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Find the remainder when 1! + 2! + 3! ... + 100! is divided by 30.
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First step: Figure out what the least \(n\) is such that \(30\) divides \(n!\). Can you tell me what this \(n\) is?
That number will be greater than the number of molecules in the universe!
5?
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.
ohhh the remainder! oops. my bad
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Ooh...that is really easy. so the answer would be 3.
Right. Oftentimes problems that look really difficult actually just require a little trick.
Thanks!
You're welcome.
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