150 points. From 2002 to 2006, the number of points for an unanswered question was 2.5 points and before 2002 it was 2 points. Students that score over 100 points or in the top 5% of the AMC 12 contest are invited to take the [[AIME]].[http://www.unl.edu/amc/e-exams/e7-aime/adminaime.html]<br /> <br /> == Curriculum ==<br /> The AMC 12 tests [[mathematical problem solving]] with [[arithmetic]], [[algebra]], [[counting]], [[geometry]], [[number theory]], and [[probability]] and other secondary school math topics. Problems are designed to be solvable by students without any background in calculus.<br /> <br /> == Resources ==<br /> === Links ===<br /> * [http://www.unl.edu/amc/ AMC homepage], their [http://www.unl.edu/amc/e-exams/e6-amc12/amc12.shtml AMC 12 page], and [http://www.unl.edu/amc/mathclub/index.html practice problems]<br /> * The [[AoPS]] [http://www.artofproblemsolving.com/Resources/AoPS_R_Contests_AMC12.php AMC 12 guide].<br /> * [http://www.artofproblemsolving.com/Forum/index.php?f=133 AMC Forum] for discussion of the AMC and problems from AMC exams.<br /> * The [http://www.artofproblemsolving.com/Forum/resources.php AoPS Contest Archive] includes problems and solutions from [http://www.artofproblemsolving.com/Forum/resources.php?c=182 past AMC exams].<br /> * [[AMC 12 Problems and Solutions]]<br /> <br /> === Recommended reading ===<br /> * [http://www.artofproblemsolving.com/Store/contests.php?contest=amc Problem and solution books for past AMC exams].<br /> <br /> * [http://www.artofproblemsolving.com/Store/viewitem.php?item=intro:algebra Introduction to Algebra] by [[Richard Rusczyk]]. <br /> * [http://www.artofproblemsolving.com/Store/viewitem.php?item=intro:counting Introduction to Counting &amp; Probability] by Dr. [[David Patrick]]. <br /> * [http://www.artofproblemsolving.com/Store/viewitem.php?item=intro:geometry Introduction to Geometry] by [[Richard Rusczyk]]. <br /> * [http://www.artofproblemsolving.com/Store/viewitem.php?item=intro:nt Introduction to Number Theory] by [[Mathew Crawford]].<br /> * [http://www.artofproblemsolving.com/Store/viewitem.php?item=interm:algebra Intermediate Algebra] by [[Richard Rusczyk]] and [[Mathew Crawford]].<br /> * [http://www.artofproblemsolving.com/Store/viewitem.php?item=interm:counting Intermediate Counting &amp; Probability] by Dr. [[David Patrick]]. <br /> * [http://www.artofproblemsolving.com/Store/viewitem.php?item=precalc Precalculus] by [[Richard Rusczyk]].<br /> * [http://www.artofproblemsolving.com/Store/viewitem.php?item=ps:aops1 The Art of Problem Solving Volume 1] by [[Sandor Lehoczky]] and [[Richard Rusczyk]].<br /> * [http://www.artofproblemsolving.com/Store/viewitem.php?item=ps:aops2 The Art of Problem Solving Volume 2] by [[Sandor Lehoczky]] and [[Richard Rusczyk]].<br /> <br /> === AMC Preparation Classes ===<br /> * [[AoPS]] hosts an [http://www.artofproblemsolving.com/School/index.php online school] teaching introductory and intermediate classes in topics covered by the AMC 12 as well as an AMC 12 preparation class.<br /> * [[AoPS]] holds many free [[Math Jams]], some of which are devoted to discussing problems on the AMC 10 and AMC 12. [http://www.artofproblemsolving.com/School/mathjams.php Math Jam Schedule]<br /> * [[EPGY]] offers an AMC 12 preparation class.<br /> <br /> == See also ==<br /> * [[Mathematics competitions]]<br /> * [[ARML]]<br /> * [[Mathematics summer programs]]<br /> <br /> <br /> <br /> [[Category:Mathematics competitions]]</div> Pieslinger https://artofproblemsolving.com/wiki/index.php?title=AMC_12&diff=39078 AMC 12 2011-06-04T00:48:24Z <p>Pieslinger: </p> <hr /> <div>The '''American Mathematics Contest 12''' ('''AMC 12''') is the first exam in the series of exams used to challenge bright students, grades 12 and below, on the path toward choosing the team that represents the United States at the [[International Mathematics Olympiad]] (IMO).<br /> <br /> High scoring AMC 12 students are invited to take the more challenging [[American Invitational Mathematics Examination]] (AIME).<br /> <br /> The AMC 12 is administered by the [[American Mathematics Competitions]] (AMC). [[Art of Problem Solving]] (AoPS) is a proud sponsor of the AMC!<br /> <br /> The AMC 12 used to be the [[American High School Mathematics Examination]] from 1951 to 1999.