Factorization and Squares

Question

Let $M = 1!\times2!\times3!\times4!\times5!\times6!\times7!\times8!\times9!$. How many factors of $M$ are perfect squares?

I tried to solve it by expanding the factorials and seeing the perfect squares, but I think this method is long. I am just looking for a better solution.

Answer 1

Hint. If $N = \prod p_i^{a_i}$ where $p_i$ are prime, is the prime factorization of $N$ then all factors of $N$ are in the form $\prod p_i^{k_i}$ where $0 \le k_i \le a_i$. And the square factory of the form of $\prod p_i^{k_1}$ where $0 \le k_i \le a_i$ and $k_i$ is even.

Hint 2: $M=1!\times ..... \times 9! =$

$1*2^8*3^7*4^6*5^5*6^4*7^3*8^2*9=$

$2^{8+12+4+6}3^{7+4+2}5^57^3 = 2^{30}*3^{13}*5^5*7^3$.

So the square factors are of the form $2^a3^b5^c7^d$ where $0 \le a \le 30; 0\le b \le 13; 0\le c\le 5; 0\le d\le 3$. And $a,b,c,d$ are even.

Hint 3: How many even $a$s are there between $0...30$? How many even $b$s are there between $0... 13$? How many combinations of the two are there.

Answer 2

I'll just give you hints:

$1)$ What are the prime factors of $M$?

$2)$ Find the prime factorisation itself

$3)$ Count the numbers having even powers of those prime numbers such that the powers aren't higher than the powers in the factorisation of $M$