My Math Forum sum factorials always divisible by 3?

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 August 3rd, 2017, 06:17 AM #1 Newbie   Joined: Aug 2017 From: Denmark Posts: 1 Thanks: 0 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.
 August 3rd, 2017, 06:25 AM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 6,972 Thanks: 2293 Math Focus: Mainly analysis and algebra 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). Thanks from agentredlum Last edited by skipjack; August 4th, 2017 at 07:13 AM.
 August 4th, 2017, 06:45 AM #3 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 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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