<br /> <br /> <br /> == Format ==<br /> <br /> The AMC 12 is a 25 question, 75 minute multiple choice test. Problems generally increase in difficulty as the exam progresses. Ever since 2008, calculators have been banned during the test.<br /> <br /> The AMC 12 is scored in a way that penalizes guesses. Correct answers are worth 6 points, incorrect questions are worth 0 points, and unanswered questions are worth 1.5 points, to give a total score out of 150 points. From 2002 to 2006, the number of points for an unanswered question was 2.5 points and before 2002 it was 2 points. Students that score over 100 points or in the top 5% of the AMC 12 contest are invited to take the [[AIME]].[http://www.unl.edu/amc/e-exams/e7-aime/adminaime.html]<br /> <br /> == Curriculum ==<br /> The AMC 12 tests [[mathematical problem solving]] with [[arithmetic]], [[algebra]], [[counting]], [[geometry]], [[number theory]], and [[probability]] and other secondary school math topics. Problems are designed to be solvable by students without any background in calculus.<br /> <br /> == Resources ==<br /> === Links ===<br /> * [http://www.unl.edu/amc/ AMC homepage], their [http://www.unl.edu/amc/e-exams/e6-amc12/amc12.shtml AMC 12 page], and [http://www.unl.edu/amc/mathclub/index.html practice problems]<br /> * The [[AoPS]] [http://www.artofproblemsolving.com/Resources/AoPS_R_Contests_AMC12.php AMC 12 guide].<br /> * [http://www.artofproblemsolving.com/Forum/index.php?f=133 AMC Forum] for discussion of the AMC and problems from AMC exams.<br /> * The [http://www.artofproblemsolving.com/Forum/resources.php AoPS Contest Archive] includes problems and solutions from [http://www.artofproblemsolving.com/Forum/resources.php?c=182 past AMC exams].<br /> * [[AMC 12 Problems and Solutions]]<br /> <br /> === Recommended reading ===<br /> * [http://www.artofproblemsolving.com/Store/contests.php?contest=amc Problem and solution books for past AMC exams].<br /> <br /> * [http://www.artofproblemsolving.com/Store/viewitem.php?item=intro:algebra Introduction to Algebra] by [[Richard Rusczyk]]. <br /> * [http://www.artofproblemsolving.com/Store/viewitem.php?item=intro:counting Introduction to Counting &amp; Probability] by Dr. [[David Patrick]]. <br /> * [http://www.artofproblemsolving.com/Store/viewitem.php?item=intro:geometry Introduction to Geometry] by [[Richard Rusczyk]]. <br /> * [http://www.artofproblemsolving.com/Store/viewitem.php?item=intro:nt Introduction to Number Theory] by [[Mathew Crawford]].<br /> * [http://www.artofproblemsolving.com/Store/viewitem.php?item=interm:algebra Intermediate Algebra] by [[Richard Rusczyk]] and [[Mathew Crawford]].<br /> * [http://www.artofproblemsolving.com/Store/viewitem.php?item=interm:counting Intermediate Counting &amp; Probability] by Dr. [[David Patrick]]. <br /> * [http://www.artofproblemsolving.com/Store/viewitem.php?item=precalc Precalculus] by [[Richard Rusczyk]].<br /> * [http://www.artofproblemsolving.com/Store/viewitem.php?item=ps:aops1 The Art of Problem Solving Volume 1] by [[Sandor Lehoczky]] and [[Richard Rusczyk]].<br /> * [http://www.artofproblemsolving.com/Store/viewitem.php?item=ps:aops2 The Art of Problem Solving Volume 2] by [[Sandor Lehoczky]] and [[Richard Rusczyk]].<br /> <br /> === AMC Preparation Classes ===<br /> * [[AoPS]] hosts an [http://www.artofproblemsolving.com/School/index.php online school] teaching introductory and intermediate classes in topics covered by the AMC 12 as well as an AMC 12 preparation class.<br /> * [[AoPS]] holds many free [[Math Jams]], some of which are devoted to discussing problems on the AMC 10 and AMC 12. [http://www.artofproblemsolving.com/School/mathjams.php Math Jam Schedule]<br /> * [[EPGY]] offers an AMC 12 preparation class.<br /> <br /> == See also ==<br /> * [[Mathematics competitions]]<br /> * [[ARML]]<br /> * [[Mathematics summer programs]]<br /> <br /> <br /> <br /> [[Category:Mathematics competitions]]</div> Pieslinger https://artofproblemsolving.com/wiki/index.php?title=AMC_12&diff=39077 AMC 12 2011-06-04T00:48:00Z <p>Pieslinger: </p> <hr /> <div>The '''American Mathematics Contest 12''' ('''AMC 12''') is the first exam in the series of exams used to challenge bright students, grades 12 and below, on the path toward choosing the team that represents the United States at the [[International Mathematics Olympiad]] (IMO).