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.
 A: I think that it is easiest to factor the expression as

 $100!(1-101+101\cdot 102-\cdots +101\cdot 102\cdots 110)$.

Then, it is a standard exercise to compute the number of trailing zeros in

 $100!$

The remainder can be computed modulo $10$, $100$, etc, to see how many trailing zeros it has.  For instance, modulo $10$, we are looking at whether

 $0!-1!+2!-3!+4!-5!+6!-7!+8!-9!+10!\stackrel{?}{\equiv} 0\pmod{10}$ 

to see if this term contributes any trailing zeros.  Since every term with both a factor of $2$ and $5$ is zero modulo $10$, we get that this simplifies to

 $0!-1!+2!-3!+4!\equiv 1-1+2-6+24\equiv 20\equiv 0\pmod{10}$

Now, computing this modulo $100$, we consider

 $0!-1!+2!-3!+4!-5!+6!-7!+8!-9!+10!\stackrel{?}{\equiv} 0\pmod{100}$

A short calculation shows that this simplifies to

 $1-1+2-6+24-20+20-40+20-80+0\equiv 20\not\equiv 0\pmod{100}$

Therefore, the number of trailing zeros is

 The number of trailing zeros of $100!$ plus $1$ since the last nonzero digits >! of $100!$ and $(1-101+101\cdot 102-\cdots +101\cdot 102\cdots 110)$ are 
 not $5$s (so no additional zeros can be created in the product).

A: The number of zero digits at the end of $n!$ can be found by looking at the factors, and decomposing them into $2$ and $5$ factors. You need one $2$ and one $5$ to produce a trailing zero.
How to combine this with that alternating sum, I have no idea yet.

 BTW: I explained it to my friend Ruby and she tells me the sum is
 $157393819470814604687108965690332604480484877$\
 $508029687469884051113407737751285106008107839$\
 $400103709226880772747397138959112221377791569$\
 $61431310006359162880000000000000000000000000$

A: $100!$ 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 $25$.
