I have no idea about how to solve this. It is impossible to solve by brute force, but only $1!,\ 2!,\ 3!$, and $4!$ leave mods of not $0$ when modded by $10$. So is it $1+2+6+24=33$?
Compute $1! + 2! + 3! + ... + 2017! \pmod{100}$
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modular-arithmetic
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2Are you modding by 10 or by 100? – 2017-01-11
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4Every number n greater then 10 is $n!\equiv 0(mod 100)$, why? – 2017-01-11
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0Ah, because 5! is 120, ending in 0, and multiplying it by 10 gives it another 0, which leaves it 0 mod 100. – 2017-01-11
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0@arberavdullahu I think you meant greater THAN, but it is actually greater THAN or EQUAL to 10. – 2017-01-11
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1@GerardL. why do you change your post? – 2017-01-11
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0@Xam I figured it out, just checking if this is correct. – 2017-01-11
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1@GerardL. Did you note that somebody spent some time answering your question *as it was then*? – 2017-01-11
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0@Did For me, his answer post did not load in and I had figured out the problem remembering another problem I solved. I did not plagiarize his answer. – 2017-01-11
1 Answers
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It is easy enough to see that since $n!=1\times2\times3\times\dots\times n$, then,
$$n!=0\pmod{100}\quad n\ge10$$
Thus, you problem reduces down to
$$1!+2!+3!+\dots+9!\equiv1+2+6+24+20+20+40+20+80\equiv13\pmod{100}$$