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Difference between revisions of "2023 AIME I Problems/Problem 4"
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==Solution==
 
==Solution==
 
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. This means <math>m</math> must contain <math>7\cdot11\cdot13</math>. Also, <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. This means <math>m</math> must contain <math>7\cdot11\cdot13</math>. Also, <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{012}</cmath>
  
 
~chem1kall
 
~chem1kall

Revision as of 14:26, 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. This means $m$ must contain $7\cdot11\cdot13$. Also, $m$ 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 $1365\cdot26\cdot270\cdot7\cdot11\cdot13 = 2\cdot3^2\cdot5\cdot7^3\cdot11\cdot13^4$. \[1+2+1+3+1+4=\boxed{012}\]

~chem1kall

Solution 2

The prime factorization of $13!$ is $2^{10} \cdot 3^5 \cdot 5^2 \cdot 7 \cdot 11 \cdot 13$. To get $\frac{13!}{m}$ a perfect square, we must have $m = 2^{2x} \cdot 3^{1 + 2y} \cdot 5^{2z} \cdot 7 \cdot 11 \cdot 13$, where $x \in \left\{ 0, 1, \cdots , 5 \right\}$, $y \in \left\{ 0, 1, 2 \right\}$, $z \in \left\{ 0, 1 \right\}$.

Hence, the sum of all feasible $m$ is \begin{align*} \sum_{x=0}^5 \sum_{y=0}^2 \sum_{z=0}^1 2^{2x} \cdot 3^{1 + 2y} \cdot 5^{2z} \cdot 7 \cdot 11 \cdot 13 & = \left( \sum_{x=0}^5 2^{2x} \right) \left( \sum_{y=0}^2 3^{1 + 2y} \right) \left( \sum_{z=0}^1 5^{2z} \right) 7 \cdot 11 \cdot 13 \\ & = \frac{4^6 - 1}{4-1} \cdot \frac{3 \cdot \left( 9^3 - 1 \right)}{9 - 1} \cdot \frac{25^2 - 1}{25 - 1} \cdot 7 \cdot 11 \cdot 13 \\ & = 2 \cdot 3^2 \cdot 5 \cdot 7^3 \cdot 11 \cdot 13^4 . \end{align*}

Therefore, the answer is \begin{align*} 1 + 2 + 1 + 3 + 1 + 4 & = \boxed{\textbf{(012) }} . \end{align*}

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

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