WMC 2018 Pursuit - Pack 2, B3

$10!= 2^8.3^4.5^2.7$

If a factor has at least two $2$ s, then the factor would be divisible by $4$ . The remainder would be zero. Such factors would be of form $2^a.3^b.5^c.7^d$ , where $2 \leq a\leq 8,0 \leq b<4, 0 \leq c \leq 2, 0\leq d \leq 1$ and $a,b,c,d$ are non-negative integers.

If the factor is of the form $2^a.3^b.5^c.7^d$ , where $a=1,0 \leq b<4, 0 \leq c \leq 2, 0\leq d \leq 1$ , then

$2.3^b.5^c.7^d \equiv 2.(-1)^b.(1)^c.(-1)^d \equiv 2.(-1)^{(b+d)} \equiv 2 \ (mod \ 4)$

If the factor is of the form $2^a.3^b.5^c.7^d$ , where $a=0,0 \leq b<4, 0 \leq c \leq 2, 0\leq d \leq 1$ , then

$3^b.5^c.7^d \equiv (-1)^b.(1)^c.(-1)^d \equiv (-1)^{(b+d)} \ (mod \ 4)$

now we have cases. If $b+d$ is odd, then the factor would be $3$ , modulo $4$ . If $b+d$ is even, the factor is $1$ , modulo $4$ , which is what we are looking for. there are $5$ cases where $b+d$ is even. the number of such factors, consequently, would be $5.3=15$ ( $3$ comes from the possibilities of $5^c$ ).

the total number of factors of $10!$ is $270$ . Therefore, the probability would be

$\frac{15}{270}=\frac{1}{18}$

and $a+b=19$