https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Tbnrfrags&feedformat=atom AoPS Wiki - User contributions [en] 2021-04-15T11:14:51Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_21&diff=114483 2018 AMC 10B Problems/Problem 21 2020-01-08T03:35:36Z <p>Tbnrfrags: Undo revision 114482 by Tbnrfrags (talk)</p> <hr /> <div>==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;\fbox{\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 Shurong.ge<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==<br /> https://www.youtube.com/watch?v=KHaLXNAkDWE<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> Tbnrfrags https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_21&diff=114482 2018 AMC 10B Problems/Problem 21 2020-01-08T03:30:17Z <p>Tbnrfrags: /* Solution 3 */</p> <hr /> <div>==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;\fbox{\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 Shurong.ge<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 /> ~bjhhar<br /> <br /> ==Video==<br /> https://www.youtube.com/watch?v=KHaLXNAkDWE<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> Tbnrfrags https://artofproblemsolving.com/wiki/index.php?title=Simon%27s_Favorite_Factoring_Trick&diff=101085 Simon's Favorite Factoring Trick 2019-01-29T23:44:03Z <p>Tbnrfrags: /* About */</p> <hr /> <div><br /> ==About==<br /> '''Dr. Simon's Favorite Factoring Trick''' (abbreviated '''SFFT''') is a special factorization first popularized by [[AoPS]] user [[user:ComplexZeta |Simon Rubinstein-Salzedo]].<br /> <br /> ==The General Statement==<br /> The general statement of SFFT is: &lt;math&gt;{xy}+{xk}+{jy}+{jk}=(x+j)(y+k)&lt;/math&gt;. Two special common cases are: &lt;math&gt;xy + x + y + 1 = (x+1)(y+1)&lt;/math&gt; and &lt;math&gt;xy - x - y +1 = (x-1)(y-1)&lt;/math&gt;.<br /> <br /> The act of adding &lt;math&gt;{jk}&lt;/math&gt; to &lt;math&gt;{xy}+{xk}+{jy}&lt;/math&gt; in order to be able to factor it could be called &quot;completing the rectangle&quot; in analogy to the more familiar &quot;completing the square.&quot;<br /> <br /> == Applications ==<br /> This factorization frequently shows up on contest problems, especially those heavy on algebraic manipulation. Usually &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; are variables and &lt;math&gt;j,k&lt;/math&gt; are known constants. Also, it is typically necessary to add the &lt;math&gt;jk&lt;/math&gt; term to both sides to perform the factorization.<br /> <br /> == Problems ==<br /> ===Introductory===<br /> *Two different [[prime number]]s between &lt;math&gt;4&lt;/math&gt; and &lt;math&gt;18&lt;/math&gt; are chosen. When their sum is subtracted from their product, which of the following numbers could be obtained?<br /> <br /> &lt;math&gt; \mathrm{(A) \ 21 } \qquad \mathrm{(B) \ 60 } \qquad \mathrm{(C) \ 119 } \qquad \mathrm{(D) \ 180 } \qquad \mathrm{(E) \ 231 } &lt;/math&gt;<br /> <br /> ([[2000 AMC 12/Problem 6|Source]])<br /> <br /> ===Intermediate===<br /> *&lt;math&gt;m, n&lt;/math&gt; are integers such that &lt;math&gt;m^2 + 3m^2n^2 = 30n^2 + 517&lt;/math&gt;. Find &lt;math&gt;3m^2n^2&lt;/math&gt;.<br /> <br /> ([[1987 AIME Problems/Problem 5|Source]])<br /> <br /> *The integer &lt;math&gt;N&lt;/math&gt; is positive. There are exactly &lt;math&gt;2005&lt;/math&gt; pairs &lt;math&gt;(x, y)&lt;/math&gt; of positive integers satisfying:<br /> <br /> &lt;cmath&gt;\frac 1x +\frac 1y = \frac 1N&lt;/cmath&gt;<br /> <br /> Prove that &lt;math&gt;N&lt;/math&gt; is a perfect square. (British Mathematical Olympiad Round 2, 2005)<br /> <br /> == See Also ==<br /> * [[Algebra]]<br /> * [[Factoring]]<br /> <br /> [[Category:Elementary algebra]]<br /> [[Category:Theorems]]</div> Tbnrfrags https://artofproblemsolving.com/wiki/index.php?title=Simon%27s_Favorite_Factoring_Trick&diff=101084 Simon's Favorite Factoring Trick 2019-01-29T23:43:40Z <p>Tbnrfrags: /* About */</p> <hr /> <div><br /> ==About==<br /> '''Dr. Simon's Favorite Factoring Trick''' (abbreviated '''SFFT''') is a special factorization first popularized by [[AoPS]] user [[user:ComplexZeta | Bob Rubinstein-Salzedo]].<br /> <br /> ==The General Statement==<br /> The general statement of SFFT is: &lt;math&gt;{xy}+{xk}+{jy}+{jk}=(x+j)(y+k)&lt;/math&gt;. Two special common cases are: &lt;math&gt;xy + x + y + 1 = (x+1)(y+1)&lt;/math&gt; and &lt;math&gt;xy - x - y +1 = (x-1)(y-1)&lt;/math&gt;.<br /> <br /> The act of adding &lt;math&gt;{jk}&lt;/math&gt; to &lt;math&gt;{xy}+{xk}+{jy}&lt;/math&gt; in order to be able to factor it could be called &quot;completing the rectangle&quot; in analogy to the more familiar &quot;completing the square.&quot;<br /> <br /> == Applications ==<br /> This factorization frequently shows up on contest problems, especially those heavy on algebraic manipulation. Usually &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; are variables and &lt;math&gt;j,k&lt;/math&gt; are known constants. Also, it is typically necessary to add the &lt;math&gt;jk&lt;/math&gt; term to both sides to perform the factorization.<br /> <br /> == Problems ==<br /> ===Introductory===<br /> *Two different [[prime number]]s between &lt;math&gt;4&lt;/math&gt; and &lt;math&gt;18&lt;/math&gt; are chosen. When their sum is subtracted from their product, which of the following numbers could be obtained?<br /> <br /> &lt;math&gt; \mathrm{(A) \ 21 } \qquad \mathrm{(B) \ 60 } \qquad \mathrm{(C) \ 119 } \qquad \mathrm{(D) \ 180 } \qquad \mathrm{(E) \ 231 } &lt;/math&gt;<br /> <br /> ([[2000 AMC 12/Problem 6|Source]])<br /> <br /> ===Intermediate===<br /> *&lt;math&gt;m, n&lt;/math&gt; are integers such that &lt;math&gt;m^2 + 3m^2n^2 = 30n^2 + 517&lt;/math&gt;. Find &lt;math&gt;3m^2n^2&lt;/math&gt;.<br /> <br /> ([[1987 AIME Problems/Problem 5|Source]])<br /> <br /> *The integer &lt;math&gt;N&lt;/math&gt; is positive. There are exactly &lt;math&gt;2005&lt;/math&gt; pairs &lt;math&gt;(x, y)&lt;/math&gt; of positive integers satisfying:<br /> <br /> &lt;cmath&gt;\frac 1x +\frac 1y = \frac 1N&lt;/cmath&gt;<br /> <br /> Prove that &lt;math&gt;N&lt;/math&gt; is a perfect square. (British Mathematical Olympiad Round 2, 2005)<br /> <br /> == See Also ==<br /> * [[Algebra]]<br /> * [[Factoring]]<br /> <br /> [[Category:Elementary algebra]]<br /> [[Category:Theorems]]</div> Tbnrfrags