What are the trailing number of the zeroes in the given integer 
Problem Statement:- The number of zeroes at the end of the integer
  $$100!-101!+\ldots-109!+110!$$

I am having a bit of a trouble in thinking how do I proceed. A little push in the right direction would be appreciated. 
And if you are posting a full solution do use the spoiler tag, as sometimes I cant stop myself from seeing the whole solution and lose the chance of thinking through it by myself with a push in the right direction from you guys.
Also, I dont know whether I am using the right tag feel free to correct it if its wrong.

$100!$ has $[\frac{100}{5}]+[\frac{100}{5^2}]=24$ trailing $0$'s.
Now,
$100!-101!+\cdots -109!+110!$
$=m\cdot 100!$ for some positive integer $m$ such that
$m\equiv 0!-1!+2!-3!+\cdots +10!$ (mod $100$)
$\equiv -\{(3!-2!)+(5!-4!)+\cdots +(9!-8!)\}+10!$ (mod $100$)
$\equiv -(2\cdot 2!+4\cdot 4!+\cdots +8\cdot 8!)+10!$ (mod $100$)
$\equiv -(4+96+20+60)$ (mod $100$)
$\equiv -180$ (mod $100$)
$\equiv 20$ (mod $100$)
Thus, we get one more 0 from the factor $m$.
Therefore, required number of trailing $0$'s is $25$.
