https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Bunny1&feedformat=atom AoPS Wiki - User contributions [en] 2021-09-23T18:25:40Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10B_Problems/Problem_20&diff=141991 2017 AMC 10B Problems/Problem 20 2021-01-13T01:59:25Z <p>Bunny1: </p> <hr /> <div>==Problem==<br /> The number &lt;math&gt;21!=51,090,942,171,709,440,000&lt;/math&gt; has over &lt;math&gt;60,000&lt;/math&gt; positive integer divisors. One of them is chosen at random. What is the probability that it is odd?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{1}{21}\qquad\textbf{(B)}\ \frac{1}{19}\qquad\textbf{(C)}\ \frac{1}{18}\qquad\textbf{(D)}\ \frac{1}{2}\qquad\textbf{(E)}\ \frac{11}{21}&lt;/math&gt;<br /> ==Solution 1==<br /> We note that the only thing that affects the parity of the factor are the powers of 2. There are &lt;math&gt;10+5+2+1 = 18&lt;/math&gt; factors of 2 in the number. Thus, there are &lt;math&gt;18&lt;/math&gt; cases in which a factor of &lt;math&gt;21!&lt;/math&gt; would be even (have a factor of &lt;math&gt;2&lt;/math&gt; in its prime factorization), and &lt;math&gt;1&lt;/math&gt; case in which a factor of &lt;math&gt;21!&lt;/math&gt; would be odd. Therefore, the answer is &lt;math&gt;\boxed{\textbf{(B)} \frac 1{19}}&lt;/math&gt;<br /> <br /> Note from Williamgolly: To see why symmetry occurs here, we group the factors of 21!into 2 groups, one with powers of 2 and the others odd factors. For each power of 2, the factors combine a certain number of 2's from the first group and numbers from the odd group. That is why symmetry occurs here.<br /> <br /> ==Solution 2: Constructive counting==<br /> Consider how to construct any divisor &lt;math&gt;D&lt;/math&gt; of &lt;math&gt;21!&lt;/math&gt;. First by Legendre's theorem for the divisors of a factorial (see here: [[Legendre's Formula]]), we have that there are a total of 18 factors of 2 in the number. &lt;math&gt;D&lt;/math&gt; can take up either 0, 1, 2, 3,..., or all 18 factors of 2, for a total of 19 possible cases. In order for &lt;math&gt;D&lt;/math&gt; to be odd, however, it must have 0 factors of 2, meaning that there is a probability of 1 case/19 cases= &lt;math&gt;\boxed{1/19, \space \text{B}}&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2017|ab=B|num-b=19|num-a=21}}<br /> {{AMC12 box|year=2017|ab=B|num-b=15|num-a=17}}<br /> {{MAA Notice}}</div> Bunny1 https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10B_Problems/Problem_20&diff=141990 2017 AMC 10B Problems/Problem 20 2021-01-13T01:58:49Z <p>Bunny1: </p> <hr /> <div>==Problem==<br /> The number &lt;math&gt;21!=51,090,942,171,709,440,000&lt;/math&gt; has over &lt;math&gt;60,000&lt;/math&gt; positive integer divisors. One of them is chosen at random. What is the probability that it is odd?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{1}{21}\qquad\textbf{(B)}\ \frac{1}{19}\qquad\textbf{(C)}\ \frac{1}{18}\qquad\textbf{(D)}\ \frac{1}{2}\qquad\textbf{(E)}\ \frac{11}{21}&lt;/math&gt;<br /> ==Solution 1==<br /> We note that the only thing that affects the parity of the factor are the powers of 2. There are &lt;math&gt;10+5+2+1 = 18&lt;/math&gt; factors of 2 in the number. Thus, there are &lt;math&gt;18&lt;/math&gt; cases in which a factor of &lt;math&gt;21!&lt;/math&gt; would be even (have a factor of &lt;math&gt;2&lt;/math&gt; in its prime factorization), and &lt;math&gt;1&lt;/math&gt; case in which a factor of &lt;math&gt;21!&lt;/math&gt; would be odd. Therefore, the answer is &lt;math&gt;\boxed{\textbf{(B)} \frac 1{19}}&lt;/math&gt;<br /> <br /> Note from Williamgolly: To see why symmetry occurs here, we group the factors of 21!into 2 groups, one with powers of 2 and the others odd factors. For each power of 2, the factors combine a certain number of 2's from the first group and numbers from the odd group. That is why symmetry occurs here.<br /> <br /> ==Solution 2: Constructive counting==<br /> Consider how to construct any divisor &lt;math&gt;D&lt;/math&gt; of &lt;math&gt;21!&lt;/math&gt;. First by Legendre's theorem for the divisors of a factorial (see here: http://www.cut-the-knot.oIlicktoesrg/blue/LegendresTheorem.shtml and here: [[Legendre's Formula]]), we have that there are a total of 18 factors of 2 in the number. &lt;math&gt;D&lt;/math&gt; can take up either 0, 1, 2, 3,..., or all 18 factors of 2, for a total of 19 possible cases. In order for &lt;math&gt;D&lt;/math&gt; to be odd, however, it must have 0 factors of 2, meaning that there is a probability of 1 case/19 cases= &lt;math&gt;\boxed{1/19, \space \text{B}}&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2017|ab=B|num-b=19|num-a=21}}<br /> {{AMC12 box|year=2017|ab=B|num-b=15|num-a=17}}<br /> {{MAA Notice}}</div> Bunny1 https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_20&diff=141267 2018 AMC 10B Problems/Problem 20 2021-01-01T19:09:39Z <p>Bunny1: </p> <hr /> <div>{{duplicate|[[2018 AMC 12B Problems|2018 AMC 12B #18]] and [[2018 AMC 10B Problems|2018 AMC 10B #20]]}}<br /> <br /> ==Problem==<br /> <br /> A function &lt;math&gt;f&lt;/math&gt; is defined recursively by &lt;math&gt;f(1)=f(2)=1&lt;/math&gt; and &lt;cmath&gt;f(n)=f(n-1)-f(n-2)+n&lt;/cmath&gt;for all integers &lt;math&gt;n \geq 3&lt;/math&gt;. What is &lt;math&gt;f(2018)&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)} \text{ 2016} \qquad \textbf{(B)} \text{ 2017} \qquad \textbf{(C)} \text{ 2018} \qquad \textbf{(D)} \text{ 2019} \qquad \textbf{(E)} \text{ 2020}&lt;/math&gt;<br /> <br /> ==Solution 1 (A Bit Bashy)==<br /> Start out by listing some terms of the sequence. <br /> &lt;cmath&gt;f(1)=1&lt;/cmath&gt;<br /> &lt;cmath&gt;f(2)=1&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;f(3)=3&lt;/cmath&gt;<br /> &lt;cmath&gt;f(4)=6&lt;/cmath&gt;<br /> &lt;cmath&gt;f(5)=8&lt;/cmath&gt;<br /> &lt;cmath&gt;f(6)=8&lt;/cmath&gt;<br /> &lt;cmath&gt;f(7)=7&lt;/cmath&gt;<br /> &lt;cmath&gt;f(8)=7&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;f(9)=9&lt;/cmath&gt;<br /> &lt;cmath&gt;f(10)=12&lt;/cmath&gt;<br /> &lt;cmath&gt;f(11)=14&lt;/cmath&gt;<br /> &lt;cmath&gt;f(12)=14&lt;/cmath&gt;<br /> &lt;cmath&gt;f(13)=13&lt;/cmath&gt;<br /> &lt;cmath&gt;f(14)=13&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;f(15)=15&lt;/cmath&gt;<br /> &lt;cmath&gt;.....&lt;/cmath&gt;<br /> Notice that &lt;math&gt;f(n)=n&lt;/math&gt; whenever &lt;math&gt;n&lt;/math&gt; is an odd multiple of &lt;math&gt;3&lt;/math&gt;, and the pattern of numbers that follow will always be &lt;math&gt;+2&lt;/math&gt;, &lt;math&gt;+3&lt;/math&gt;, &lt;math&gt;+2&lt;/math&gt;, &lt;math&gt;+0&lt;/math&gt;, &lt;math&gt;-1&lt;/math&gt;, &lt;math&gt;+0&lt;/math&gt;.<br /> The largest odd multiple of &lt;math&gt;3&lt;/math&gt; smaller than &lt;math&gt;2018&lt;/math&gt; is &lt;math&gt;2013&lt;/math&gt;, so we have<br /> &lt;cmath&gt;f(2013)=2013&lt;/cmath&gt;<br /> &lt;cmath&gt;f(2014)=2016&lt;/cmath&gt;<br /> &lt;cmath&gt;f(2015)=2018&lt;/cmath&gt;<br /> &lt;cmath&gt;f(2016)=2018&lt;/cmath&gt;<br /> &lt;cmath&gt;f(2017)=2017&lt;/cmath&gt;<br /> &lt;cmath&gt;f(2018)=\boxed{(B) 2017}.&lt;/cmath&gt;<br /> minor edits by bunny1<br /> <br /> ==Solution 2 (Bashy Pattern Finding)==<br /> Writing out the first few values, we get:<br /> &lt;math&gt;1,1,3,6,8,8,7,7,9,12,14,14,13,13,15,18,20,20,19,19...&lt;/math&gt;. Examining, we see that every number &lt;math&gt;x&lt;/math&gt; where &lt;math&gt;x \equiv 1\pmod 6&lt;/math&gt; has &lt;math&gt;f(x)=x&lt;/math&gt;, &lt;math&gt;f(x+1)=f(x)=x&lt;/math&gt;, and &lt;math&gt;f(x-1)=f(x-2)=x+1&lt;/math&gt;. The greatest number that's &lt;math&gt;1\pmod{6}&lt;/math&gt; and less &lt;math&gt;2018&lt;/math&gt; is &lt;math&gt;2017&lt;/math&gt;, so we have &lt;math&gt;f(2017)=f(2018)=2017.&lt;/math&gt; &lt;math&gt;\boxed B&lt;/math&gt;<br /> <br /> ==Solution 3 (Algebra)==<br /> &lt;cmath&gt;f(n)=f(n-1)-f(n-2)+n.&lt;/cmath&gt;<br /> &lt;cmath&gt;f(n-1)=f(n-2)-f(n-3)+n-1.&lt;/cmath&gt;<br /> Adding the two equations, we have that &lt;cmath&gt;f(n)=2n-1-f(n-3).&lt;/cmath&gt; <br /> Hence, &lt;math&gt;f(n)+f(n-3)=2n-1&lt;/math&gt;.<br /> After plugging in &lt;math&gt;n-3&lt;/math&gt; to the equation above and doing some algebra, we have that &lt;math&gt;f(n)-f(n-6)=6&lt;/math&gt;.<br /> Consequently,<br /> &lt;cmath&gt;f(2018)-f(2012)=6.&lt;/cmath&gt;<br /> &lt;cmath&gt;f(2012)-f(2006)=6.&lt;/cmath&gt;<br /> &lt;cmath&gt;\ldots&lt;/cmath&gt;<br /> &lt;cmath&gt;f(8)-f(2)=6.&lt;/cmath&gt;<br /> Adding these &lt;math&gt;336&lt;/math&gt; equations up, we have that &lt;math&gt;f(2018)-f(2)=6 \cdot 336&lt;/math&gt; and &lt;math&gt;f(2018)=\boxed{2017}&lt;/math&gt;.<br /> <br /> ~AopsUser101<br /> <br /> ==Video Solution==<br /> <br /> https://www.youtube.com/watch?v=aubDsjVFFTc<br /> <br /> ~bunny1<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2018|ab=B|num-b=19|num-a=21}}<br /> {{AMC12 box|year=2018|ab=B|num-b=17|num-a=19}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Introductory Algebra Problems]]</div> Bunny1 https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_20&diff=141266 2018 AMC 10B Problems/Problem 20 2021-01-01T19:08:57Z <p>Bunny1: </p> <hr /> <div>{{duplicate|[[2018 AMC 12B Problems|2018 AMC 12B #18]] and [[2018 AMC 10B Problems|2018 AMC 10B #20]]}}<br /> <br /> ==Problem==<br /> <br /> A function &lt;math&gt;f&lt;/math&gt; is defined recursively by &lt;math&gt;f(1)=f(2)=1&lt;/math&gt; and &lt;cmath&gt;f(n)=f(n-1)-f(n-2)+n&lt;/cmath&gt;for all integers &lt;math&gt;n \geq 3&lt;/math&gt;. What is &lt;math&gt;f(2018)&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)} \text{ 2016} \qquad \textbf{(B)} \text{ 2017} \qquad \textbf{(C)} \text{ 2018} \qquad \textbf{(D)} \text{ 2019} \qquad \textbf{(E)} \text{ 2020}&lt;/math&gt;<br /> <br /> ==Solution 1 (A Bit Bashy)==<br /> Start out by listing some terms of the sequence. <br /> &lt;cmath&gt;f(1)=1&lt;/cmath&gt;<br /> &lt;cmath&gt;f(2)=1&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;f(3)=3&lt;/cmath&gt;<br /> &lt;cmath&gt;f(4)=6&lt;/cmath&gt;<br /> &lt;cmath&gt;f(5)=8&lt;/cmath&gt;<br /> &lt;cmath&gt;f(6)=8&lt;/cmath&gt;<br /> &lt;cmath&gt;f(7)=7&lt;/cmath&gt;<br /> &lt;cmath&gt;f(8)=7&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;f(9)=9&lt;/cmath&gt;<br /> &lt;cmath&gt;f(10)=12&lt;/cmath&gt;<br /> &lt;cmath&gt;f(11)=14&lt;/cmath&gt;<br /> &lt;cmath&gt;f(12)=14&lt;/cmath&gt;<br /> &lt;cmath&gt;f(13)=13&lt;/cmath&gt;<br /> &lt;cmath&gt;f(14)=13&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;f(15)=15&lt;/cmath&gt;<br /> &lt;cmath&gt;.....&lt;/cmath&gt;<br /> Notice that &lt;math&gt;f(n)=n&lt;/math&gt; whenever &lt;math&gt;n&lt;/math&gt; is an odd multiple of &lt;math&gt;3&lt;/math&gt;, and the pattern of numbers that follow will always be &lt;math&gt;+2&lt;/math&gt;, &lt;math&gt;+3&lt;/math&gt;, &lt;math&gt;+2&lt;/math&gt;, &lt;math&gt;+0&lt;/math&gt;, &lt;math&gt;-1&lt;/math&gt;, &lt;math&gt;+0&lt;/math&gt;.<br /> The largest odd multiple of &lt;math&gt;3&lt;/math&gt; smaller than &lt;math&gt;2018&lt;/math&gt; is &lt;math&gt;2013&lt;/math&gt;, so we have<br /> &lt;cmath&gt;f(2013)=2013&lt;/cmath&gt;<br /> &lt;cmath&gt;f(2014)=2016&lt;/cmath&gt;<br /> &lt;cmath&gt;f(2015)=2018&lt;/cmath&gt;<br /> &lt;cmath&gt;f(2016)=2018&lt;/cmath&gt;<br /> &lt;cmath&gt;f(2017)=2017&lt;/cmath&gt;<br /> &lt;cmath&gt;f(2018)=\boxed{(B) 2017}.