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August 3rd, 2017, 06:17 AM   #1
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sum factorials always divisible by 3?

So I was messing around in python, defining factorials.

And I did that, and I tried summing up all the factorials up to a certain n.

n! + (n-1)! + (n-2)! ... + 1!.

With 5, it would be

5! + 4! + 3! + 2! + 1! = 153.

I tried messing around, and looked at the divisors, and it seemed that 3 would always be a divisor.

Except from the trivial case of 1, where 1! is of course not divisible by 3.

But I tried with 2,3,4,5... 57.. 100, and it seems to work for them.
Is there any way to prove, that the sum of factorials up to n is always divisible by 3, if n is not 1? If not, what are the counterexamples and what's the system between these (because they seem very rare, if they exist).

I'm not very good at math, so I thought it would be fun, if any of you could help.
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August 3rd, 2017, 06:25 AM   #2
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1! + 2! = 3
All other factorials are divisible by 3 by definition. Therefore (1! + 2!) + 3! + 4! +... + n! is divisible by 3 because 3 is a factor of every term in the sum (where the first term is the whole parenthesis).
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Last edited by skipjack; August 4th, 2017 at 07:13 AM.
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August 4th, 2017, 06:45 AM   #3
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By the same reasoning , if $n \ge 5 $ the sum of factorials is divisible by 9

$(1! + 2! + 3! + 4! + 5!) + 6! + 7! \ \ + \ ... \ + \ \ n!$

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