Total number of divisors of $9!$ which are is in the form of $3m+2$, where $m\in \mathbb{N}$
My Try: Let $ N = 9! = 1\times 2 \times 3 \times 2^2 \times 5 \times 2 \times 3 \times 7 \times 2^3 \times 3^2 = 2^7 \times 3^4 \times 5 \times 7$
Now If Here $N$ must be a mutiple of $3m+2$, means when $N$ is divided by $3$ It will gave a remainder $2$
But I did not understand how can i proceed further, thanks in advance
Hint: If the factor is of the form $3m+2$, then the prime factorization must be of the form
$$ 2^a \times 3^0 \times 5^b \times 7^c, $$
where $a+b \equiv 1 \pmod{2} $ and $ c= 0$ or $1$.
Count the number of possibilities.
There are $8 \times 2 \times 2=32$ divisors of $2^7 \times 5 \times 7$, and each is congruent to either $1 \pmod 3$ or $2 \pmod 3$.
If $2^i$ exactly divides the divisor $d$, for $i \in \{0,2,4,6\}$, we pair it up with the divisor $2d$. Each pair consists of one divisor congruent to $1 \pmod 3$ and one congruent to $2 \pmod 3$.
So exactly $16$ of the divisors of $2^7 \times 5 \times 7$ are of the form $2 \pmod 3$. These are the only divisors of $9!$ congruent to $2 \pmod 3$.