<br /> <br /> High scoring AMC 12 students are invited to take the more challenging [[American Invitational Mathematics Examination]] (AIME).<br /> <br /> The AMC 12 is administered by the [[American Mathematics Competitions]] (AMC). [[Art of Problem Solving]] (AoPS) is a proud sponsor of the AMC!<br /> <br /> The AMC 12 used to be the [[American High School Mathematics Examination]] from 1951 to 1999.<br /> <br /> <br /> == Format ==<br /> <br /> The AMC 12 is a 25 question, 75 minute multiple choice test. Problems generally increase in difficulty as the exam progresses. Ever since 2008, calculators have been banned.<br /> <br /> The AMC 12 is scored in a way that penalizes guesses. Correct answers are worth 6 points, incorrect questions are worth 0 points, and unanswered questions are worth 1.5 points, to give a total score out of 150 points. From 2002 to 2006, the number of points for an unanswered question was 2.5 points and before 2002 it was 2 points. Students that score over 100 points or in the top 5% of the AMC 12 contest are invited to take the [[AIME]].[http://www.unl.edu/amc/e-exams/e7-aime/adminaime.html]<br /> <br /> == Curriculum ==<br /> The AMC 12 tests [[mathematical problem solving]] with [[arithmetic]], [[algebra]], [[counting]], [[geometry]], [[number theory]], and [[probability]] and other secondary school math topics. Problems are designed to be solvable by students without any background in calculus.<br /> <br /> == Resources ==<br /> === Links ===<br /> * [http://www.unl.edu/amc/ AMC homepage], their [http://www.unl.edu/amc/e-exams/e6-amc12/amc12.shtml AMC 12 page], and [http://www.unl.edu/amc/mathclub/index.html practice problems]<br /> * The [[AoPS]] [http://www.artofproblemsolving.com/Resources/AoPS_R_Contests_AMC12.php AMC 12 guide].<br /> * [http://www.artofproblemsolving.com/Forum/index.php?f=133 AMC Forum] for discussion of the AMC and problems from AMC exams.<br /> * The [http://www.artofproblemsolving.com/Forum/resources.php AoPS Contest Archive] includes problems and solutions from [http://www.artofproblemsolving.com/Forum/resources.php?c=182 past AMC exams].<br /> * [[AMC 12 Problems and Solutions]]<br /> <br /> === Recommended reading ===<br /> * [http://www.artofproblemsolving.com/Store/contests.php?contest=amc Problem and solution books for past AMC exams].<br /> <br /> * [http://www.artofproblemsolving.com/Store/viewitem.php?item=intro:algebra Introduction to Algebra] by [[Richard Rusczyk]]. <br /> * [http://www.artofproblemsolving.com/Store/viewitem.php?item=intro:counting Introduction to Counting &amp; Probability] by Dr. [[David Patrick]]. <br /> * [http://www.artofproblemsolving.com/Store/viewitem.php?item=intro:geometry Introduction to Geometry] by [[Richard Rusczyk]]. <br /> * [http://www.artofproblemsolving.com/Store/viewitem.php?item=intro:nt Introduction to Number Theory] by [[Mathew Crawford]].<br /> * [http://www.artofproblemsolving.com/Store/viewitem.php?item=interm:algebra Intermediate Algebra] by [[Richard Rusczyk]] and [[Mathew Crawford]].<br /> * [http://www.artofproblemsolving.com/Store/viewitem.php?item=interm:counting Intermediate Counting &amp; Probability] by Dr. [[David Patrick]]. <br /> * [http://www.artofproblemsolving.com/Store/viewitem.php?item=precalc Precalculus] by [[Richard Rusczyk]].<br /> * [http://www.artofproblemsolving.com/Store/viewitem.php?item=ps:aops1 The Art of Problem Solving Volume 1] by [[Sandor Lehoczky]] and [[Richard Rusczyk]].<br /> * [http://www.artofproblemsolving.com/Store/viewitem.php?item=ps:aops2 The Art of Problem Solving Volume 2] by [[Sandor Lehoczky]] and [[Richard Rusczyk]].<br /> <br /> === AMC Preparation Classes ===<br /> * [[AoPS]] hosts an [http://www.artofproblemsolving.com/School/index.php online school] teaching introductory and intermediate classes in topics covered by the AMC 12 as well as an AMC 12 preparation class.<br /> * [[AoPS]] holds many free [[Math Jams]], some of which are devoted to discussing problems on the AMC 10 and AMC 12. [http://www.artofproblemsolving.com/School/mathjams.php Math Jam Schedule]<br /> * [[EPGY]] offers an AMC 12 preparation class.<br /> <br /> == See also ==<br /> * [[Mathematics competitions]]<br /> * [[ARML]]<br /> * [[Mathematics summer programs]]<br /> <br /> <br /> <br /> [[Category:Mathematics competitions]]</div> Pieslinger