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Difference between revisions of "2023 AIME I Problems/Problem 4"

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Unofficial problem: The sum of all positive integers <math>m</math> such that <math>\frac{13!}{m}</math> is a perfect square can be written as <math>2^a3^b5^c7^d11^e13^f</math>, where <math>a,b,c,d,e,</math> and <math>f</math> are positive integers. Find <math>a+b+c+d+e+f.</math>
 
Unofficial problem: The sum of all positive integers <math>m</math> such that <math>\frac{13!}{m}</math> is a perfect square can be written as <math>2^a3^b5^c7^d11^e13^f</math>, where <math>a,b,c,d,e,</math> and <math>f</math> are positive integers. Find <math>a+b+c+d+e+f.</math>
 
==Solution==
 
==Solution==
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===Solution 1===
 
 
We first rewrite 13! as a prime factorization, which is <math>2^{10}\cdot3^5\cdot5^2\cdot7\cdot11\cdot13.</math>
 
We first rewrite 13! as a prime factorization, which is <math>2^{10}\cdot3^5\cdot5^2\cdot7\cdot11\cdot13.</math>
 
For the fraction to be a square, it needs each prime to be an even power. <math>m</math> must contain <math>7\cdot\cdot11\cdot13</math>. <math>m</math> can contain any even power of 2 up to 10, any odd power of 3 up to 5, and any even power of 5 up to 2. The sum of <math>m</math> is <math>(7\cdot11\cdot13)(2^0+2^2+2^4+2^6+2^8+2^{10})(3^1+3^3+3^5)(5^0+5^2)</math>, which simplifies to <math>1365\cdot26\cdot270\cdot7\cdot11\cdot13 = 2\cdot3^2\cdot5\cdot7^3\cdot11\cdot13^4</math>. <cmath>1+2+1+3+1+4=\boxed{12}</cmath>
 
For the fraction to be a square, it needs each prime to be an even power. <math>m</math> must contain <math>7\cdot\cdot11\cdot13</math>. <math>m</math> can contain any even power of 2 up to 10, any odd power of 3 up to 5, and any even power of 5 up to 2. The sum of <math>m</math> is <math>(7\cdot11\cdot13)(2^0+2^2+2^4+2^6+2^8+2^{10})(3^1+3^3+3^5)(5^0+5^2)</math>, which simplifies to <math>1365\cdot26\cdot270\cdot7\cdot11\cdot13 = 2\cdot3^2\cdot5\cdot7^3\cdot11\cdot13^4</math>. <cmath>1+2+1+3+1+4=\boxed{12}</cmath>
  
 
~chem1kall
 
~chem1kall

Revision as of 13:25, 8 February 2023

Problem 4

Unofficial problem: The sum of all positive integers $m$ such that $\frac{13!}{m}$ is a perfect square can be written as $2^a3^b5^c7^d11^e13^f$, where $a,b,c,d,e,$ and $f$ are positive integers. Find $a+b+c+d+e+f.$

Solution

We first rewrite 13! as a prime factorization, which is $2^{10}\cdot3^5\cdot5^2\cdot7\cdot11\cdot13.$ For the fraction to be a square, it needs each prime to be an even power. $m$ must contain $7\cdot\cdot11\cdot13$. $m$ can contain any even power of 2 up to 10, any odd power of 3 up to 5, and any even power of 5 up to 2. The sum of $m$ is $(7\cdot11\cdot13)(2^0+2^2+2^4+2^6+2^8+2^{10})(3^1+3^3+3^5)(5^0+5^2)$, which simplifies to $1365\cdot26\cdot270\cdot7\cdot11\cdot13 = 2\cdot3^2\cdot5\cdot7^3\cdot11\cdot13^4$. \[1+2+1+3+1+4=\boxed{12}\]

~chem1kall

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