https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Tbnrfrags&feedformat=atomAoPS Wiki - User contributions [en]2024-03-28T19:18:38ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_21&diff=1144832018 AMC 10B Problems/Problem 212020-01-08T03:35:36Z<p>Tbnrfrags: Undo revision 114482 by Tbnrfrags (talk)</p>
<hr />
<div>==Problem==<br />
Mary chose an even <math>4</math>-digit number <math>n</math>. She wrote down all the divisors of <math>n</math> in increasing order from left to right: <math>1,2,...,\dfrac{n}{2},n</math>. At some moment Mary wrote <math>323</math> as a divisor of <math>n</math>. What is the smallest possible value of the next divisor written to the right of <math>323</math>?<br />
<br />
<math>\textbf{(A) } 324 \qquad \textbf{(B) } 330 \qquad \textbf{(C) } 340 \qquad \textbf{(D) } 361 \qquad \textbf{(E) } 646</math><br />
<br />
==Solution 1==<br />
<br />
Since prime factorizing <math>323</math> gives you <math>17 \cdot 19</math>, the desired answer needs to be a multiple of <math>17</math> or <math>19</math>, this is because if it is not a multiple of <math>17</math> or <math>19</math>, <math>n</math> will be more than a <math>4</math> digit number. For example, if the answer were to instead be <math>324</math>, <math>n</math> would have to be a multiple of <math>2^2 * 3^4 * 17 * 19</math> for both <math>323</math> and <math>324</math> to be a valid factor, meaning <math>n</math> would have to be at least <math>104652</math>, which is too big. Looking at the answer choices, <math>\text{(A) }324</math> and <math>\text{(B) }330</math> are both not a multiple of neither 17 nor 19, <math>\text{(C) }340</math> is divisible by <math>17</math>. <math>\text{(D) }361</math> is divisible by <math>19</math>, and <math>\text{(E) }646</math> is divisible by both <math>17</math> and <math>19</math>. Since <math>\fbox{\text{(C) }340}</math> is the smallest number divisible by either <math>17</math> or <math>19</math> it is the answer. Checking, we can see that <math>n</math> would be <math>6460</math>, a four-digit number. Note that <math>n</math> is also divisible by <math>2</math>, one of the listed divisors of <math>n</math>. (If <math>n</math> was not divisible by <math>2</math>, 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 <math>k</math>. Suppose <math>\gcd(k,323)=1</math>. Then, as <math>323|n, k|n</math>, therefore, <math>323\cdot k|n.</math> However, because <math>k>323</math>, <math>323k>323\cdot 324>9999</math>. Therefore, <math>\gcd(k,323)>1</math>. Note that <math>323=17\cdot 19</math>. Therefore, the smallest the GCD can be is <math>17</math> and our answer is <math>323+17=\boxed{\text{(C) }340}</math>.<br />
<br />
==Solution 3==<br />
Again, recognize <math>323=17 \cdot 19</math>. The 4-digit number is even, so its prime factorization must then be <math>17 \cdot 19 \cdot 2 \cdot n</math>. Also, <math>1000\leq 646n \leq 9998</math>, so <math>2 \leq n \leq 15</math>. Since <math>15 \cdot 2=30</math>, the prime factorization of the number after <math>323</math> needs to have either <math>17</math> or <math>19</math>. The next highest product after <math>17 \cdot 19</math> is <math>17 \cdot 2 \cdot 10 =340</math> or <math>19 \cdot 2 \cdot 9 =342</math> <math>\implies \boxed{\text{(C) }340}</math>. <br />
<br />
<br />
You can also tell by inspection that <math>19\cdot18 > 20\cdot17</math>, because <math>19\cdot18</math> 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>Tbnrfragshttps://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_21&diff=1144822018 AMC 10B Problems/Problem 212020-01-08T03:30:17Z<p>Tbnrfrags: /* Solution 3 */</p>
<hr />
<div>==Problem==<br />
Mary chose an even <math>4</math>-digit number <math>n</math>. She wrote down all the divisors of <math>n</math> in increasing order from left to right: <math>1,2,...,\dfrac{n}{2},n</math>. At some moment Mary wrote <math>323</math> as a divisor of <math>n</math>. What is the smallest possible value of the next divisor written to the right of <math>323</math>?<br />
<br />
<math>\textbf{(A) } 324 \qquad \textbf{(B) } 330 \qquad \textbf{(C) } 340 \qquad \textbf{(D) } 361 \qquad \textbf{(E) } 646</math><br />
<br />
==Solution 1==<br />
<br />
Since prime factorizing <math>323</math> gives you <math>17 \cdot 19</math>, the desired answer needs to be a multiple of <math>17</math> or <math>19</math>, this is because if it is not a multiple of <math>17</math> or <math>19</math>, <math>n</math> will be more than a <math>4</math> digit number. For example, if the answer were to instead be <math>324</math>, <math>n</math> would have to be a multiple of <math>2^2 * 3^4 * 17 * 19</math> for both <math>323</math> and <math>324</math> to be a valid factor, meaning <math>n</math> would have to be at least <math>104652</math>, which is too big. Looking at the answer choices, <math>\text{(A) }324</math> and <math>\text{(B) }330</math> are both not a multiple of neither 17 nor 19, <math>\text{(C) }340</math> is divisible by <math>17</math>. <math>\text{(D) }361</math> is divisible by <math>19</math>, and <math>\text{(E) }646</math> is divisible by both <math>17</math> and <math>19</math>. Since <math>\fbox{\text{(C) }340}</math> is the smallest number divisible by either <math>17</math> or <math>19</math> it is the answer. Checking, we can see that <math>n</math> would be <math>6460</math>, a four-digit number. Note that <math>n</math> is also divisible by <math>2</math>, one of the listed divisors of <math>n</math>. (If <math>n</math> was not divisible by <math>2</math>, 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 <math>k</math>. Suppose <math>\gcd(k,323)=1</math>. Then, as <math>323|n, k|n</math>, therefore, <math>323\cdot k|n.