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Mathematics 27 Online
OpenStudy (anonymous):

Find the remainder when 1! + 2! + 3! ... + 100! is divided by 30.

OpenStudy (kinggeorge):

First step: Figure out what the least \(n\) is such that \(30\) divides \(n!\). Can you tell me what this \(n\) is?

OpenStudy (anonymous):

That number will be greater than the number of molecules in the universe!

OpenStudy (anonymous):

5?

OpenStudy (kinggeorge):

Bingo. So we can rewrite is as\[1!+2!+3!+4!+30(\text{huge number})\]So the question is as simple as finding the remainder when you divide \(1!+2!+3!+4!\) by 30.

OpenStudy (anonymous):

ohhh the remainder! oops. my bad

OpenStudy (anonymous):

Ooh...that is really easy. so the answer would be 3.

OpenStudy (kinggeorge):

Right. Oftentimes problems that look really difficult actually just require a little trick.

OpenStudy (anonymous):

Thanks!

OpenStudy (kinggeorge):

You're welcome.

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