&lt;/cmath&gt;<br /> <br /> ==Solution 2 (Bashy Pattern Finding)==<br /> Writing out the first few values, we get:<br /> &lt;math&gt;1,1,3,6,8,8,7,7,9,12,14,14,13,13,15,18,20,20,19,19...&lt;/math&gt;. Examining, we see that every number &lt;math&gt;x&lt;/math&gt; where &lt;math&gt;x \equiv 1\pmod 6&lt;/math&gt; has &lt;math&gt;f(x)=x&lt;/math&gt;, &lt;math&gt;f(x+1)=f(x)=x&lt;/math&gt;, and &lt;math&gt;f(x-1)=f(x-2)=x+1&lt;/math&gt;. The greatest number that's &lt;math&gt;1\pmod{6}&lt;/math&gt; and less &lt;math&gt;2018&lt;/math&gt; is &lt;math&gt;2017&lt;/math&gt;, so we have &lt;math&gt;f(2017)=f(2018)=2017.&lt;/math&gt; &lt;math&gt;\boxed B&lt;/math&gt;<br /> <br /> ==Solution 3 (Algebra)==<br /> &lt;cmath&gt;f(n)=f(n-1)-f(n-2)+n.&lt;/cmath&gt;<br /> &lt;cmath&gt;f(n-1)=f(n-2)-f(n-3)+n-1.&lt;/cmath&gt;<br /> Adding the two equations, we have that &lt;cmath&gt;f(n)=2n-1-f(n-3).&lt;/cmath&gt; <br /> Hence, &lt;math&gt;f(n)+f(n-3)=2n-1&lt;/math&gt;.<br /> After plugging in &lt;math&gt;n-3&lt;/math&gt; to the equation above and doing some algebra, we have that &lt;math&gt;f(n)-f(n-6)=6&lt;/math&gt;.<br /> Consequently,<br /> &lt;cmath&gt;f(2018)-f(2012)=6.&lt;/cmath&gt;<br /> &lt;cmath&gt;f(2012)-f(2006)=6.&lt;/cmath&gt;<br /> &lt;cmath&gt;\ldots&lt;/cmath&gt;<br /> &lt;cmath&gt;f(8)-f(2)=6.&lt;/cmath&gt;<br /> Adding these &lt;math&gt;336&lt;/math&gt; equations up, we have that &lt;math&gt;f(2018)-f(2)=6 \cdot 336&lt;/math&gt; and &lt;math&gt;f(2018)=\boxed{2017}&lt;/math&gt;.<br /> <br /> ~AopsUser101<br /> <br /> ==Video Solution==<br /> <br /> https://www.youtube.com/watch?v=aubDsjVFFTc<br /> <br /> ~bunny1<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2018|ab=B|num-b=19|num-a=21}}<br /> {{AMC12 box|year=2018|ab=B|num-b=17|num-a=19}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Introductory Algebra Problems]]</div> Bunny1 https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_23&diff=141230 2018 AMC 10B Problems/Problem 23 2021-01-01T02:02:15Z <p>Bunny1: </p> <hr /> <div>==Problem==<br /> <br /> How many ordered pairs &lt;math&gt;(a, b)&lt;/math&gt; of positive integers satisfy the equation <br /> &lt;cmath&gt;a\cdot b + 63 = 20\cdot \text{lcm}(a, b) + 12\cdot\text{gcd}(a,b),&lt;/cmath&gt;<br /> where &lt;math&gt;\text{gcd}(a,b)&lt;/math&gt; denotes the greatest common divisor of &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;\text{lcm}(a,b)&lt;/math&gt; denotes their least common multiple?<br /> <br /> &lt;math&gt;\textbf{(A)} \text{ 0} \qquad \textbf{(B)} \text{ 2} \qquad \textbf{(C)} \text{ 4} \qquad \textbf{(D)} \text{ 6} \qquad \textbf{(E)} \text{ 8}&lt;/math&gt;<br /> <br /> <br /> ==Solution==<br /> Let &lt;math&gt;x = lcm(a, b)&lt;/math&gt;, and &lt;math&gt;y = gcd(a, b)&lt;/math&gt;. Therefore, &lt;math&gt;a\cdot b = lcm(a, b)\cdot gcd(a, b) = x\cdot y&lt;/math&gt;. Thus, the equation becomes<br /> <br /> &lt;cmath&gt;x\cdot y + 63 = 20x + 12y&lt;/cmath&gt;<br /> &lt;cmath&gt;x\cdot y - 20x - 12y + 63 = 0&lt;/cmath&gt;<br /> <br /> Using [[Simon's Favorite Factoring Trick]], we rewrite this equation as<br /> <br /> &lt;cmath&gt;(x - 12)(y - 20) - 240 + 63 = 0&lt;/cmath&gt;<br /> &lt;cmath&gt;(x - 12)(y - 20) = 177&lt;/cmath&gt;<br /> <br /> Since &lt;math&gt;177 = 3\cdot 59&lt;/math&gt; and &lt;math&gt;x &gt; y&lt;/math&gt;, we have &lt;math&gt;x - 12 = 59&lt;/math&gt; and &lt;math&gt;y - 20 = 3&lt;/math&gt;, or &lt;math&gt;x - 12 = 177&lt;/math&gt; and &lt;math&gt;y - 20 = 1&lt;/math&gt;. This gives us the solutions &lt;math&gt;(71, 23)&lt;/math&gt; and &lt;math&gt;(189, 21)&lt;/math&gt;. Since the &lt;math&gt;\text{GCD}&lt;/math&gt; must be a divisor of the &lt;math&gt;\text{LCM}&lt;/math&gt;, the first pair does not work. Assume &lt;math&gt;a&gt;b&lt;/math&gt;. We must have &lt;math&gt;a = 21 \cdot 9&lt;/math&gt; and &lt;math&gt;b = 21&lt;/math&gt;, and we could then have &lt;math&gt;a&lt;b&lt;/math&gt;, so there are &lt;math&gt;\boxed{2}&lt;/math&gt; solutions.<br /> (awesomeag)<br /> <br /> Edited by IronicNinja, Firebolt360, and mprincess0229~<br /> <br /> ==Video Solution==<br /> <br /> https://www.youtube.com/watch?v=JWGHYUeOx-k<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2018|ab=B|num-b=22|num-a=24}}<br /> {{MAA Notice}}</div> Bunny1 https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_23&diff=141229 2018 AMC 10B Problems/Problem 23 2021-01-01T02:02:00Z <p>Bunny1: </p> <hr /> <div>==Problem==<br /> <br /> How many ordered pairs &lt;math&gt;(a, b)&lt;/math&gt; of positive integers satisfy the equation <br /> &lt;cmath&gt;a\cdot b + 63 = 20\cdot \text{lcm}(a, b) + 12\cdot\text{gcd}(a,b),&lt;/cmath&gt;<br /> where &lt;math&gt;\text{gcd}(a,b)&lt;/math&gt; denotes the greatest common divisor of &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;\text{lcm}(a,b)&lt;/math&gt; denotes their least common multiple?<br /> <br /> &lt;math&gt;\textbf{(A)} \text{ 0} \qquad \textbf{(B)} \text{ 2} \qquad \textbf{(C)} \text{ 4} \qquad \textbf{(D)} \text{ 6} \qquad \textbf{(E)} \text{ 8}&lt;/math&gt;<br /> <br /> <br /> ==Solution==<br /> Let &lt;math&gt;x = lcm(a, b)&lt;/math&gt;, and &lt;math&gt;y = gcd(a, b)&lt;/math&gt;. Therefore, &lt;math&gt;a\cdot b = lcm(a, b)\cdot gcd(a, b) = x\cdot y&lt;/math&gt;. Thus, the equation becomes<br /> <br /> &lt;cmath&gt;x\cdot y + 63 = 20x + 12y&lt;/cmath&gt;<br /> &lt;cmath&gt;x\cdot y - 20x - 12y + 63 = 0&lt;/cmath&gt;<br /> <br /> Using [[Simon's Favorite Factoring Trick]], we rewrite this equation as<br /> <br /> &lt;cmath&gt;(x - 12)(y - 20) - 240 + 63 = 0&lt;/cmath&gt;<br /> &lt;cmath&gt;(x - 12)(y - 20) = 177&lt;/cmath&gt;<br /> <br /> Since &lt;math&gt;177 = 3\cdot 59&lt;/math&gt; and &lt;math&gt;x &gt; y&lt;/math&gt;, we have &lt;math&gt;x - 12 = 59&lt;/math&gt; and &lt;math&gt;y - 20 = 3&lt;/math&gt;, or &lt;math&gt;x - 12 = 177&lt;/math&gt; and &lt;math&gt;y - 20 = 1&lt;/math&gt;. This gives us the solutions &lt;math&gt;(71, 23)&lt;/math&gt; and &lt;math&gt;(189, 21)&lt;/math&gt;. Since the &lt;math&gt;\text{GCD}&lt;/math&gt; must be a divisor of the &lt;math&gt;\text{LCM}&lt;/math&gt;, the first pair does not work. Assume &lt;math&gt;a&gt;b&lt;/math&gt;. We must have &lt;math&gt;a = 21 \cdot 9&lt;/math&gt; and &lt;math&gt;b = 21&lt;/math&gt;, and we could then have &lt;math&gt;a&lt;b&lt;/math&gt;, so there are &lt;math&gt;\boxed{2}&lt;/math&gt; solutions.<br /> (awesomeag)<br /> <br /> Edited by IronicNinja, Firebolt360, and mprincess0229~<br /> <br /> ==Video Solution 1==<br /> <br /> https://www.youtube.com/watch?v=JWGHYUeOx-k<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2018|ab=B|num-b=22|num-a=24}}<br /> {{MAA Notice}}</div> Bunny1 https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_21&diff=141228 2018 AMC 10B Problems/Problem 21 2021-01-01T01:59:27Z <p>Bunny1: </p> <hr /> <div>{{duplicate|[[2018 AMC 12B Problems|2018 AMC 12B #19]] and [[2018 AMC 10B Problems|2018 AMC 10B #21]]}}<br /> <br /> ==Problem==<br /> Mary chose an even &lt;math&gt;4&lt;/math&gt;-digit number &lt;math&gt;n&lt;/math&gt;. She wrote down all the divisors of &lt;math&gt;n&lt;/math&gt; in increasing order from left to right: &lt;math&gt;1,2,...,\dfrac{n}{2},n&lt;/math&gt;. At some moment Mary wrote &lt;math&gt;323&lt;/math&gt; as a divisor of &lt;math&gt;n&lt;/math&gt;. What is the smallest possible value of the next divisor written to the right of &lt;math&gt;323&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 324 \qquad \textbf{(B) } 330 \qquad \textbf{(C) } 340 \qquad \textbf{(D) } 361 \qquad \textbf{(E) } 646&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> Since prime factorizing &lt;math&gt;323&lt;/math&gt; gives you &lt;math&gt;17 \cdot 19&lt;/math&gt;, the desired answer needs to be a multiple of &lt;math&gt;17&lt;/math&gt; or &lt;math&gt;19&lt;/math&gt;, this is because if it is not a multiple of &lt;math&gt;17&lt;/math&gt; or &lt;math&gt;19&lt;/math&gt;, &lt;math&gt;n&lt;/math&gt; will be more than a &lt;math&gt;4&lt;/math&gt; digit number. For example, if the answer were to instead be &lt;math&gt;324&lt;/math&gt;, &lt;math&gt;n&lt;/math&gt; would have to be a multiple of &lt;math&gt;2^2 * 3^4 * 17 * 19&lt;/math&gt; for both &lt;math&gt;323&lt;/math&gt; and &lt;math&gt;324&lt;/math&gt; to be a valid factor, meaning &lt;math&gt;n&lt;/math&gt; would have to be at least &lt;math&gt;104652&lt;/math&gt;, which is too big. Looking at the answer choices, &lt;math&gt;\text{(A) }324&lt;/math&gt; and &lt;math&gt;\text{(B) }330&lt;/math&gt; are both not a multiple of neither 17 nor 19, &lt;math&gt;\text{(C) }340&lt;/math&gt; is divisible by &lt;math&gt;17&lt;/math&gt;. &lt;math&gt;\text{(D) }361&lt;/math&gt; is divisible by &lt;math&gt;19&lt;/math&gt;, and &lt;math&gt;\text{(E) }646&lt;/math&gt; is divisible by both &lt;math&gt;17&lt;/math&gt; and &lt;math&gt;19&lt;/math&gt;. Since &lt;math&gt;\boxed{\text{(C) }340}&lt;/math&gt; is the smallest number divisible by either &lt;math&gt;17&lt;/math&gt; or &lt;math&gt;19&lt;/math&gt; it is the answer. Checking, we can see that &lt;math&gt;n&lt;/math&gt; would be &lt;math&gt;6460&lt;/math&gt;, a four-digit number. Note that &lt;math&gt;n&lt;/math&gt; is also divisible by &lt;math&gt;2&lt;/math&gt;, one of the listed divisors of &lt;math&gt;n&lt;/math&gt;. (If &lt;math&gt;n&lt;/math&gt; was not divisible by &lt;math&gt;2&lt;/math&gt;, we would need to look for a different divisor)<br /> <br /> -Edited by Mathandski<br /> <br /> ==Solution 2==<br /> Let the next largest divisor be &lt;math&gt;k&lt;/math&gt;. Suppose &lt;math&gt;\gcd(k,323)=1&lt;/math&gt;. Then, as &lt;math&gt;323|n, k|n&lt;/math&gt;, therefore, &lt;math&gt;323\cdot k|n.&lt;/math&gt; However, because &lt;math&gt;k&gt;323&lt;/math&gt;, &lt;math&gt;323k&gt;323\cdot 324&gt;9999&lt;/math&gt;. Therefore, &lt;math&gt;\gcd(k,323)&gt;1&lt;/math&gt;. Note that &lt;math&gt;323=17\cdot 19&lt;/math&gt;. Therefore, the smallest the GCD can be is &lt;math&gt;17&lt;/math&gt; and our answer is &lt;math&gt;323+17=\boxed{\text{(C) }340}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> Again, recognize &lt;math&gt;323=17 \cdot 19&lt;/math&gt;. The 4-digit number is even, so its prime factorization must then be &lt;math&gt;17 \cdot 19 \cdot 2 \cdot n&lt;/math&gt;. Also, &lt;math&gt;1000\leq 646n \leq 9998&lt;/math&gt;, so &lt;math&gt;2 \leq n \leq 15&lt;/math&gt;. Since &lt;math&gt;15 \cdot 2=30&lt;/math&gt;, the prime factorization of the number after &lt;math&gt;323&lt;/math&gt; needs to have either &lt;math&gt;17&lt;/math&gt; or &lt;math&gt;19&lt;/math&gt;. The next highest product after &lt;math&gt;17 \cdot 19&lt;/math&gt; is &lt;math&gt;17 \cdot 2 \cdot 10 =340&lt;/math&gt; or &lt;math&gt;19 \cdot 2 \cdot 9 =342&lt;/math&gt; &lt;math&gt;\implies \boxed{\text{(C) }340}&lt;/math&gt;. <br /> <br /> <br /> You can also tell by inspection that &lt;math&gt;19\cdot18 &gt; 20\cdot17&lt;/math&gt;, because &lt;math&gt;19\cdot18&lt;/math&gt; is closer to the side lengths of a square, which maximizes the product.