I was able to get to a theorem saying "that a|b if and only if for any prime in the factorization of a or b, its exponent in the factorization of a is less than or equal to its exponent in the factorization of b."
I tried to use this theorem where k <= m, k <= n, where m and n are exponents of 2 and 5, respectively, in the factorization of 100! but I was not able to work out the math so far.
Any help will be greatly appreciated!
This looks like a homework problem so I'll not give the complete answer.
Instead I invite you to ask:
How many times does the factor $5$ occur in the first 100 natural numbers? (NB remember that $25$, $50$ and $100$ are each divisible by $5^2$.) What about the factor 2?
What power of 10 can you make out of those?
10 and 100 divide by 10, as do 2*5, 12*15, 22*25, 32*35, 42*45, 52*55, 62*65, 72*75, 82*85, and 92*95. Multiplying these numbers may also produce more pairs of numbers that divide by 10 when multiplied together, but there won't be any that aren't in this list.
Hint:
You use Legendre's formula: if $p$ is a prime number and $v_p(n)$ denotes the $p$-adic valuation of the natural number $n$, then
$$v_p(n!)=\biggl\lfloor\frac np\biggr\rfloor+\biggl\lfloor\frac n{p^2}\biggr\rfloor+\dotsm $$
