Find the number of trailing zeros of $ 100! - 101! + ... - 109! + 110!$

Question

My try : I know number of trailing zeros of individual factorials, owing to their numbers of factors divisible by $5, 25$.

 100! till 104! : 24 zeros
 105! till 109! : 25 zeros
 110! : 26 zeros

So minimum number of trailing zeros will be $24$. But when subtracting, the number of trailing can increase. Without doing too many addition and multiplications, how to easily find this out?

Answer 1

Try by this method

$100 ! - 101 ! + 102 ! - \ldots -109 ! + 110!$
= $100 ! - 101 ! (1 - 102) - 103 ! ( 1 - 104) - \ldots - 109 ! ( 1 - 110)$
=$100 ! + 101(101 !) + 103(103 !) + 105(105 !) + 107(107 !) + 109(109 !)$

Answer 2

Modulo $100(100!)$, it is $100!(0!-1!+2!-3!\ldots +10!)$ and the factor is a multiple of $10$ but not $100$