<br /> <br /> ~bjhhar<br /> <br /> ==Video Solution 1==<br /> <br /> https://www.youtube.com/watch?v=qlHE_sAXiY8<br /> <br /> ==Video Solution 2==<br /> <br /> https://www.youtube.com/watch?v=KHaLXNAkDWE<br /> <br /> ==Video Solution 3==<br /> <br /> https://www.youtube.com/watch?v=vc1FHO9YYKQ<br /> <br /> ~bunny1<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2018|ab=B|num-b=20|num-a=22}}<br /> {{AMC12 box|year=2018|ab=B|num-b=18|num-a=20}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Intermediate Number Theory Problems]]</div> Bunny1 https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_20&diff=141227 2018 AMC 10B Problems/Problem 20 2021-01-01T01:57:16Z <p>Bunny1: </p> <hr /> <div>{{duplicate|[[2018 AMC 12B Problems|2018 AMC 12B #18]] and [[2018 AMC 10B Problems|2018 AMC 10B #20]]}}<br /> <br /> ==Problem==<br /> <br /> A function &lt;math&gt;f&lt;/math&gt; is defined recursively by &lt;math&gt;f(1)=f(2)=1&lt;/math&gt; and &lt;cmath&gt;f(n)=f(n-1)-f(n-2)+n&lt;/cmath&gt;for all integers &lt;math&gt;n \geq 3&lt;/math&gt;. What is &lt;math&gt;f(2018)&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)} \text{ 2016} \qquad \textbf{(B)} \text{ 2017} \qquad \textbf{(C)} \text{ 2018} \qquad \textbf{(D)} \text{ 2019} \qquad \textbf{(E)} \text{ 2020}&lt;/math&gt;<br /> <br /> ==Solution 1 (A Bit Bashy)==<br /> Start out by listing some terms of the sequence. <br /> &lt;cmath&gt;f(1)=1&lt;/cmath&gt;<br /> &lt;cmath&gt;f(2)=1&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;f(3)=3&lt;/cmath&gt;<br /> &lt;cmath&gt;f(4)=6&lt;/cmath&gt;<br /> &lt;cmath&gt;f(5)=8&lt;/cmath&gt;<br /> &lt;cmath&gt;f(6)=8&lt;/cmath&gt;<br /> &lt;cmath&gt;f(7)=7&lt;/cmath&gt;<br /> &lt;cmath&gt;f(8)=7&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;f(9)=9&lt;/cmath&gt;<br /> &lt;cmath&gt;f(10)=12&lt;/cmath&gt;<br /> &lt;cmath&gt;f(11)=14&lt;/cmath&gt;<br /> &lt;cmath&gt;f(12)=14&lt;/cmath&gt;<br /> &lt;cmath&gt;f(13)=13&lt;/cmath&gt;<br /> &lt;cmath&gt;f(14)=13&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;f(15)=15&lt;/cmath&gt;<br /> &lt;cmath&gt;.....&lt;/cmath&gt;<br /> Notice that &lt;math&gt;f(n)=n&lt;/math&gt; whenever &lt;math&gt;n&lt;/math&gt; is an odd multiple of &lt;math&gt;3&lt;/math&gt;, and the pattern of numbers that follow will always be &lt;math&gt;+3&lt;/math&gt;, &lt;math&gt;+2&lt;/math&gt;, &lt;math&gt;+0&lt;/math&gt;, &lt;math&gt;-1&lt;/math&gt;, &lt;math&gt;+0&lt;/math&gt;.<br /> The largest odd multiple of &lt;math&gt;3&lt;/math&gt; smaller than &lt;math&gt;2018&lt;/math&gt; is &lt;math&gt;2013&lt;/math&gt;, so we have<br /> &lt;cmath&gt;f(2013)=2013&lt;/cmath&gt;<br /> &lt;cmath&gt;f(2014)=2016&lt;/cmath&gt;<br /> &lt;cmath&gt;f(2015)=2018&lt;/cmath&gt;<br /> &lt;cmath&gt;f(2016)=2018&lt;/cmath&gt;<br /> &lt;cmath&gt;f(2017)=2017&lt;/cmath&gt;<br /> &lt;cmath&gt;f(2018)=\boxed{(B) 2017}.&lt;/cmath&gt;<br /> <br /> ==Solution 2 (Bashy Pattern Finding)==<br /> Writing out the first few values, we get:<br /> &lt;math&gt;1,1,3,6,8,8,7,7,9,12,14,14,13,13,15,18,20,20,19,19...&lt;/math&gt;. Examining, we see that every number &lt;math&gt;x&lt;/math&gt; where &lt;math&gt;x \equiv 1\pmod 6&lt;/math&gt; has &lt;math&gt;f(x)=x&lt;/math&gt;, &lt;math&gt;f(x+1)=f(x)=x&lt;/math&gt;, and &lt;math&gt;f(x-1)=f(x-2)=x+1&lt;/math&gt;. The greatest number that's &lt;math&gt;1\pmod{6}&lt;/math&gt; and less &lt;math&gt;2018&lt;/math&gt; is &lt;math&gt;2017&lt;/math&gt;, so we have &lt;math&gt;f(2017)=f(2018)=2017.&lt;/math&gt; &lt;math&gt;\boxed B&lt;/math&gt;<br /> <br /> ==Solution 3 (Algebra)==<br /> &lt;cmath&gt;f(n)=f(n-1)-f(n-2)+n.&lt;/cmath&gt;<br /> &lt;cmath&gt;f(n-1)=f(n-2)-f(n-3)+n-1.&lt;/cmath&gt;<br /> Adding the two equations, we have that &lt;cmath&gt;f(n)=2n-1-f(n-3).&lt;/cmath&gt; <br /> Hence, &lt;math&gt;f(n)+f(n-3)=2n-1&lt;/math&gt;.<br /> After plugging in &lt;math&gt;n-3&lt;/math&gt; to the equation above and doing some algebra, we have that &lt;math&gt;f(n)-f(n-6)=6&lt;/math&gt;.<br /> Consequently,<br /> &lt;cmath&gt;f(2018)-f(2012)=6.&lt;/cmath&gt;<br /> &lt;cmath&gt;f(2012)-f(2006)=6.&lt;/cmath&gt;<br /> &lt;cmath&gt;\ldots&lt;/cmath&gt;<br /> &lt;cmath&gt;f(8)-f(2)=6.&lt;/cmath&gt;<br /> Adding these &lt;math&gt;336&lt;/math&gt; equations up, we have that &lt;math&gt;f(2018)-f(2)=6 \cdot 336&lt;/math&gt; and &lt;math&gt;f(2018)=\boxed{2017}&lt;/math&gt;.<br /> <br /> ~AopsUser101<br /> <br /> ==Video Solution==<br /> <br /> https://www.youtube.com/watch?v=aubDsjVFFTc<br /> <br /> ~bunny1<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2018|ab=B|num-b=19|num-a=21}}<br /> {{AMC12 box|year=2018|ab=B|num-b=17|num-a=19}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Introductory Algebra Problems]]</div> Bunny1 https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_16&diff=141226 2018 AMC 10B Problems/Problem 16 2021-01-01T01:17:28Z <p>Bunny1: </p> <hr /> <div>==Problem==<br /> <br /> Let &lt;math&gt;a_1,a_2,\dots,a_{2018}&lt;/math&gt; be a strictly increasing sequence of positive integers such that &lt;cmath&gt;a_1+a_2+\cdots+a_{2018}=2018^{2018}.&lt;/cmath&gt;<br /> What is the remainder when &lt;math&gt;a_1^3+a_2^3+\cdots+a_{2018}^3&lt;/math&gt; is divided by &lt;math&gt;6&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 /> ==Solution 1==<br /> <br /> One could simply list out all the residues to the third power &lt;math&gt;\mod 6&lt;/math&gt;. (Edit: Euler's totient theorem is not a valid approach to showing that they are all congruent &lt;math&gt;\mod 6&lt;/math&gt;. This is due to the fact that &lt;math&gt;a_k&lt;/math&gt; need not be relatively prime to &lt;math&gt;6&lt;/math&gt;.)<br /> <br /> Therefore the answer is congruent to &lt;math&gt;2018^{2018}\equiv 2^{2018} \pmod{6} = \boxed{ (E)4}&lt;/math&gt;<br /> <br /> Note from Williamgolly: We can WLOG assume &lt;math&gt;a_1,a_2... a_{2017} \equiv 0 \pmod 6&lt;/math&gt; and have &lt;math&gt;a_{2018} \equiv 2 \pmod 6&lt;/math&gt; to make life easier.<br /> <br /> ==Solution 2==<br /> <br /> Note that &lt;math&gt;\left(a_1+a_2+\cdots+a_{2018}\right)^3=a_1^3+a_2^3+\cdots+a_{2018}^3+3a_1^2\left(a_1+a_2+\cdots+a_{2018}-a_1\right)+3a_2^2\left(a_1+a_2+\cdots+a_{2018}-a_2\right)+\cdots+3a_{2018}^2\left(a_1+a_2+\cdots+a_{2018}-a_{2018}\right)+6\sum_{i\neq j\neq k}^{2018} a_ia_ja_k&lt;/math&gt;<br /> <br /> Note that &lt;math&gt;<br /> a_1^3+a_2^3+\cdots+a_{2018}^3+3a_1^2\left(a_1+a_2+\cdots+a_{2018}-a_1\right)+3a_2^2\left(a_1+a_2+\cdots+a_{2018}-a_2\right)+\cdots+3a_{2018}^2\left(a_1+a_2+\cdots+a_{2018}-a_{2018}\right)+6\sum_{i\neq j\neq k}^{2018} a_ia_ja_k\equiv a_1^3+a_2^3+\cdots+a_{2018}^3+3a_1^2({2018}^{2018}-a_1)+3a_2^2({2018}^{2018}-a_2)+\cdots+3a_{2018}^2({2018}^{2018}-a_{2018})<br /> \equiv -2(a_1^3+a_2^3+\cdots+a_{2018}^3)\pmod 6<br /> &lt;/math&gt;<br /> Therefore, &lt;math&gt;-2(a_1^3+a_2^3+\cdots+a_{2018}^3)\equiv \left(2018^{2018}\right)^3\equiv\left( 2^{2018}\right)^3\equiv 4^3\equiv 4\pmod{6}&lt;/math&gt;.<br /> <br /> Thus, &lt;math&gt;a_1^3+a_2^3+\cdots+a_{2018}^3\equiv 1\pmod 3&lt;/math&gt;. However, since cubing preserves parity, and the sum of the individual terms is even, the some of the cubes is also even, and our answer is &lt;math&gt;\boxed{\text{(E) }4}&lt;/math&gt;<br /> <br /> ==Solution 3 (Partial Proof)==<br /> First, we can assume that the problem will have a consistent answer for all possible values of &lt;math&gt;a_1&lt;/math&gt;. For the purpose of this solution, we will assume that &lt;math&gt;a_1 = 1&lt;/math&gt;.<br /> <br /> We first note that &lt;math&gt;1^3+2^3+...+n^3 = (1+2+...+n)^2&lt;/math&gt;. So what we are trying to find is what &lt;math&gt;\left(2018^{2018}\right)^2=\left(2018^{4036}\right)&lt;/math&gt; mod &lt;math&gt;6&lt;/math&gt;. We start by noting that &lt;math&gt;2018&lt;/math&gt; is congruent to &lt;math&gt;2 \pmod{6}&lt;/math&gt;. So we are trying to find &lt;math&gt;\left(2^{4036}\right) \pmod{6}&lt;/math&gt;. Instead of trying to do this with some number theory skills, we could just look for a pattern. We start with small powers of &lt;math&gt;2&lt;/math&gt; and see that &lt;math&gt;2^1&lt;/math&gt; is &lt;math&gt;2&lt;/math&gt; mod &lt;math&gt;6&lt;/math&gt;, &lt;math&gt;2^2&lt;/math&gt; is &lt;math&gt;4&lt;/math&gt; mod &lt;math&gt;6&lt;/math&gt;, &lt;math&gt;2^3&lt;/math&gt; is &lt;math&gt;2&lt;/math&gt; mod &lt;math&gt;6&lt;/math&gt;, &lt;math&gt;2^4&lt;/math&gt; is &lt;math&gt;4&lt;/math&gt; mod &lt;math&gt;6&lt;/math&gt;, and so on... So we see that since &lt;math&gt;\left(2^{4036}\right)&lt;/math&gt; has an even power, it must be congruent to &lt;math&gt;4 \pmod{6}&lt;/math&gt;, thus giving our answer &lt;math&gt;\boxed{\text{(E) }4}&lt;/math&gt;. You can prove this pattern using mods. But I thought this was easier.<br /> <br /> -TheMagician<br /> <br /> ==Solution 4 (Lazy solution)==<br /> First, we can assume that the problem will have a consistent answer for all possible values of &lt;math&gt;a_1&lt;/math&gt;. For the purpose of this solution, assume &lt;math&gt;a_1, a_2, ... a_{2017}&lt;/math&gt; are multiples of 6 and find &lt;math&gt;2018^{2018} \pmod{6}&lt;/math&gt; (which happens to be &lt;math&gt;4&lt;/math&gt;). Then &lt;math&gt;{a_1}^3 + ... + {a_{2018}}^3&lt;/math&gt; is congruent to &lt;math&gt;64 \pmod{6}&lt;/math&gt; or just &lt;math&gt;4&lt;/math&gt;. <br /> <br /> -Patrick4President<br /> <br /> ==Solution 5 (Nichomauss' Theorem)==<br /> <br /> Seeing the cubes of numbers, we think of Nichomauss's theorem, which states that &lt;math&gt;(a_1^3 + a_2^3 + ... + a_n^3) = (a_1 + a_2 + ... + a_n)^2&lt;/math&gt;. We can do this and deduce that &lt;math&gt;(a_1^3 + a_2^3 + ... + a_{2018}^3) = 2018^{2018}&lt;/math&gt; squared. <br /> <br /> Now, we find &lt;math&gt;2018\mod 6&lt;/math&gt;, which is 2. This means that we need to find &lt;math&gt;2^{2018} \mod {6}&lt;/math&gt;, which we can find using a pattern to be &lt;math&gt;4&lt;/math&gt;. Therefore, the answer is &lt;math&gt;4^2\mod 6&lt;/math&gt;, which is congruent to &lt;math&gt;\boxed{\textbf{(E)} 4}&lt;/math&gt;<br /> <br /> -ericshi1685<br /> &lt;br&gt;<br /> Minor edits by fasterthanlight<br /> <br /> ==Algebraic Insight into Given Property==<br /> Mods is a good way to prove &lt;math&gt;a^3 \equiv a \pmod6&lt;/math&gt;: residues are simply &lt;math&gt;3, \pm 2, \ pm 1, 0&lt;/math&gt;. Only &lt;math&gt;2&lt;/math&gt; and &lt;math&gt;3&lt;/math&gt; are necessary to check.<br /> Another way is to observe that &lt;math&gt;a^3-a&lt;/math&gt; factors into &lt;math&gt;(a-1)*a*(a+1)&lt;/math&gt;. Any &lt;math&gt;k&lt;/math&gt; consecutive numbers must be a multiple of &lt;math&gt;k&lt;/math&gt;, so &lt;math&gt;a^3-a&lt;/math&gt; is both divisible by &lt;math&gt;2&lt;/math&gt; and &lt;math&gt;3&lt;/math&gt;. This provides an algebraic method for proving &lt;math&gt;a^3 \equiv a \pmod6&lt;/math&gt; for all &lt;math&gt;a&lt;/math&gt;.<br /> <br /> <br /> ==Video Solution 1==<br /> With Modular Arithmetic Intro<br /> https://www.youtube.com/watch?v=wbv3TArroSs<br /> <br /> ~IceMatrix<br /> <br /> ==Video Solution 2==<br /> https://www.youtube.com/watch?v=SRjZ6B5DR74<br /> <br /> ~bunny1<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2018|ab=B|num-b=15|num-a=17}}<br /> {{MAA Notice}}</div> Bunny1 https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_16&diff=141224 2018 AMC 10B Problems/Problem 16 2021-01-01T01:13:45Z <p>Bunny1: </p> <hr /> <div>==Problem==<br /> <br /> Let &lt;math&gt;a_1,a_2,\dots,a_{2018}&lt;/math&gt; be a strictly increasing sequence of positive integers such that &lt;cmath&gt;a_1+a_2+\cdots+a_{2018}=2018^{2018}.