Difference between revisions of "2023 AIME I Problems/Problem 4"
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+ | ==Solution 2== | ||
+ | |||
+ | The prime factorization of <math>13!</math> is <math>2^{10} \cdot 3^5 \cdot 5^2 \cdot 7 \cdot 11 \cdot 13</math>. | ||
+ | To get <math>\frac{13!}{m}</math> a perfect square, we must have <math>m = 2^{2x} \cdot 3^{1 + 2y} \cdot 5^{2z} \cdot 7 \cdot 11 \cdot 13</math>, where <math>x \in \left\{ 0, 1, \cdots , 5 \right\}</math>, <math>y \in \left\{ 0, 1, 2 \right\}</math>, <math>z \in \left\{ 0, 1 \right\}</math>. | ||
+ | |||
+ | Hence, the sum of all feasible <math>m</math> is | ||
+ | <cmath> | ||
+ | \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*} | ||
+ | </cmath> | ||
+ | |||
+ | Therefore, the answer is | ||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | 1 + 2 + 1 + 3 + 1 + 4 | ||
+ | & = \boxed{\textbf{(012) }} . | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | |||
+ | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) |
Revision as of 13:35, 8 February 2023
Problem 4
Unofficial problem: The sum of all positive integers such that is a perfect square can be written as , where and are positive integers. Find
Solution
We first rewrite 13! as a prime factorization, which is For the fraction to be a square, it needs each prime to be an even power. This means must contain . Also, 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 is , which simplifies to .
~chem1kall
Solution 2
The prime factorization of is . To get a perfect square, we must have , where , , .
Hence, the sum of all feasible is
Therefore, the answer is
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)