factorial of no. which are is in the form of $5m+1,5m+3$ [a] The no. of positive integer divisers of $10!$ which are is in the form of $5m+1\; \forall m\in \mathbb{N}$
[b] The no. of positive integer divisers of $10!$ which are is in the form of $5m+2\; \forall m\in \mathbb{N}$
[c] The no. of positive integer divisers of $10!$ which are is in the form of $5m+1\; \forall m\in \mathbb{N}$
My Try:: We Can write $10! = 2^8.3^4.5^2.7 = 2^x.3^y.5^z.7^t$ where $x\in\{0,1,2,......,8\}$ and $y\in \{0,1,2,.......4\}$ and $z\in\{0,1,2\}$ and $t\in {0,1}$
[a] If The no. is of the form $5m+1$. Then $z=0$ So divisers must be in the form of $=2^x.3^y.7^t$
Now How can I write the divisers which is in the form of $5m+1$
Thanks in advance
 A: Some work may be inevitable. We suggest a useful shortcut. This is first discussed in the $5k+1$ context, but also works for the others.  
We can take advantage of the fact that the powers of $2$ available run, modulo $5$, through all possibilities not congruent to $0$. In fact they run through them twice, and (sadly), $2^8$ gives the same result as $2^0$ and $2^4$. So forget about $2^0$ for a while.  
Take any divisor $d$ of $3^4 7^1$. There are $10$ of these. For any of them, there are  exactly two powers of $2$ among $2^1$ to $2^8$  that when multiplied by $d$ give us shape  $5k+1$. So we have found $20$ divisors of the right shape. Now we need to count the divisors of $3^47^1$ that have the right shape. We can use the same trick, or count directly. 
The same idea exactly works for $5k+2$, $5k+3$, $5k+4$. For each there is a guaranteed $20$, plus "extras" that come from the divisors of $5^47^1$. So for these, one might as well draw up a table and do all congruence classes mod $5$ at the same time. 