&lt;/cmath&gt;<br /> What is the remainder when &lt;math&gt;a_1^3+a_2^3+\cdots+a_{2018}^3&lt;/math&gt; is divided by &lt;math&gt;6&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 /> ==Solution 1==<br /> <br /> One could simply list out all the residues to the third power &lt;math&gt;\mod 6&lt;/math&gt;. (Edit: Euler's totient theorem is not a valid approach to showing that they are all congruent &lt;math&gt;\mod 6&lt;/math&gt;. This is due to the fact that &lt;math&gt;a_k&lt;/math&gt; need not be relatively prime to &lt;math&gt;6&lt;/math&gt;.)<br /> <br /> Therefore the answer is congruent to &lt;math&gt;2018^{2018}\equiv 2^{2018} \pmod{6} = \boxed{ (E)4}&lt;/math&gt;<br /> <br /> Note from Williamgolly: We can WLOG assume &lt;math&gt;a_1,a_2... a_{2017} \equiv 0 \pmod 6&lt;/math&gt; and have &lt;math&gt;a_{2018} \equiv 2 \pmod 6&lt;/math&gt; to make life easier.<br /> <br /> ==Solution 2==<br /> <br /> Note that &lt;math&gt;\left(a_1+a_2+\cdots+a_{2018}\right)^3=a_1^3+a_2^3+\cdots+a_{2018}^3+3a_1^2\left(a_1+a_2+\cdots+a_{2018}-a_1\right)+3a_2^2\left(a_1+a_2+\cdots+a_{2018}-a_2\right)+\cdots+3a_{2018}^2\left(a_1+a_2+\cdots+a_{2018}-a_{2018}\right)+6\sum_{i\neq j\neq k}^{2018} a_ia_ja_k&lt;/math&gt;<br /> <br /> Note that &lt;math&gt;<br /> a_1^3+a_2^3+\cdots+a_{2018}^3+3a_1^2\left(a_1+a_2+\cdots+a_{2018}-a_1\right)+3a_2^2\left(a_1+a_2+\cdots+a_{2018}-a_2\right)+\cdots+3a_{2018}^2\left(a_1+a_2+\cdots+a_{2018}-a_{2018}\right)+6\sum_{i\neq j\neq k}^{2018} a_ia_ja_k\equiv a_1^3+a_2^3+\cdots+a_{2018}^3+3a_1^2({2018}^{2018}-a_1)+3a_2^2({2018}^{2018}-a_2)+\cdots+3a_{2018}^2({2018}^{2018}-a_{2018})<br /> \equiv -2(a_1^3+a_2^3+\cdots+a_{2018}^3)\pmod 6<br /> &lt;/math&gt;<br /> Therefore, &lt;math&gt;-2(a_1^3+a_2^3+\cdots+a_{2018}^3)\equiv \left(2018^{2018}\right)^3\equiv\left( 2^{2018}\right)^3\equiv 4^3\equiv 4\pmod{6}&lt;/math&gt;.<br /> <br /> Thus, &lt;math&gt;a_1^3+a_2^3+\cdots+a_{2018}^3\equiv 1\pmod 3&lt;/math&gt;. However, since cubing preserves parity, and the sum of the individual terms is even, the some of the cubes is also even, and our answer is &lt;math&gt;\boxed{\text{(E) }4}&lt;/math&gt;<br /> <br /> ==Solution 3 (Partial Proof)==<br /> First, we can assume that the problem will have a consistent answer for all possible values of &lt;math&gt;a_1&lt;/math&gt;. For the purpose of this solution, we will assume that &lt;math&gt;a_1 = 1&lt;/math&gt;.<br /> <br /> We first note that &lt;math&gt;1^3+2^3+...+n^3 = (1+2+...+n)^2&lt;/math&gt;. So what we are trying to find is what &lt;math&gt;\left(2018^{2018}\right)^2=\left(2018^{4036}\right)&lt;/math&gt; mod &lt;math&gt;6&lt;/math&gt;. We start by noting that &lt;math&gt;2018&lt;/math&gt; is congruent to &lt;math&gt;2 \pmod{6}&lt;/math&gt;. So we are trying to find &lt;math&gt;\left(2^{4036}\right) \pmod{6}&lt;/math&gt;. Instead of trying to do this with some number theory skills, we could just look for a pattern. We start with small powers of &lt;math&gt;2&lt;/math&gt; and see that &lt;math&gt;2^1&lt;/math&gt; is &lt;math&gt;2&lt;/math&gt; mod &lt;math&gt;6&lt;/math&gt;, &lt;math&gt;2^2&lt;/math&gt; is &lt;math&gt;4&lt;/math&gt; mod &lt;math&gt;6&lt;/math&gt;, &lt;math&gt;2^3&lt;/math&gt; is &lt;math&gt;2&lt;/math&gt; mod &lt;math&gt;6&lt;/math&gt;, &lt;math&gt;2^4&lt;/math&gt; is &lt;math&gt;4&lt;/math&gt; mod &lt;math&gt;6&lt;/math&gt;, and so on... So we see that since &lt;math&gt;\left(2^{4036}\right)&lt;/math&gt; has an even power, it must be congruent to &lt;math&gt;4 \pmod{6}&lt;/math&gt;, thus giving our answer &lt;math&gt;\boxed{\text{(E) }4}&lt;/math&gt;. You can prove this pattern using mods. But I thought this was easier.<br /> <br /> -TheMagician<br /> <br /> ==Solution 4 (Lazy solution)==<br /> First, we can assume that the problem will have a consistent answer for all possible values of &lt;math&gt;a_1&lt;/math&gt;. For the purpose of this solution, assume &lt;math&gt;a_1, a_2, ... a_{2017}&lt;/math&gt; are multiples of 6 and find &lt;math&gt;2018^{2018} \pmod{6}&lt;/math&gt; (which happens to be &lt;math&gt;4&lt;/math&gt;). Then &lt;math&gt;{a_1}^3 + ... + {a_{2018}}^3&lt;/math&gt; is congruent to &lt;math&gt;64 \pmod{6}&lt;/math&gt; or just &lt;math&gt;4&lt;/math&gt;. <br /> <br /> -Patrick4President<br /> <br /> ==Solution 5 (Nichomauss' Theorem)==<br /> <br /> Seeing the cubes of numbers, we think of Nichomauss's theorem, which states that &lt;math&gt;(a_1^3 + a_2^3 + ... + a_n^3) = (a_1 + a_2 + ... + a_n)^2&lt;/math&gt;. We can do this and deduce that &lt;math&gt;(a_1^3 + a_2^3 + ... + a_{2018}^3) = 2018^{2018}&lt;/math&gt; squared. <br /> <br /> Now, we find &lt;math&gt;2018\mod 6&lt;/math&gt;, which is 2. This means that we need to find &lt;math&gt;2^{2018} \mod {6}&lt;/math&gt;, which we can find using a pattern to be &lt;math&gt;4&lt;/math&gt;. Therefore, the answer is &lt;math&gt;4^2\mod 6&lt;/math&gt;, which is congruent to &lt;math&gt;\boxed{\textbf{(E)} 4}&lt;/math&gt;<br /> <br /> -ericshi1685<br /> &lt;br&gt;<br /> Minor edits by fasterthanlight<br /> <br /> ==Algebraic Insight into Given Property==<br /> Mods is a good way to prove &lt;math&gt;a^3 \equiv a \pmod6&lt;/math&gt;: residues are simply &lt;math&gt;3, \pm 2, \ pm 1, 0&lt;/math&gt;. Only &lt;math&gt;2&lt;/math&gt; and &lt;math&gt;3&lt;/math&gt; are necessary to check.<br /> Another way is to observe that &lt;math&gt;a^3-a&lt;/math&gt; factors into &lt;math&gt;(a-1)*a*(a+1)&lt;/math&gt;. Any &lt;math&gt;k&lt;/math&gt; consecutive numbers must be a multiple of &lt;math&gt;k&lt;/math&gt;, so &lt;math&gt;a^3-a&lt;/math&gt; is both divisible by &lt;math&gt;2&lt;/math&gt; and &lt;math&gt;3&lt;/math&gt;. This provides an algebraic method for proving &lt;math&gt;a^3 \equiv a \pmod6&lt;/math&gt; for all &lt;math&gt;a&lt;/math&gt;.<br /> <br /> <br /> ==Excellent Video Solution==<br /> With Modular Arithmetic Intro<br /> https://www.youtube.com/watch?v=wbv3TArroSs<br /> <br /> ~IceMatrix<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2018|ab=B|num-b=15|num-a=17}}<br /> {{MAA Notice}}</div> Bunny1 https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_25&diff=141223 2018 AMC 10B Problems/Problem 25 2021-01-01T01:12:00Z <p>Bunny1: </p> <hr /> <div>{{duplicate|[[2018 AMC 12B Problems|2018 AMC 12B #25]] and [[2018 AMC 10B Problems|2018 AMC 10B #24]]}}<br /> <br /> == Problem ==<br /> Let &lt;math&gt;\lfloor x \rfloor&lt;/math&gt; denote the greatest integer less than or equal to &lt;math&gt;x&lt;/math&gt;. How many real numbers &lt;math&gt;x&lt;/math&gt; satisfy the equation &lt;math&gt;x^2 + 10,000\lfloor x \rfloor = 10,000x&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 197 \qquad \textbf{(B) } 198 \qquad \textbf{(C) } 199 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 201&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> This rewrites itself to &lt;math&gt;x^2=10,000\{x\}&lt;/math&gt;.<br /> <br /> Graphing &lt;math&gt;y=10,000\{x\}&lt;/math&gt; and &lt;math&gt;y=x^2&lt;/math&gt; we see that the former is a set of line segments with slope &lt;math&gt;10,000&lt;/math&gt; from &lt;math&gt;0&lt;/math&gt; to &lt;math&gt;1&lt;/math&gt; with a hole at &lt;math&gt;x=1&lt;/math&gt;, then &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;2&lt;/math&gt; with a hole at &lt;math&gt;x=2&lt;/math&gt; etc.<br /> Here is a graph of &lt;math&gt;y=x^2&lt;/math&gt; and &lt;math&gt;y=16\{x\}&lt;/math&gt; for visualization.<br /> <br /> &lt;asy&gt;<br /> import graph;<br /> size(400);<br /> xaxis(&quot;$x$&quot;,Ticks(Label(fontsize(8pt)),new real[]{-5,-4,-3, -2, -1,0,1 2,3, 4,5}));<br /> yaxis(&quot;$y$&quot;,Ticks(Label(fontsize(8pt)),new real[]{0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18}));<br /> real y(real x) {return x^2;}<br /> draw(circle((-4,16), 0.1));<br /> draw(circle((-3,16), 0.1));<br /> draw(circle((-2,16), 0.1));<br /> draw(circle((-1,16), 0.1));<br /> draw(circle((0,16), 0.1));<br /> draw(circle((1,16), 0.1));<br /> draw(circle((2,16), 0.1));<br /> draw(circle((3,16), 0.1));<br /> draw(circle((4,16), 0.1));<br /> draw((-5,0)--(-4,16), black);<br /> draw((-4,0)--(-3,16), black);<br /> draw((-3,0)--(-2,16), black);<br /> draw((-2,0)--(-1,16), black);<br /> draw((-1,0)--(-0,16), black);<br /> draw((0,0)--(1,16), black);<br /> draw((1,0)--(2,16), black);<br /> draw((2,0)--(3,16), black);<br /> draw((3,0)--(4,16), black);<br /> draw(graph(y,-4.2,4.2),green);<br /> &lt;/asy&gt;<br /> <br /> Now notice that when &lt;math&gt;x=\pm 100&lt;/math&gt; then graph has a hole at &lt;math&gt;(\pm 100,10,000)&lt;/math&gt; which the equation &lt;math&gt;y=x^2&lt;/math&gt; passes through and then continues upwards. Thus our set of possible solutions is bounded by &lt;math&gt;(-100,100)&lt;/math&gt;. We can see that &lt;math&gt;y=x^2&lt;/math&gt; intersects each of the lines once and there are &lt;math&gt;99-(-99)+1=199&lt;/math&gt; lines for an answer of &lt;math&gt;\boxed{\text{(C)}~199}&lt;/math&gt;.<br /> <br /> Note: From the graph, we can clearly see there are &lt;math&gt;4&lt;/math&gt; solutions on the negative side of the &lt;math&gt;x&lt;/math&gt;-axis and only &lt;math&gt;2&lt;/math&gt; on the positive side of the &lt;math&gt;x&lt;/math&gt;-axis. So the solution really should be from &lt;math&gt;-100&lt;/math&gt; to &lt;math&gt;98&lt;/math&gt;, which still counts to &lt;math&gt;199&lt;/math&gt;. A couple of the alternative solutions also seem to have the same flaw.<br /> <br /> ==Solution 2==<br /> <br /> Same as the first solution, &lt;math&gt;x^2=10,000\{x\} &lt;/math&gt;.<br /> <br /> <br /> We can write &lt;math&gt;x&lt;/math&gt; as &lt;math&gt;\lfloor x \rfloor+\{x\}&lt;/math&gt;. Expanding everything, we get a quadratic in &lt;math&gt;\{x\}&lt;/math&gt; in terms of &lt;math&gt;\lfloor x \rfloor&lt;/math&gt;:<br /> &lt;cmath&gt; \{x\}^2+ (2\lfloor x \rfloor -10,000)\{x\} + \lfloor x \rfloor ^2 = 0&lt;/cmath&gt;<br /> <br /> <br /> We use the quadratic formula to solve for &lt;math&gt;\{x\}&lt;/math&gt; :<br /> &lt;cmath&gt; \{x\} = \frac {-2\lfloor x \rfloor + 10,000 \pm \sqrt{ ( 2\lfloor x \rfloor - 10,000 )^2 - 4\lfloor x \rfloor^2 }}{2} = \frac {-2\lfloor x \rfloor + 10,000 \pm \sqrt{ 4\lfloor x \rfloor^2 -40,000 \lfloor x \rfloor + 10,000^2- 4\lfloor x \rfloor^2 }}{2} &lt;/cmath&gt;<br /> <br /> <br /> Since &lt;math&gt; 0 \leq \{x\} &lt; 1 &lt;/math&gt;, we get an inequality which we can then solve. After simplifying a lot, we get that &lt;math&gt;\lfloor x \rfloor^2 + 2\lfloor x \rfloor - 9999 &lt; 0&lt;/math&gt;.<br /> <br /> <br /> Solving over the integers, &lt;math&gt;-101 &lt; \lfloor x \rfloor &lt; 99 &lt;/math&gt;, and since &lt;math&gt;\lfloor x \rfloor&lt;/math&gt; is an integer, there are &lt;math&gt;\boxed{\text{(C)}~199}&lt;/math&gt; solutions. Each value of &lt;math&gt; \lfloor x \rfloor&lt;/math&gt; should correspond to one value of &lt;math&gt;x&lt;/math&gt;, so we are done.<br /> <br /> ==Solution 3==<br /> <br /> Let &lt;math&gt;x = a+k&lt;/math&gt; where &lt;math&gt;a&lt;/math&gt; is the integer part of &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;k&lt;/math&gt; is the fractional part of &lt;math&gt;x&lt;/math&gt;.