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.

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
 $157393819470814604687108965690332604480484877$\
 $508029687469884051113407737751285106008107839$\
 $400103709226880772747397138959112221377791569$\
 $61431310006359162880000000000000000000000000$

