AIME 2023 I · 第 4 题

Solution 1

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 $2^{10}$, any odd power of $3$ up to $3^{5}$, and any even power of $5$ up to $5^{2}$. The sum of $m$ is

$$(2^0+2^2+2^4+2^6+2^8+2^{10})(3^1+3^3+3^5)(5^0+5^2)(7^1)(11^1)(13^1) =$$ $$1365\cdot273\cdot26\cdot7\cdot11\cdot13 = 2\cdot3^2\cdot5\cdot7^3\cdot11\cdot13^4.$$ Therefore, the answer is $1+2+1+3+1+4=\boxed{012}$.

~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

$$\begin{aligned} \sum_{x=0}^5 \sum_{y=0}^2 \sum_{z=0}^1 2^{2x} \cdot 3^{1 + 2y} \cdot 5^{2z} \cdot 7 \cdot 11 \cdot 13 & = \left( \sum_{x=0}^5 2^{2x} \right) \left( \sum_{y=0}^2 3^{1 + 2y} \right) \left( \sum_{z=0}^1 5^{2z} \right) 7 \cdot 11 \cdot 13 \\ & = \frac{4^6 - 1}{4-1} \cdot \frac{3 \cdot \left( 9^3 - 1 \right)}{9 - 1} \cdot \frac{25^2 - 1}{25 - 1} \cdot 7 \cdot 11 \cdot 13 \\ & = 2 \cdot 3^2 \cdot 5 \cdot 7^3 \cdot 11 \cdot 13^4 . \end{aligned}$$ Therefore, the answer is

$$\begin{aligned} 1 + 2 + 1 + 3 + 1 + 4 & = \boxed{012} . \end{aligned}$$ ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Solution 3 (Educated Guess and Engineer's Induction (Fake solve))

Try smaller cases. There is clearly only one $m$ that makes $\frac{2!}{m}$ a square, and this is $m=2$. Here, the sum of the exponents in the prime factorization is just $1$. Furthermore, the only $m$ that makes $\frac{3!}{m}$ a square is $m = 6 = 2^13^1$, and the sum of the exponents is $2$ here. Trying $\frac{4!}{m}$ and $\frac{5!}{m}$, the sums of the exponents are $3$ and $4$. Based on this, we (incorrectly!) conclude that, when we are given $\frac{n!}{m}$, the desired sum is $n-1$. The problem gives us $\frac{13!}{m}$, so the answer is $13-1 = \boxed{012}$.

-InsetIowa9

However...

The induction fails starting at $n = 9$ ! The actual answers $f(n)$ for small $n$ are: $0, 1, 2, 3, 4, 5, 6, 7, 7, 7, 8, 11, 12$ In general, $f(p) = f(p-1)+1$ if p is prime, $n=4,6,8$ are "lucky", and the pattern breaks down after $n=8$

-"fake" warning by oinava

Video Solutions

I also somewhat try to explain how the formula for sum of the divisors works, not sure I succeeded. Was 3 AM lol. https://youtube.com/MUYC2fBF2U4

~IceMatrix