<br /> We can then rewrite the problem below:<br /> <br /> &lt;math&gt;(a+k)^2 + 10000a = 10000(a+k)&lt;/math&gt;<br /> <br /> From here, we get<br /> <br /> &lt;math&gt;(a+k)^2 + 10000a = 10000a + 10000k&lt;/math&gt;<br /> <br /> Solving for &lt;math&gt;a+k = x&lt;/math&gt;<br /> <br /> &lt;math&gt;(a+k)^2 = 10000k&lt;/math&gt;<br /> <br /> &lt;math&gt;x = a+k = \pm100\sqrt{k}&lt;/math&gt;<br /> <br /> Because &lt;math&gt;0 \leq k &lt; 1&lt;/math&gt;, we know that &lt;math&gt;a+k&lt;/math&gt; cannot be less than or equal to &lt;math&gt;-100&lt;/math&gt; nor greater than or equal to &lt;math&gt;100&lt;/math&gt;. Therefore:<br /> <br /> &lt;math&gt;-99 \leq x \leq 99&lt;/math&gt;<br /> <br /> There are &lt;math&gt;199&lt;/math&gt; elements in this range, so the answer is &lt;math&gt;\boxed{\textbf{(C)} \text{ 199}}&lt;/math&gt;.<br /> <br /> Note (not by author): this solution seems to be invalid at first, because one can not determine whether &lt;math&gt;x&lt;/math&gt; is an integer or not. However, it actually works because although &lt;math&gt;x&lt;/math&gt; itself might not be an integer, it is very close to one, so there are 199 potential &lt;math&gt;x&lt;/math&gt;.<br /> <br /> ==Solution 4==<br /> <br /> Notice the given equation is equivilent to &lt;math&gt;(\lfloor x \rfloor+\{x\})^2=10,000\{x\} &lt;/math&gt;<br /> <br /> Now we now that &lt;math&gt;\{x\} &lt; 1&lt;/math&gt; so plugging in &lt;math&gt;1&lt;/math&gt; for &lt;math&gt;\{x\}&lt;/math&gt; we can find the upper and lower bounds for the values.<br /> <br /> &lt;math&gt;(\lfloor x \rfloor +1)^2 = 10,000(1)&lt;/math&gt;<br /> <br /> &lt;math&gt;(\lfloor x \rfloor +1) = \pm 100&lt;/math&gt;<br /> <br /> &lt;math&gt;\lfloor x \rfloor = 99, -101&lt;/math&gt;<br /> <br /> And just like &lt;math&gt;\textbf{Solution 2}&lt;/math&gt;, we see that &lt;math&gt;-101 &lt; \lfloor x \rfloor &lt; 99 &lt;/math&gt;, and since &lt;math&gt;\lfloor x \rfloor&lt;/math&gt; is an integer, there are &lt;math&gt;\boxed{\text{(C)}~199}&lt;/math&gt; solutions. Each value of &lt;math&gt; \lfloor x \rfloor&lt;/math&gt; should correspond to one value of &lt;math&gt;x&lt;/math&gt;, so we are done.<br /> <br /> ==Solution 5==<br /> <br /> First, we can let &lt;math&gt;\{x\} = b, \lfloor x \rfloor = a&lt;/math&gt;. We know that &lt;math&gt;a + b = x&lt;/math&gt; by definition. We can rearrange the equation to obtain <br /> <br /> &lt;math&gt;x^2 = 10^4(x - a)&lt;/math&gt;. <br /> <br /> By taking square root on both sides, we obtain &lt;math&gt;x = \pm 100 \sqrt{b}&lt;/math&gt; (because &lt;math&gt;x - a = b&lt;/math&gt;). We know since &lt;math&gt;b&lt;/math&gt; is the fractional part of &lt;math&gt;x&lt;/math&gt;, it must be that &lt;math&gt;0 \leq b &lt; 1&lt;/math&gt;. Thus, &lt;math&gt;x&lt;/math&gt; may take any value in the interval &lt;math&gt;-100 &lt; x &lt; 100&lt;/math&gt;. Hence, we know that there are &lt;math&gt;\boxed{\text{(C)}~199}&lt;/math&gt; potential values for &lt;math&gt;\lfloor x \rfloor&lt;/math&gt; in that range and we are done. <br /> <br /> ~awesome1st<br /> <br /> ==Solution 6==<br /> <br /> Firstly, we can rearrange to get &lt;math&gt;\lfloor x \rfloor = x-\frac{x^2}{10,000}&lt;/math&gt;<br /> <br /> Rearranging, we get &lt;math&gt;\frac{x^2}{10,000} &lt; 1&lt;/math&gt;<br /> <br /> Noticing that &lt;math&gt;10,000 = 100^2&lt;/math&gt;, we know that x can only be within the boundaries of &lt;math&gt;-100&lt;x&lt;100&lt;/math&gt; and hence, we know that there are &lt;math&gt;\boxed{\text{(C)}~199}&lt;/math&gt; potential values.<br /> <br /> ==Solution 7==<br /> <br /> Firstly, if &lt;math&gt;x&lt;/math&gt; is an integer, then &lt;math&gt;10,000\lfloor x \rfloor=10,000x&lt;/math&gt;, so &lt;math&gt;x&lt;/math&gt; must be &lt;math&gt;0&lt;/math&gt;.<br /> <br /> If &lt;math&gt;0&lt;x&lt;1&lt;/math&gt;, then we know the following:<br /> <br /> &lt;math&gt;0&lt;x^2&lt;1&lt;/math&gt;<br /> <br /> &lt;math&gt;10,000\lfloor x \rfloor =0&lt;/math&gt;<br /> <br /> &lt;math&gt;0&lt;10,000x&lt;10,000&lt;/math&gt;<br /> <br /> Therefore, &lt;math&gt;0&lt;x^2+10,000\lfloor x \rfloor &lt;1&lt;/math&gt;, which overlaps with &lt;math&gt;0&lt;10,000x&lt;10,000&lt;/math&gt;. This means that there is at least one real solution between &lt;math&gt;0&lt;/math&gt; and &lt;math&gt;1&lt;/math&gt;. Since &lt;math&gt;x^2+10,000\lfloor x \rfloor &lt;/math&gt; increases exponentially and &lt;math&gt;10,000x&lt;/math&gt; increases linearly, there is only one solution for this case. <br /> <br /> Similarly, if &lt;math&gt;1&lt;x&lt;2&lt;/math&gt;, then we know the following:<br /> <br /> &lt;math&gt;1&lt;x^2&lt;4&lt;/math&gt;<br /> <br /> &lt;math&gt;10,000\lfloor x \rfloor =10,000&lt;/math&gt;<br /> <br /> &lt;math&gt;&lt;10,000&lt;10,000x&lt;20,000&lt;/math&gt;<br /> <br /> By following similar logic, we can find that there is one solution between &lt;math&gt;1&lt;/math&gt; ad &lt;math&gt;2&lt;/math&gt;. <br /> <br /> We can also follow the same process to find that there are negative solutions for &lt;math&gt;x&lt;/math&gt; as well.<br /> <br /> There are not an infinite amount of solutions, so at one point there will be no solutions when &lt;math&gt;n&lt;x&lt;n+1&lt;/math&gt; for some integer &lt;math&gt;n&lt;/math&gt;. For there to be no solutions in a given range means that the range of &lt;math&gt;10,000\lfloor x \rfloor + x^2&lt;/math&gt; does not intersect the range of &lt;math&gt;10,000x&lt;/math&gt;. &lt;math&gt;x^2&lt;/math&gt; will always be positive, and &lt;math&gt;10,000\lfloor x \rfloor&lt;/math&gt; is less than &lt;math&gt;10,000&lt;/math&gt; less than &lt;math&gt;10,000x&lt;/math&gt;, so when &lt;math&gt;x^2 &gt;= 10,000&lt;/math&gt;, the equation will have no solutions. This means that there are &lt;math&gt;99&lt;/math&gt; positive solutions, &lt;math&gt;99&lt;/math&gt; negative solutions, and &lt;math&gt;0&lt;/math&gt; for a total of &lt;math&gt;\boxed{\text{(C)}~199}&lt;/math&gt; solutions.<br /> <br /> ~Owen1204<br /> <br /> ==Video Solution==<br /> https://www.youtube.com/watch?v=vHKPbaXwJUE<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2018|ab=B|num-b=24|after=Last Problem}}<br /> {{AMC12 box|year=2018|ab=B|num-b=23|num-a=25}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Intermediate Algebra Problems]]</div> Bunny1 https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_24&diff=141222 2018 AMC 10B Problems/Problem 24 2021-01-01T01:11:35Z <p>Bunny1: </p> <hr /> <div>{{duplicate|[[2018 AMC 12B Problems|2018 AMC 12B #20]] and [[2018 AMC 10B Problems|2018 AMC 10B #24]]}}<br /> <br /> ==Problem==<br /> <br /> Let &lt;math&gt;ABCDEF&lt;/math&gt; be a regular hexagon with side length &lt;math&gt;1&lt;/math&gt;. Denote by &lt;math&gt;X&lt;/math&gt;, &lt;math&gt;Y&lt;/math&gt;, and &lt;math&gt;Z&lt;/math&gt; the midpoints of sides &lt;math&gt;\overline {AB}&lt;/math&gt;, &lt;math&gt;\overline{CD}&lt;/math&gt;, and &lt;math&gt;\overline{EF}&lt;/math&gt;, respectively. What is the area of the convex hexagon whose interior is the intersection of the interiors of &lt;math&gt;\triangle ACE&lt;/math&gt; and &lt;math&gt;\triangle XYZ&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)} \frac {3}{8}\sqrt{3} \qquad \textbf{(B)} \frac {7}{16}\sqrt{3} \qquad \textbf{(C)} \frac {15}{32}\sqrt{3} \qquad \textbf{(D)} \frac {1}{2}\sqrt{3} \qquad \textbf{(E)} \frac {9}{16}\sqrt{3} \qquad &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> &lt;asy&gt;<br /> pair A,B,C,D,E,F,W,X,Y,Z,M,N,O,P,Q,R;<br /> A=(0,sqrt(3));<br /> B=(1,sqrt(3));<br /> C=(3/2,sqrt(3)/2);<br /> D=(1,0);<br /> E=(0,0);<br /> F=(-1/2,sqrt(3)/2);<br /> X=(1/2, sqrt(3));<br /> Y=(5/4, sqrt(3)/4);<br /> Z=(-1/4, sqrt(3)/4); <br /> M=(0,sqrt(3)/2);<br /> N=(3/4,3sqrt(3)/4);<br /> O=(3/4,sqrt(3)/4);<br /> P=(3/8,7sqrt(3)/8);<br /> Q=(9/8, 3sqrt(3)/8);<br /> R=(0,sqrt(3)/4);<br /> label(&quot;$A$&quot;,A,NW);<br /> label(&quot;$B$&quot;,B,NE);<br /> label(&quot;$C$&quot;,C,ESE);<br /> label(&quot;$D$&quot;,D,SE);<br /> label(&quot;$E$&quot;,E,SW);<br /> label(&quot;$F$&quot;,F,WSW);<br /> label(&quot;$X$&quot;, X, N);<br /> label(&quot;$Y$&quot;, Y, ESE);<br /> label(&quot;$Z$&quot;, Z, WSW);<br /> label(&quot;$M$&quot;, M, NW);<br /> label(&quot;$N$&quot;, N, NE);<br /> label(&quot;$O$&quot;, O, SE);<br /> label(&quot;$P$&quot;, P, NNW);<br /> label(&quot;$Q$&quot;, Q, ESE);<br /> label(&quot;$R$&quot;, R, SW);<br /> fill((0,sqrt(3)/2)--(3/8,7sqrt(3)/8)--(3/4,3sqrt(3)/4)--(9/8, 3sqrt(3)/8)--(3/4,sqrt(3)/4)--(0,sqrt(3)/4)--cycle,gray);<br /> draw(A--B--C--D--E--F--cycle);<br /> draw(A--C--E--cycle);<br /> draw(X--Y--Z--cycle);<br /> draw(M--N--O--cycle);<br /> <br /> &lt;/asy&gt;<br /> <br /> The desired area (hexagon &lt;math&gt;MPNQOR&lt;/math&gt;) consists of an equilateral triangle (&lt;math&gt;\triangle MNO&lt;/math&gt;) and three right triangles (&lt;math&gt;\triangle MPN&lt;/math&gt;, &lt;math&gt;\triangle NQO&lt;/math&gt;, and &lt;math&gt;\triangle ORM&lt;/math&gt;).<br /> <br /> Notice that &lt;math&gt;\overline {AD}&lt;/math&gt; (not shown) and &lt;math&gt;\overline {BC}&lt;/math&gt; are parallel. &lt;math&gt;\overline {XY}&lt;/math&gt; divides transversals &lt;math&gt;\overline {AB}&lt;/math&gt; and &lt;math&gt;\overline {CD}&lt;/math&gt; into a &lt;math&gt;1:1&lt;/math&gt; ratio. (Note from Williamgolly: you can see this with similar triangles.) Thus, it must also divide transversal &lt;math&gt;\overline {AC}&lt;/math&gt; and transversal &lt;math&gt;\overline {CO}&lt;/math&gt; into a &lt;math&gt;1:1&lt;/math&gt; ratio. By symmetry, the same applies for &lt;math&gt;\overline {CE}&lt;/math&gt; and &lt;math&gt;\overline {EA}&lt;/math&gt; as well as &lt;math&gt;\overline {EM}&lt;/math&gt; and &lt;math&gt;\overline {AN}&lt;/math&gt;.<br /> <br /> <br /> In &lt;math&gt;\triangle ACE&lt;/math&gt;, we see that &lt;math&gt;\frac{[MNO]}{[ACE]} = \frac{1}{4}&lt;/math&gt; and &lt;math&gt;\frac{[MPN]}{[ACE]} = \frac{1}{8}&lt;/math&gt;. Our desired area becomes <br /> <br /> &lt;cmath&gt;(\frac{1}{4}+3 \cdot \frac{1}{8}) \cdot \frac{(\sqrt{3})^2 \cdot \sqrt{3}}{4} = \frac {15}{32}\sqrt{3} = \boxed {C}&lt;/cmath&gt;<br /> <br /> ==Solution 2 (Alternate Geometrical Approach to 1)==<br /> Instead of directly finding the desired hexagonal area, &lt;math&gt;\triangle XYZ&lt;/math&gt; can be found. It consists of three triangles and the desired hexagon. Given triangle rotational symmetry, the three triangles are congruent. See that &lt;math&gt;\triangle XYZ&lt;/math&gt; and &lt;math&gt;\triangle ACE&lt;/math&gt; are equilateral, so &lt;math&gt;m\angle PXN=60&lt;/math&gt;, so &lt;math&gt;m\angle AXP = \frac{180-60}{2}=60&lt;/math&gt;. As &lt;math&gt;\overline {AC}&lt;/math&gt; is a transversal running through &lt;math&gt;\overline {FC}&lt;/math&gt; (use your imagination) and &lt;math&gt;\overline {AB}&lt;/math&gt;, &lt;math&gt;m\angle BAC=m\angle FCA = \frac{m\angle ACE}{2}=30&lt;/math&gt;. <br /> <br /> <br /> Then, &lt;math&gt;\triangle APX&lt;/math&gt; is a &lt;math&gt;30&lt;/math&gt;-&lt;math&gt;60&lt;/math&gt;-&lt;math&gt;90&lt;/math&gt; triangle. By HL congruence, &lt;math&gt;\triangle APX \cong \triangle NPX&lt;/math&gt;. &lt;math&gt;AX=\frac{1}{2}&lt;/math&gt;. Then, the area of &lt;math&gt;\triangle PXN&lt;/math&gt; is &lt;math&gt;\frac{\sqrt{3}}{32}&lt;/math&gt;. There are three such triangles for a total area of &lt;math&gt;\triangle XYZ&lt;/math&gt; is &lt;math&gt;\frac{3\sqrt{3}}{32}&lt;/math&gt;. Find the side of &lt;math&gt;\triangle XYZ&lt;/math&gt; to be &lt;math&gt;\frac{3}{2}&lt;/math&gt;, so the area is &lt;math&gt;\frac{9\sqrt{3}}{16}&lt;/math&gt;. <br /> <br /> <br /> &lt;math&gt;\frac{9\sqrt{3}}{16}-\frac{3\sqrt{3}}{32}=\frac{15\sqrt{3}}{32} \implies \boxed{C}&lt;/math&gt;<br /> <br /> <br /> ~BJHHar<br /> <br /> ==Solution 3 ==<br /> <br /> <br /> &lt;asy&gt;<br /> pair A,B,C,D,E,F,W,X,Y,Z,M,N,O,P,Q,R;<br /> A=(0,sqrt(3));<br /> B=(1,sqrt(3));<br /> C=(3/2,sqrt(3)/2);<br /> D=(1,0);<br /> E=(0,0);<br /> F=(-1/2,sqrt(3)/2);<br /> X=(1/2, sqrt(3));<br /> Y=(5/4, sqrt(3)/4);<br /> Z=(-1/4, sqrt(3)/4); <br /> <br /> P=(3/8,7sqrt(3)/8);<br /> Q=(9/8, 3sqrt(3)/8);<br /> R=(0,sqrt(3)/4);<br /> label(&quot;$A$&quot;,A,NW);<br /> label(&quot;$B$&quot;,B,NE);<br /> label(&quot;$C$&quot;,C,ESE);<br /> label(&quot;$D$&quot;,D,SE);<br /> label(&quot;$E$&quot;,E,SW);<br /> label(&quot;$F$&quot;,F,WSW);<br /> label(&quot;$X$&quot;, X, SE);<br /> label(&quot;$Y$&quot;, Y, ESE);<br /> label(&quot;$Z$&quot;, Z, WSW);<br /> label(&quot;$P$&quot;, P, NNW);<br /> label(&quot;$Q$&quot;, Q, ESE);<br /> label(&quot;$R$&quot;, R, SW);<br /> fill((0,sqrt(3)/2)--(3/8,7sqrt(3)/8)--(3/4,3sqrt(3)/4)--(9/8, 3sqrt(3)/8)--(3/4,sqrt(3)/4)--(0,sqrt(3)/4)--cycle,gray);<br /> draw(A--B--C--D--E--F--cycle);<br /> draw(A--C--E--cycle);<br /> draw(X--Y--Z--cycle);<br /> draw(M--N--O--cycle);<br /> <br /> &lt;/asy&gt;<br /> <br /> <br /> Now, if we look at the figure, we can see that the complement of the hexagon we are trying to find is composed of &lt;math&gt;3&lt;/math&gt; isosceles trapezoids (&lt;math&gt;AXFZ&lt;/math&gt;, &lt;math&gt;XBCY&lt;/math&gt;, and &lt;math&gt;ZYED&lt;/math&gt;), and &lt;math&gt;3&lt;/math&gt; right triangles, with one right angle on each of &lt;math&gt;X&lt;/math&gt;, &lt;math&gt;Y&lt;/math&gt;, and &lt;math&gt;Z&lt;/math&gt;. <br /> Finding the trapezoid's area, we know that one base of each trapezoid is just the side length of the hexagon, which is &lt;math&gt;1&lt;/math&gt;, and the other base is &lt;math&gt;\frac{3}{2}&lt;/math&gt; (it is halfway in between the side and the longest diagonal, which has length &lt;math&gt;2&lt;/math&gt;) with a height of &lt;math&gt;\frac{\sqrt{3}}{4}&lt;/math&gt; (by using the Pythagorean Theorem and the fact that it is an isosceles trapezoid) to give each trapezoid having an area of &lt;math&gt;\frac{5\sqrt{3}}{16}&lt;/math&gt; for a total area of &lt;math&gt;\frac{15\sqrt{3}}{16}.&lt;/math&gt; (Alternatively, we could have calculated the area of hexagon &lt;math&gt;ABCDEF&lt;/math&gt; and subtracted the area of &lt;math&gt;\triangle XYZ&lt;/math&gt;, which, as we showed before, had a side length of &lt;math&gt;\frac{3}{2}&lt;/math&gt;). <br /> Now, we need to find the area of each of the small triangles, which, if we look at the triangle that has a vertex on &lt;math&gt;X&lt;/math&gt;, is similar to the triangle with a base of &lt;math&gt;YC = 1/2.&lt;/math&gt; Using similar triangles, we calculate the base to be &lt;math&gt;\frac{1}{4}&lt;/math&gt; and the height to be &lt;math&gt;\frac{\sqrt{3}}{4}&lt;/math&gt; giving us an area of &lt;math&gt;\frac{\sqrt{3}}{32}&lt;/math&gt; per triangle, and a total area of &lt;math&gt;3\frac{\sqrt{3}}{32}&lt;/math&gt;. Adding the two areas together, we get &lt;math&gt;\frac{15\sqrt{3}}{16} + \frac{3\sqrt{3}}{32} = \frac{33\sqrt{3}}{32}&lt;/math&gt;. Finding the total area, we get &lt;math&gt;6 \cdot 1^2 \cdot \frac{\sqrt{3}}{4}=\frac{3\sqrt{3}}{2}&lt;/math&gt;. Taking the complement, we get &lt;math&gt;\frac{3\sqrt{3}}{2} - \frac{33\sqrt{3}}{32} = \frac{15\sqrt{3}}{32} = (C)\frac{15}{32}\sqrt{3}&lt;/math&gt;<br /> <br /> ==Solution 4 (Trig)==<br /> Notice, the area of the convex hexagon formed through the intersection of the &lt;math&gt;2&lt;/math&gt; triangles can be found by finding the area of the triangle formed by the midpoints of the sides and subtracting the smaller triangles that are formed by the region inside this triangle but outside the other triangle. First, let's find the area of the triangle formed by the midpoint of the sides. Notice, this is an equilateral triangle, thus all we need is to find the length of its side. <br /> To do this, we look at the isosceles trapezoid outside this triangle but inside the outer hexagon. Since the interior angle of a regular hexagon is &lt;math&gt;120^{\textrm{o}}&lt;/math&gt; and the trapezoid is isosceles, we know that the angle opposite is &lt;math&gt;60^{\textrm{o}}&lt;/math&gt;, and thus the side length of this triangle is &lt;math&gt;1+2(\frac{1}{2}\cos(60^{\textrm{o}})=1+\frac{1}{2}=\frac{3}{2}&lt;/math&gt;. So the area of this triangle is &lt;math&gt;\frac{\sqrt{3}}{4}s^2=\frac{9\sqrt{3}}{16}&lt;/math&gt;<br /> Now let's find the area of the smaller triangles. Notice, triangle &lt;math&gt;ACE&lt;/math&gt; cuts off smaller isosceles triangles from the outer hexagon. The base of these isosceles triangles is perpendicular to the base of the isosceles trapezoid mentioned before, thus we can use trigonometric ratios to find the base and height of these smaller triangles, which are all congruent due to the rotational symmetry of a regular hexagon. The area is then &lt;math&gt;\frac{1}{2}(\frac{1}{2})\cos(60^{\textrm{o}}))(\frac{1}{2}\sin(60^{\textrm{o}}))=\frac{\sqrt{3}}{32}&lt;/math&gt; and the sum of the areas is &lt;math&gt;3\cdot \frac{\sqrt{3}}{32}=\frac{3\sqrt{3}}{32}&lt;/math&gt;<br /> Therefore, the area of the convex hexagon is &lt;math&gt;\frac{9\sqrt{3}}{16}-\frac{3\sqrt{3}}{32}=\frac{18\sqrt{3}}{32}-\frac{3\sqrt{3}}{32}=\boxed{\frac{15\sqrt{3}}{32}}\implies \boxed{C}&lt;/math&gt;<br /> <br /> ==Solution 5==<br /> Dividing &lt;math&gt;\triangle MNO&lt;/math&gt; into two right triangles congruent to &lt;math&gt;\triangle PMN&lt;/math&gt;, we see that &lt;math&gt;[MPNQOR]=\dfrac{5}{8}[ACE]&lt;/math&gt;. Because &lt;math&gt;[ACE] = \dfrac{1}{2}[ABCDEF]&lt;/math&gt;, we have &lt;math&gt;[MPNQOR]=\dfrac{5}{16}[ABCDEF]&lt;/math&gt;. From here, you should be able to tell that the answer will have a factor of &lt;math&gt;5&lt;/math&gt;, and &lt;math&gt;\boxed{\textbf{(C)} \frac {15}{32}\sqrt{3}}&lt;/math&gt; is the only answer that has a factor of &lt;math&gt;5&lt;/math&gt;. However, if you want to actually calculate the area, you would calculate &lt;math&gt;[ABCDEF]&lt;/math&gt; to be &lt;math&gt;6 \cdot \dfrac{\sqrt{3}}{2 \cdot 2} = \dfrac{3\sqrt{3}}{2}&lt;/math&gt;, so &lt;math&gt;[MPNQOR] = \dfrac{5}{16} \cdot \dfrac{3\sqrt{3}}{2} = \boxed{\textbf{(C)} \frac {15}{32}\sqrt{3}}&lt;/math&gt;.<br /> <br /> ==Solution 6 (Least Algebra Needed)==<br /> [[Image:billybobjoe.png|thumb|center|300px]]<br /> <br /> We can see by the picture that there are a total of 24 small equilateral triangles in the hexagon, each with the same area(Look at the black lines). In the yellow region, there are 6 full triangles, and 3 half triangles, giving us 7.5 triangles worth of area. Thus, area of the region we want is &lt;math&gt;\frac{7.5}{24}&lt;/math&gt; of the entire hexagon. The total area of the hexagon is &lt;math&gt;\frac{3\sqrt{3}}{2}&lt;/math&gt;, so our answer is &lt;math&gt;\frac{7.5}{24} \cdot \frac{3\sqrt{3}}{2}&lt;/math&gt; = &lt;math&gt;\boxed{(C) \frac{15\sqrt{3}}{32}}&lt;/math&gt; <br /> <br /> -AlexLikeMath<br /> <br /> ==Solution 7==<br /> If we try to coordinate bash this problem, it's going to look very ugly with a lot of radicals. However, we can alter and skew the diagram in such a way that all ratios of lengths and areas stay the same while making it a lot easier to work with. Then, we can find the ratio of the area of the wanted region to the area of &lt;math&gt;ABCDEF&lt;/math&gt; then apply it to the old diagram.<br /> &lt;asy&gt;<br /> unitsize(1cm);<br /> draw((0,0)--(4,0)--(6,3.464)--(2,3.464)--(0,0));<br /> draw((2,0)--(1,1.732));<br /> draw((5,1.732)--(4,3.464));<br /> draw((1.5, 0.866)--(3, 3.464)--(4.5, 0.866)--cycle);<br /> draw((2,0)--(2,3.464)--(5,1.732)--cycle);<br /> &lt;/asy&gt;<br /> <br /> &lt;asy&gt;<br /> unitsize(1cm);<br /> draw((1,0)--(1,4),gray(.7));<br /> draw((2,0)--(2,4),gray(.7));<br /> draw((3,0)--(3,4),gray(.7));<br /> draw((0,1)--(4,1),gray(.7));<br /> draw((0,2)--(4,2),gray(.7));<br /> draw((0,3)--(4,3),gray(.7));<br /> draw((0,0)--(4,0)--(4,4)--(0,4)--(0,0));<br /> draw((2,0)--(0,2));<br /> draw((4,2)--(2,4));<br /> draw((1,1)--(1,4)--(4,1)--cycle);<br /> draw((0,4)--(2,0)--(4,2)--cycle);<br /> fill((1,4)--(1,3.5)--(2,3)--cycle,red);<br /> fill((1,1)--(1.5,1)--(1,2)--cycle,red);<br /> fill((3,1)--(3.5,1.5)--(4,1)--cycle,red);<br /> &lt;/asy&gt;<br /> <br /> The isosceles right triangle with a leg length of &lt;math&gt;3&lt;/math&gt; in the new diagram is &lt;math&gt;\triangle XYZ&lt;/math&gt; in the old diagram. We see that if we want to take the area of the new hexagon, we must subtract &lt;math&gt;\frac{3}{4}&lt;/math&gt; from the area of &lt;math&gt;\triangle XYZ&lt;/math&gt; (the red triangles), giving us &lt;math&gt;\frac{15}{4}&lt;/math&gt;. However, we need to take the ratio of this area to the area of &lt;math&gt;ABCDEF&lt;/math&gt;, which is &lt;math&gt;\frac{\frac{15}{4}}{12}=\frac{5}{16}&lt;/math&gt;. Now we know that our answer is &lt;math&gt;\frac{5}{16} \cdot \frac{3\sqrt{3}}{2}=\boxed{\frac{15}{32}\sqrt{3}}&lt;/math&gt;.<br /> <br /> ==Video Solution==<br /> https://www.youtube.com/watch?v=yDbn9Mx2myw<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2018|ab=B|num-b=23|num-a=25}}<br /> {{AMC12 box|year=2018|ab=B|num-b=19|num-a=21}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Intermediate Geometry Problems]]</div> Bunny1 https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_21&diff=141220 2018 AMC 10B Problems/Problem 21 2021-01-01T00:45:38Z <p>Bunny1: </p> <hr /> <div>{{duplicate|[[2018 AMC 12B Problems|2018 AMC 12B #19]] and [[2018 AMC 10B Problems|2018 AMC 10B #21]]}}<br /> <br /> ==Problem==<br /> Mary chose an even &lt;math&gt;4&lt;/math&gt;-digit number &lt;math&gt;n&lt;/math&gt;. She wrote down all the divisors of &lt;math&gt;n&lt;/math&gt; in increasing order from left to right: &lt;math&gt;1,2,...,\dfrac{n}{2},n&lt;/math&gt;. At some moment Mary wrote &lt;math&gt;323&lt;/math&gt; as a divisor of &lt;math&gt;n&lt;/math&gt;. What is the smallest possible value of the next divisor written to the right of &lt;math&gt;323&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 324 \qquad \textbf{(B) } 330 \qquad \textbf{(C) } 340 \qquad \textbf{(D) } 361 \qquad \textbf{(E) } 646&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> Since prime factorizing &lt;math&gt;323&lt;/math&gt; gives you &lt;math&gt;17 \cdot 19&lt;/math&gt;, the desired answer needs to be a multiple of &lt;math&gt;17&lt;/math&gt; or &lt;math&gt;19&lt;/math&gt;, this is because if it is not a multiple of &lt;math&gt;17&lt;/math&gt; or &lt;math&gt;19&lt;/math&gt;, &lt;math&gt;n&lt;/math&gt; will be more than a &lt;math&gt;4&lt;/math&gt; digit number. For example, if the answer were to instead be &lt;math&gt;324&lt;/math&gt;, &lt;math&gt;n&lt;/math&gt; would have to be a multiple of &lt;math&gt;2^2 * 3^4 * 17 * 19&lt;/math&gt; for both &lt;math&gt;323&lt;/math&gt; and &lt;math&gt;324&lt;/math&gt; to be a valid factor, meaning &lt;math&gt;n&lt;/math&gt; would have to be at least &lt;math&gt;104652&lt;/math&gt;, which is too big. Looking at the answer choices, &lt;math&gt;\text{(A) }324&lt;/math&gt; and &lt;math&gt;\text{(B) }330&lt;/math&gt; are both not a multiple of neither 17 nor 19, &lt;math&gt;\text{(C) }340&lt;/math&gt; is divisible by &lt;math&gt;17&lt;/math&gt;. &lt;math&gt;\text{(D) }361&lt;/math&gt; is divisible by &lt;math&gt;19&lt;/math&gt;, and &lt;math&gt;\text{(E) }646&lt;/math&gt; is divisible by both &lt;math&gt;17&lt;/math&gt; and &lt;math&gt;19&lt;/math&gt;. Since &lt;math&gt;\boxed{\text{(C) }340}&lt;/math&gt; is the smallest number divisible by either &lt;math&gt;17&lt;/math&gt; or &lt;math&gt;19&lt;/math&gt; it is the answer. Checking, we can see that &lt;math&gt;n&lt;/math&gt; would be &lt;math&gt;6460&lt;/math&gt;, a four-digit number. Note that &lt;math&gt;n&lt;/math&gt; is also divisible by &lt;math&gt;2&lt;/math&gt;, one of the listed divisors of &lt;math&gt;n&lt;/math&gt;. (If &lt;math&gt;n&lt;/math&gt; was not divisible by &lt;math&gt;2&lt;/math&gt;, we would need to look for a different divisor)<br /> <br /> -Edited by Mathandski<br /> <br /> ==Solution 2==<br /> Let the next largest divisor be &lt;math&gt;k&lt;/math&gt;. Suppose &lt;math&gt;\gcd(k,323)=1&lt;/math&gt;. Then, as &lt;math&gt;323|n, k|n&lt;/math&gt;, therefore, &lt;math&gt;323\cdot k|n.