Only The Last Two...

Since, only the last 2 digits matter, we only need to consider the factorials that do not have zeroes at the last two place values.
10!(3628800) is the first term to have 2 zeroes at the end. Therefore we only consider 1! to 9!.

Notice that all (odd number)! is positive and all (even number)! is negative.

9!=362880 In this term we only need to consider the last two digits i.e. 80. Continuing like this, if we add the last two digits of all the odd number terms (from 1 to 9), we get 147 . Similarly, the sum of the last two digits of all even numbered terms is 66 .

The final answer would be $\rightarrow$ Sum of (last 2 digits) of odd terms - Sum of (last 2 digits) of even terms = 147-66 = $\boxed{81}$

The expression can be written as $\sum_{n=1}^{2017} (-1)^{n+1} n! = \sum_{n=0}^{1008} (2n+1)! -\sum_{n=1}^{100}(2n)!$ we note that $n!$ factorial has trailling zeros and we need to find last 2 zeros so we just need to focus the value of $n$ such $n!$ or has 2 trailing zeros which is vivid that $n=10$ . Now $\text{last two digits} = \left(\left(\sum_{n=0}^{5}(2n+1)!\right)\mod(100) -\left(\sum_{n=1}^{5}(2n)! \right)\mod(100)\right)\mod(100) \\ L_d = \left(1+6+20+40+80+00) -(2+24+20+20+00\right)\mod(100) \\ L_d =\boxed{81}$

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Similar problem

using $Mathematica$

Mod[Sum[(-1)^(n+1)n!,{n,2017}],100]

returns $\boxed{81}$