Croatia_2018

For $n\ge 10$, $n!$ is divisible by $10$, so $(n!{)}^2$ is divisible by $100$. Therefore, only the terms $(1!{)}^2,(2!{)}^2,\dots ,(9!{)}^2$ contribute to the last two digits.

Compute each term:

$(1!{)}^2=1^2=1$

$(2!{)}^2=2^2=4$

$(3!{)}^2=6^2=36$

$(4!{)}^2=24^2=576$

$(5!{)}^2=120^2=14400$

$(6!{)}^2=720^2=518400$

$(7!{)}^2=5040^2=25401600$

$(8!{)}^2=40320^2=1625702400$

$(9!{)}^2=362880^2=131681894400$

Now, take the last two digits of each term:

$(1!{)}^2=1$ (last two digits: $01$)

$(2!{)}^2=4$ (last two digits: $04$)

$(3!{)}^2=36$ (last two digits: $36$)

$(4!{)}^2=576$ (last two digits: $76$)

$(5!{)}^2=14400$ (last two digits: $00$)

$(6!{)}^2=518400$ (last two digits: $00$)

$(7!{)}^2=25401600$ (last two digits: $00$)

$(8!{)}^2=1625702400$ (last two digits: $00$)

$(9!{)}^2=131681894400$ (last two digits: $00$)

Sum the last two digits: $01+04+36+76+00+00+00+00+00=117$

The last two digits of $117$ are $17$.

Answer: $17$