</math> However, because <math>k>323</math>, <math>323k>323\cdot 324>9999</math>. Therefore, <math>\gcd(k,323)>1</math>. Note that <math>323=17\cdot 19</math>. Therefore, the smallest the GCD can be is <math>17</math> and our answer is <math>323+17=\boxed{\text{(C) }340}</math>.<br />
<br />
==Solution 3==<br />
Again, recognize <math>323=17 \cdot 19</math>. The 4-digit number is even, so its prime factorization must then be <math>17 \cdot 19 \cdot 2 \cdot n</math>. Also, <math>1000\leq 646n \leq 9998</math>, so <math>2 \leq n \leq 15</math>. Since <math>15 \cdot 2=30</math>, the prime factorization of the number after <math>323</math> needs to have either <math>17</math> or <math>19</math>. The next highest product after <math>17 \cdot 19</math> is <math>17 \cdot 2 \cdot 10 =340</math> or <math>19 \cdot 2 \cdot 9 =342</math> <math>\implies \boxed{\text{(C) }340}</math>. <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>Tbnrfragshttps://artofproblemsolving.com/wiki/index.php?title=Simon%27s_Favorite_Factoring_Trick&diff=101085Simon's Favorite Factoring Trick2019-01-29T23:44:03Z<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: <math>{xy}+{xk}+{jy}+{jk}=(x+j)(y+k)</math>. Two special common cases are: <math>xy + x + y + 1 = (x+1)(y+1)</math> and <math>xy - x - y +1 = (x-1)(y-1)</math>.<br />
<br />
The act of adding <math>{jk}</math> to <math>{xy}+{xk}+{jy}</math> in order to be able to factor it could be called "completing the rectangle" in analogy to the more familiar "completing the square."<br />
<br />
== Applications ==<br />
This factorization frequently shows up on contest problems, especially those heavy on algebraic manipulation. Usually <math>x</math> and <math>y</math> are variables and <math>j,k</math> are known constants. Also, it is typically necessary to add the <math>jk</math> term to both sides to perform the factorization.<br />
<br />
== Problems ==<br />
===Introductory===<br />
*Two different [[prime number]]s between <math>4</math> and <math>18</math> are chosen. When their sum is subtracted from their product, which of the following numbers could be obtained?<br />
<br />
<math> \mathrm{(A) \ 21 } \qquad \mathrm{(B) \ 60 } \qquad \mathrm{(C) \ 119 } \qquad \mathrm{(D) \ 180 } \qquad \mathrm{(E) \ 231 } </math><br />
<br />
([[2000 AMC 12/Problem 6|Source]])<br />
<br />
===Intermediate===<br />
*<math>m, n</math> are integers such that <math>m^2 + 3m^2n^2 = 30n^2 + 517</math>. Find <math>3m^2n^2</math>.<br />
<br />
([[1987 AIME Problems/Problem 5|Source]])<br />
<br />
*The integer <math>N</math> is positive. There are exactly <math>2005</math> pairs <math>(x, y)</math> of positive integers satisfying:<br />
<br />
<cmath>\frac 1x +\frac 1y = \frac 1N</cmath><br />
<br />
Prove that <math>N</math> 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>Tbnrfragshttps://artofproblemsolving.com/wiki/index.php?title=Simon%27s_Favorite_Factoring_Trick&diff=101084Simon's Favorite Factoring Trick2019-01-29T23:43:40Z<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: <math>{xy}+{xk}+{jy}+{jk}=(x+j)(y+k)</math>. Two special common cases are: <math>xy + x + y + 1 = (x+1)(y+1)</math> and <math>xy - x - y +1 = (x-1)(y-1)</math>.<br />
<br />
The act of adding <math>{jk}</math> to <math>{xy}+{xk}+{jy}</math> in order to be able to factor it could be called "completing the rectangle" in analogy to the more familiar "completing the square."<br />
<br />
== Applications ==<br />
This factorization frequently shows up on contest problems, especially those heavy on algebraic manipulation. Usually <math>x</math> and <math>y</math> are variables and <math>j,k</math> are known constants. Also, it is typically necessary to add the <math>jk</math> term to both sides to perform the factorization.<br />
<br />
== Problems ==<br />
===Introductory===<br />
*Two different [[prime number]]s between <math>4</math> and <math>18</math> are chosen. When their sum is subtracted from their product, which of the following numbers could be obtained?<br />
<br />
<math> \mathrm{(A) \ 21 } \qquad \mathrm{(B) \ 60 } \qquad \mathrm{(C) \ 119 } \qquad \mathrm{(D) \ 180 } \qquad \mathrm{(E) \ 231 } </math><br />
<br />
([[2000 AMC 12/Problem 6|Source]])<br />
<br />
===Intermediate===<br />
*<math>m, n</math> are integers such that <math>m^2 + 3m^2n^2 = 30n^2 + 517</math>. Find <math>3m^2n^2</math>.<br />
<br />
([[1987 AIME Problems/Problem 5|Source]])<br />
<br />
*The integer <math>N</math> is positive. There are exactly <math>2005</math> pairs <math>(x, y)</math> of positive integers satisfying:<br />
<br />
<cmath>\frac 1x +\frac 1y = \frac 1N</cmath><br />
<br />
Prove that <math>N</math> 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