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!
On
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.
On
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?
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
$$v_p(n!)=\biggl\lfloor\frac np\biggr\rfloor+\biggl\lfloor\frac n{p^2}\biggr\rfloor+\dotsm $$