&lt;/math&gt; However, because &lt;math&gt;k&gt;323&lt;/math&gt;, &lt;math&gt;323k&gt;323\cdot 324&gt;9999&lt;/math&gt;. Therefore, &lt;math&gt;\gcd(k,323)&gt;1&lt;/math&gt;. Note that &lt;math&gt;323=17\cdot 19&lt;/math&gt;. Therefore, the smallest the GCD can be is &lt;math&gt;17&lt;/math&gt; and our answer is &lt;math&gt;323+17=\boxed{\text{(C) }340}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> Again, recognize &lt;math&gt;323=17 \cdot 19&lt;/math&gt;. The 4-digit number is even, so its prime factorization must then be &lt;math&gt;17 \cdot 19 \cdot 2 \cdot n&lt;/math&gt;. Also, &lt;math&gt;1000\leq 646n \leq 9998&lt;/math&gt;, so &lt;math&gt;2 \leq n \leq 15&lt;/math&gt;. Since &lt;math&gt;15 \cdot 2=30&lt;/math&gt;, the prime factorization of the number after &lt;math&gt;323&lt;/math&gt; needs to have either &lt;math&gt;17&lt;/math&gt; or &lt;math&gt;19&lt;/math&gt;. The next highest product after &lt;math&gt;17 \cdot 19&lt;/math&gt; is &lt;math&gt;17 \cdot 2 \cdot 10 =340&lt;/math&gt; or &lt;math&gt;19 \cdot 2 \cdot 9 =342&lt;/math&gt; &lt;math&gt;\implies \boxed{\text{(C) }340}&lt;/math&gt;. <br /> <br /> <br /> You can also tell by inspection that &lt;math&gt;19\cdot18 &gt; 20\cdot17&lt;/math&gt;, because &lt;math&gt;19\cdot18&lt;/math&gt; is closer to the side lengths of a square, which maximizes the product.<br /> <br /> ~bjhhar<br /> <br /> ==Videos==<br /> https://youtu.be/gMSkM6PLDwk<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2018|ab=B|num-b=20|num-a=22}}<br /> {{AMC12 box|year=2018|ab=B|num-b=18|num-a=20}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Intermediate Number Theory Problems]]</div> Bunny1 https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_20&diff=141219 2018 AMC 10B Problems/Problem 20 2021-01-01T00:44:53Z <p>Bunny1: </p> <hr /> <div>{{duplicate|[[2018 AMC 12B Problems|2018 AMC 12B #18]] and [[2018 AMC 10B Problems|2018 AMC 10B #20]]}}<br /> <br /> ==Problem==<br /> <br /> A function &lt;math&gt;f&lt;/math&gt; is defined recursively by &lt;math&gt;f(1)=f(2)=1&lt;/math&gt; and &lt;cmath&gt;f(n)=f(n-1)-f(n-2)+n&lt;/cmath&gt;for all integers &lt;math&gt;n \geq 3&lt;/math&gt;. What is &lt;math&gt;f(2018)&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)} \text{ 2016} \qquad \textbf{(B)} \text{ 2017} \qquad \textbf{(C)} \text{ 2018} \qquad \textbf{(D)} \text{ 2019} \qquad \textbf{(E)} \text{ 2020}&lt;/math&gt;<br /> <br /> ==Solution 1 (A Bit Bashy)==<br /> Start out by listing some terms of the sequence. <br /> &lt;cmath&gt;f(1)=1&lt;/cmath&gt;<br /> &lt;cmath&gt;f(2)=1&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;f(3)=3&lt;/cmath&gt;<br /> &lt;cmath&gt;f(4)=6&lt;/cmath&gt;<br /> &lt;cmath&gt;f(5)=8&lt;/cmath&gt;<br /> &lt;cmath&gt;f(6)=8&lt;/cmath&gt;<br /> &lt;cmath&gt;f(7)=7&lt;/cmath&gt;<br /> &lt;cmath&gt;f(8)=7&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;f(9)=9&lt;/cmath&gt;<br /> &lt;cmath&gt;f(10)=12&lt;/cmath&gt;<br /> &lt;cmath&gt;f(11)=14&lt;/cmath&gt;<br /> &lt;cmath&gt;f(12)=14&lt;/cmath&gt;<br /> &lt;cmath&gt;f(13)=13&lt;/cmath&gt;<br /> &lt;cmath&gt;f(14)=13&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;f(15)=15&lt;/cmath&gt;<br /> &lt;cmath&gt;.....&lt;/cmath&gt;<br /> Notice that &lt;math&gt;f(n)=n&lt;/math&gt; whenever &lt;math&gt;n&lt;/math&gt; is an odd multiple of &lt;math&gt;3&lt;/math&gt;, and the pattern of numbers that follow will always be &lt;math&gt;+3&lt;/math&gt;, &lt;math&gt;+2&lt;/math&gt;, &lt;math&gt;+0&lt;/math&gt;, &lt;math&gt;-1&lt;/math&gt;, &lt;math&gt;+0&lt;/math&gt;.<br /> The largest odd multiple of &lt;math&gt;3&lt;/math&gt; smaller than &lt;math&gt;2018&lt;/math&gt; is &lt;math&gt;2013&lt;/math&gt;, so we have<br /> &lt;cmath&gt;f(2013)=2013&lt;/cmath&gt;<br /> &lt;cmath&gt;f(2014)=2016&lt;/cmath&gt;<br /> &lt;cmath&gt;f(2015)=2018&lt;/cmath&gt;<br /> &lt;cmath&gt;f(2016)=2018&lt;/cmath&gt;<br /> &lt;cmath&gt;f(2017)=2017&lt;/cmath&gt;<br /> &lt;cmath&gt;f(2018)=\boxed{(B) 2017}.&lt;/cmath&gt;<br /> <br /> ==Solution 2 (Bashy Pattern Finding)==<br /> Writing out the first few values, we get:<br /> &lt;math&gt;1,1,3,6,8,8,7,7,9,12,14,14,13,13,15,18,20,20,19,19...&lt;/math&gt;. Examining, we see that every number &lt;math&gt;x&lt;/math&gt; where &lt;math&gt;x \equiv 1\pmod 6&lt;/math&gt; has &lt;math&gt;f(x)=x&lt;/math&gt;, &lt;math&gt;f(x+1)=f(x)=x&lt;/math&gt;, and &lt;math&gt;f(x-1)=f(x-2)=x+1&lt;/math&gt;. The greatest number that's &lt;math&gt;1\pmod{6}&lt;/math&gt; and less &lt;math&gt;2018&lt;/math&gt; is &lt;math&gt;2017&lt;/math&gt;, so we have &lt;math&gt;f(2017)=f(2018)=2017.&lt;/math&gt; &lt;math&gt;\boxed B&lt;/math&gt;<br /> <br /> ==Solution 3 (Algebra)==<br /> &lt;cmath&gt;f(n)=f(n-1)-f(n-2)+n.&lt;/cmath&gt;<br /> &lt;cmath&gt;f(n-1)=f(n-2)-f(n-3)+n-1.&lt;/cmath&gt;<br /> Adding the two equations, we have that &lt;cmath&gt;f(n)=2n-1-f(n-3).&lt;/cmath&gt; <br /> Hence, &lt;math&gt;f(n)+f(n-3)=2n-1&lt;/math&gt;.<br /> After plugging in &lt;math&gt;n-3&lt;/math&gt; to the equation above and doing some algebra, we have that &lt;math&gt;f(n)-f(n-6)=6&lt;/math&gt;.<br /> Consequently,<br /> &lt;cmath&gt;f(2018)-f(2012)=6.&lt;/cmath&gt;<br /> &lt;cmath&gt;f(2012)-f(2006)=6.&lt;/cmath&gt;<br /> &lt;cmath&gt;\ldots&lt;/cmath&gt;<br /> &lt;cmath&gt;f(8)-f(2)=6.&lt;/cmath&gt;<br /> Adding these &lt;math&gt;336&lt;/math&gt; equations up, we have that &lt;math&gt;f(2018)-f(2)=6 \cdot 336&lt;/math&gt; and &lt;math&gt;f(2018)=\boxed{2017}&lt;/math&gt;.<br /> <br /> ~AopsUser101<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2018|ab=B|num-b=19|num-a=21}}<br /> {{AMC12 box|year=2018|ab=B|num-b=17|num-a=19}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Introductory Algebra Problems]]</div> Bunny1 https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_19&diff=141218 2018 AMC 10B Problems/Problem 19 2021-01-01T00:44:24Z <p>Bunny1: </p> <hr /> <div>{{duplicate|[[2018 AMC 12B Problems|2018 AMC 12B #14]] and [[2018 AMC 10B Problems|2018 AMC 10B #19]]}}<br /> <br /> ==Problem==<br /> Joey and Chloe and their daughter Zoe all have the same birthday. Joey is &lt;math&gt;1&lt;/math&gt; year older than Chloe, and Zoe is exactly &lt;math&gt;1&lt;/math&gt; year old today. Today is the first of the &lt;math&gt;9&lt;/math&gt; birthdays on which Chloe's age will be an integral multiple of Zoe's age. What will be the sum of the two digits of Joey's age the next time his age is a multiple of Zoe's age?<br /> <br /> &lt;math&gt;\textbf{(A) } 7 \qquad \textbf{(B) } 8 \qquad \textbf{(C) } 9 \qquad \textbf{(D) } 10 \qquad \textbf{(E) } 11 &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Let Joey's age be &lt;math&gt;j&lt;/math&gt;, Chloe's age be &lt;math&gt;c&lt;/math&gt;, and we know that Zoe's age is &lt;math&gt;1&lt;/math&gt;. <br /> <br /> We know that there must be &lt;math&gt;9&lt;/math&gt; values &lt;math&gt;k\in\mathbb{Z}&lt;/math&gt; such that &lt;math&gt;c+k=a(1+k)&lt;/math&gt; where &lt;math&gt;a&lt;/math&gt; is an integer.<br /> <br /> Therefore, &lt;math&gt;c-1+(1+k)=a(1+k)&lt;/math&gt; and &lt;math&gt;c-1=(1+k)(a-1)&lt;/math&gt;. Therefore, we know that, as there are &lt;math&gt;9&lt;/math&gt; solutions for &lt;math&gt;k&lt;/math&gt;, there must be &lt;math&gt;9&lt;/math&gt; solutions for &lt;math&gt;c-1&lt;/math&gt;. We know that this must be a perfect square. Testing perfect squares, we see that &lt;math&gt;c-1=36&lt;/math&gt;, so &lt;math&gt;c=37&lt;/math&gt;. Therefore, &lt;math&gt;j=38&lt;/math&gt;. Now, since &lt;math&gt;j-1=37&lt;/math&gt;, by similar logic, &lt;math&gt;37=(1+k)(a-1)&lt;/math&gt;, so &lt;math&gt;k=36&lt;/math&gt; and Joey will be &lt;math&gt;38+36=74&lt;/math&gt; and the sum of the digits is &lt;math&gt;\boxed{\text{(E) }11}&lt;/math&gt;<br /> <br /> == Solution 2 ==<br /> <br /> Here's a different way of saying the above solution:<br /> <br /> If a number is a multiple of both Chloe's age and Zoe's age, then it is a multiple of their difference. Since the difference between their ages does not change, then that means the difference between their ages has &lt;math&gt;9&lt;/math&gt; factors. Therefore, the difference between Chloe and Zoe's age is &lt;math&gt;36&lt;/math&gt;, so Chloe is &lt;math&gt;37&lt;/math&gt;, and Joey is &lt;math&gt;38&lt;/math&gt;. The common factor that will divide both of their ages is &lt;math&gt;37&lt;/math&gt;, so Joey will be &lt;math&gt;74&lt;/math&gt;. &lt;math&gt;7 + 4 = \boxed{\text{(E) }11}&lt;/math&gt;<br /> <br /> ==Solution 3==<br /> Similar approach to above, just explained less concisely and more in terms of the problem (less algebra-y)<br /> <br /> Let &lt;math&gt;C+n&lt;/math&gt; denote Chloe's age, &lt;math&gt;J+n&lt;/math&gt; denote Joey's age, and &lt;math&gt;Z+n&lt;/math&gt; denote Zoe's age, where &lt;math&gt;n&lt;/math&gt; is the number of years from now. We are told that &lt;math&gt;C+n&lt;/math&gt; is a multiple of &lt;math&gt;Z+n&lt;/math&gt; exactly nine times. Because &lt;math&gt;Z+n&lt;/math&gt; is &lt;math&gt;1&lt;/math&gt; at &lt;math&gt;n=0&lt;/math&gt; and will increase until greater than &lt;math&gt;C-Z&lt;/math&gt;, it will hit every natural number less than &lt;math&gt;C-Z&lt;/math&gt;, including every factor of &lt;math&gt;C-Z&lt;/math&gt;. For &lt;math&gt;C+n&lt;/math&gt; to be an integral multiple of &lt;math&gt;Z+n&lt;/math&gt;, the difference &lt;math&gt;C-Z&lt;/math&gt; must also be a multiple of &lt;math&gt;Z&lt;/math&gt;, which happens if &lt;math&gt;Z&lt;/math&gt; is a factor of &lt;math&gt;C-Z&lt;/math&gt;. Therefore, &lt;math&gt;C-Z&lt;/math&gt; has nine factors. The smallest number that has nine positive factors is &lt;math&gt;2^23^2=36&lt;/math&gt; . (We want it to be small so that Joey will not have reached three digits of age before his age is a multiple of Zoe's.) We also know &lt;math&gt;Z=1&lt;/math&gt; and &lt;math&gt;J=C+1&lt;/math&gt;. Thus, &lt;cmath&gt;C-Z=36&lt;/cmath&gt; &lt;cmath&gt;J-Z=37&lt;/cmath&gt; By our above logic, the next time &lt;math&gt;J-Z&lt;/math&gt; is a multiple of &lt;math&gt;Z+n&lt;/math&gt; will occur when &lt;math&gt;Z+n&lt;/math&gt; is a factor of &lt;math&gt;J-Z&lt;/math&gt;. Because &lt;math&gt;37&lt;/math&gt; is prime, the next time this happens is at &lt;math&gt;Z+n=37&lt;/math&gt;, when &lt;math&gt;J+n=74&lt;/math&gt;. &lt;math&gt;7+4=\boxed{(\textbf{E}) 11}&lt;/math&gt;<br /> <br /> ==Solution 4==<br /> Denote Zoe's age with &lt;math&gt;n&lt;/math&gt;, then Chloe's age is &lt;math&gt;C-1+n&lt;/math&gt; where &lt;math&gt;C&lt;/math&gt; represents Chloe's age when Zoe is one. We must have &lt;math&gt;n | C-1+n&lt;/math&gt;. Obviously &lt;math&gt;n|n&lt;/math&gt;, therefore, &lt;math&gt;n | C-1&lt;/math&gt; for 9 values of &lt;math&gt;n&lt;/math&gt;, and therefore, &lt;math&gt;C-1&lt;/math&gt; has &lt;math&gt;9&lt;/math&gt; factors. &lt;math&gt;C-1&lt;/math&gt; either takes the form of &lt;math&gt;a^8&lt;/math&gt; (which is too large) or &lt;math&gt;a^2 b^2&lt;/math&gt;. &lt;math&gt;C&lt;/math&gt; must be less than &lt;math&gt;100&lt;/math&gt; and &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; must be prime, therefore the only answer is &lt;math&gt;C-1 = 36&lt;/math&gt;. Joey's age is &lt;math&gt;37+n&lt;/math&gt;, which is divisible by &lt;math&gt;n&lt;/math&gt; when &lt;math&gt;n|37&lt;/math&gt;, therefore the answer occurs when &lt;math&gt;n=37&lt;/math&gt; and Joey is &lt;math&gt;74&lt;/math&gt;. &lt;math&gt;7+4=\boxed{(\textbf{E}) 11}&lt;/math&gt;<br /> <br /> ==Video Solution==<br /> https://youtu.be/E2SbkCQ1V84<br /> <br /> ~savannahsolver<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2018|ab=B|num-b=18|num-a=20}}<br /> {{AMC12 box|year=2018|ab=B|num-b=13|num-a=15}}<br /> <br /> {{MAA Notice}}<br /> <br /> [[Category:Introductory Number Theory Problems]]</div> Bunny1 https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_15&diff=141217 2018 AMC 10B Problems/Problem 15 2021-01-01T00:43:42Z <p>Bunny1: </p> <hr /> <div>{{duplicate|[[2018 AMC 12B Problems|2018 AMC 12B #11]] and [[2018 AMC 10B Problems|2018 AMC 10B #15]]}}<br /> <br /> == Problem ==<br /> <br /> A closed box with a square base is to be wrapped with a square sheet of wrapping paper. The box is centered on the wrapping paper with the vertices of the base lying on the midlines of the square sheet of paper, as shown in the figure on the left. The four corners of the wrapping paper are to be folded up over the sides and brought together to meet at the center of the top of the box, point &lt;math&gt;A&lt;/math&gt; in the figure on the right. The box has base length &lt;math&gt;w&lt;/math&gt; and height &lt;math&gt;h&lt;/math&gt;. What is the area of the sheet of wrapping paper?<br /> &lt;asy&gt;defaultpen(fontsize(10pt));<br /> filldraw(((3,3)--(-3,3)--(-3,-3)--(3,-3)--cycle),lightgrey);<br /> dot((-3,3));<br /> label(&quot;$A$&quot;,(-3,3),NW);<br /> draw((1,3)--(-3,-1),dashed+linewidth(.5));<br /> draw((-1,3)--(3,-1),dashed+linewidth(.5));<br /> draw((-1,-3)--(3,1),dashed+linewidth(.5));<br /> draw((1,-3)--(-3,1),dashed+linewidth(.5));<br /> draw((0,2)--(2,0)--(0,-2)--(-2,0)--cycle,linewidth(.5));<br /> draw((0,3)--(0,-3),linetype(&quot;2.5 2.5&quot;)+linewidth(.5));<br /> draw((3,0)--(-3,0),linetype(&quot;2.5 2.5&quot;)+linewidth(.5));<br /> label('$w$',(-1,-1),SW);<br /> label('$w$',(1,-1),SE);<br /> draw((4.5,0)--(6.5,2)--(8.5,0)--(6.5,-2)--cycle);<br /> draw((4.5,0)--(8.5,0));<br /> draw((6.5,2)--(6.5,-2));<br /> label(&quot;$A$&quot;,(6.5,0),NW);<br /> dot((6.5,0));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 2(w+h)^2 \qquad \textbf{(B) } \frac{(w+h)^2}2 \qquad \textbf{(C) } 2w^2+4wh \qquad \textbf{(D) } 2w^2 \qquad \textbf{(E) } w^2h &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Consider one-quarter of the image (the wrapping paper is divided up into 4 congruent squares). The length of each dotted line is &lt;math&gt;h&lt;/math&gt;. The area of the rectangle that is &lt;math&gt;w&lt;/math&gt; by &lt;math&gt;h&lt;/math&gt; is &lt;math&gt;wh&lt;/math&gt;. The combined figure of the two triangles with base &lt;math&gt;h&lt;/math&gt; is a square with &lt;math&gt;h&lt;/math&gt; as its diagonal. Using the Pythagorean Theorem, each side of this square is &lt;math&gt;\sqrt{\frac{h^2}{2}}&lt;/math&gt;. Thus, the area is the side length squared which is &lt;math&gt;\frac{h^2}{2}&lt;/math&gt;. Similarly, the combined figure of the two triangles with base &lt;math&gt;w&lt;/math&gt; is a square with area &lt;math&gt;\frac{w^2}{2}&lt;/math&gt;. Adding all of these together, we get &lt;math&gt;\frac{w^2}{2} + \frac{h^2}{2} + wh&lt;/math&gt;. Since we have four of these areas in the entire wrapping paper, we multiply this by 4, getting &lt;math&gt;4(\frac{w^2}{2} + \frac{h^2}{2} + wh) = 2(w^2 + h^2 + 2wh) = \boxed{\textbf{(A) } 2(w+h)^2} \qquad&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> The sheet of paper is made out of the surface area of the box plus the sum of the four triangles. The surface area is &lt;math&gt;2w^2 + 2wh + 2wh&lt;/math&gt; which equals &lt;math&gt;2w^2 + 4wh&lt;/math&gt;.The four triangles each have a height and a base of &lt;math&gt;h&lt;/math&gt;, so they each have an area of &lt;math&gt;\frac{h^2}{2}&lt;/math&gt;. There are four of them, so multiplied by four is &lt;math&gt;2h^2&lt;/math&gt;. Together, paper's area is &lt;math&gt;2w^2 + 4wh + 2h^2&lt;/math&gt;. This can be factored and written as &lt;math&gt;\boxed{\textbf{(A) } 2(w+h)^2} \qquad&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> The sheet of paper is made out of &lt;math&gt;4&lt;/math&gt; squares. Each square has a side length of &lt;math&gt;\frac{w}{\sqrt{2}} + \frac{h}{\sqrt{2}}&lt;/math&gt;, which we get from the Pythagorean Theorem (a &lt;math&gt;45-45-90&lt;/math&gt; triangle's legs is the hypotenuse divided by &lt;math&gt;\sqrt2&lt;/math&gt;). Thus, to find the area of the entire paper, we square our side length and multiply by 4. So, &lt;math&gt;4\cdot\left(\frac{w}{\sqrt{2}} + \frac{h}{\sqrt{2}}\right)^2 \Rightarrow \boxed{\textbf{(A) } 2(w+h)^2} \qquad&lt;/math&gt;, which is the answer.<br /> <br /> Sol by IronicNinja<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2018|ab=B|num-b=14|num-a=16}}<br /> {{AMC12 box|year=2018|ab=B|num-b=10|num-a=12}}<br /> <br /> {{MAA Notice}}<br /> <br /> [[Category:Introductory Geometry Problems]]</div> Bunny1 https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_14&diff=141216 2018 AMC 10B Problems/Problem 14 2021-01-01T00:43:16Z <p>Bunny1: </p> <hr /> <div>{{duplicate|[[2018 AMC 12B Problems|2018 AMC 12B #10]] and [[2018 AMC 10B Problems|2018 AMC 10B #14]]}}<br /> <br /> == Problem ==<br /> <br /> A list of &lt;math&gt;2018&lt;/math&gt; positive integers has a unique mode, which occurs exactly &lt;math&gt;10&lt;/math&gt; times. What is the least number of distinct values that can occur in the list?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 202\qquad\textbf{(B)}\ 223\qquad\textbf{(C)}\ 224\qquad\textbf{(D)}\ 225\qquad\textbf{(E)}\ 234&lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> To minimize the number of distinct values, we want to maximize the number of times they appear. So, we could have &lt;math&gt;223&lt;/math&gt; numbers appear &lt;math&gt;9&lt;/math&gt; times, &lt;math&gt;1&lt;/math&gt; number appear once, and the mode appear &lt;math&gt;10&lt;/math&gt; times, giving us a total of &lt;math&gt;223 + 1 + 1&lt;/math&gt; = &lt;math&gt;\boxed{\textbf{(D) } 225}&lt;/math&gt;<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2018|ab=B|num-b=13|num-a=15}}<br /> {{AMC12 box|year=2018|ab=B|num-b=9|num-a=11}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Introductory Combinatorics Problems]]</div> Bunny1 https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_2&diff=141215 2018 AMC 10B Problems/Problem 2 2021-01-01T00:41:53Z <p>Bunny1: </p> <hr /> <div>{{duplicate|[[2018 AMC 12B Problems|2018 AMC 12B #2]] and [[2018 AMC 10B Problems|2018 AMC 10B #2]]}}<br /> <br /> ==Problem==<br /> Sam drove 96 miles in 90 minutes. His average speed during the first 30 minutes was 60 mph (miles per hour), and his average speed during the second 30 minutes was 65 mph. What was his average speed, in mph, during the last 30 minutes?<br /> <br /> &lt;math&gt;\textbf{(A) } 64 \qquad \textbf{(B) } 65 \qquad \textbf{(C) } 66 \qquad \textbf{(D) } 67 \qquad \textbf{(E) } 68&lt;/math&gt;<br /> <br /> <br /> ==Solution 1 ==<br /> <br /> Let Sam drive at exactly &lt;math&gt;60&lt;/math&gt; mph in the first half hour, &lt;math&gt;65&lt;/math&gt; mph in the second half hour, and &lt;math&gt;x&lt;/math&gt; mph in the third half hour.<br /> <br /> Due to &lt;math&gt;rt = d&lt;/math&gt;, and that &lt;math&gt;30&lt;/math&gt; min is half an hour, he covered &lt;math&gt;60 \cdot \frac{1}{2} = 30&lt;/math&gt; miles in the first &lt;math&gt;30&lt;/math&gt; mins.<br /> <br /> SImilarly, he covered &lt;math&gt;\frac{65}{2}&lt;/math&gt; miles in the &lt;math&gt;2&lt;/math&gt;nd half hour period. <br /> <br /> The problem states that Sam drove &lt;math&gt;96&lt;/math&gt; miles in &lt;math&gt;90&lt;/math&gt; min, so that means that he must have covered &lt;math&gt;96 - \left(30 + \frac{65}{2}\right) = 33 \frac{1}{2}&lt;/math&gt; miles in the third half hour period.<br /> <br /> &lt;math&gt;rt = d&lt;/math&gt;, so &lt;math&gt;x \cdot \frac{1}{2} = 33 \frac{1}{2}&lt;/math&gt;.<br /> <br /> Therefore, Sam was driving &lt;math&gt;\boxed{\textbf{(D) } 67}&lt;/math&gt; miles per hour in the third half hour.<br /> <br /> ==Solution 2 (Faster)==<br /> <br /> The average speed for the total trip is &lt;cmath&gt;\text{avg. speed} = \frac{96}{\frac{3}{2}} = 64.&lt;/cmath&gt; Therefore the average speed for the total trip is the average of the average speeds of the three intrevals. So we have &lt;math&gt;64 = \frac{60 + 65 + x}{3}&lt;/math&gt; and solving for &lt;math&gt;x = 67&lt;/math&gt;. So the answer is &lt;math&gt;\boxed{\textbf{(D) } 67}&lt;/math&gt;. <br /> ~coolmath_2018<br /> <br /> ==Video Solution==<br /> https://youtu.be/77dDIzKprzA<br /> <br /> ~savannahsolver<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2018|ab=B|num-b=1|num-a=3}}<br /> {{AMC12 box|year=2018|ab=B|num-b=1|num-a=3}}<br /> {{MAA Notice}}</div> Bunny1 https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_1&diff=141214 2018 AMC 10B Problems/Problem 1 2021-01-01T00:41:18Z <p>Bunny1: </p> <hr /> <div>{{duplicate|[[2018 AMC 12B Problems|2018 AMC 12B #1]] and [[2018 AMC 10B Problems|2018 AMC 10B #1]]}}<br /> <br /> ==Problem==<br /> Kate bakes a &lt;math&gt;20&lt;/math&gt;-inch by &lt;math&gt;18&lt;/math&gt;-inch pan of cornbread. The cornbread is cut into pieces that measure &lt;math&gt;2&lt;/math&gt; inches by &lt;math&gt;2&lt;/math&gt; inches. How many pieces of cornbread does the pan contain?<br /> <br /> &lt;math&gt;\textbf{(A) } 90 \qquad \textbf{(B) } 100 \qquad \textbf{(C) } 180 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 360&lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> <br /> The area of the pan is &lt;math&gt;20\cdot18&lt;/math&gt; = &lt;math&gt;360&lt;/math&gt;. Since the area of each piece is &lt;math&gt;4&lt;/math&gt;, there are &lt;math&gt;\frac{360}{4} = 90&lt;/math&gt; pieces. Thus, the answer is &lt;math&gt;\boxed{A}&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> <br /> By dividing each of the dimensions by &lt;math&gt;2&lt;/math&gt;, we get a &lt;math&gt;10\times9&lt;/math&gt; grid which makes &lt;math&gt;90&lt;/math&gt; pieces. Thus, the answer is &lt;math&gt;\boxed{A}&lt;/math&gt;.<br /> <br /> ==Video Solution==<br /> https://youtu.be/o5MUHOmF1zo<br /> <br /> ~savannahsolver<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2018|ab=B|before=First Problem|num-a=2}}<br /> {{AMC12 box|year=2018|ab=B|before=First Problem|num-a=2}}<br /> {{MAA Notice}}</div> Bunny1