Date: 09/11/2003 at 05:35:24 From: Inari Subject: divisibility Prove: 1. that (n^2 - n) is divisible by 2 for every integer n 2. that (n^3 - n) is divisible by 6 3. that (n^5 - n) is divisible by 30 I know that n^2 - n = n(n-1) Since n and n-1 are consecutive integers, one of them must be divisible by 2, so their product is divisible by 2. Likewise, n^3 - n = (n-1)n(n+1). Therefore, one of these factors is divisble by 2, one of the others is divisible by 3, which implies that (n^3 - n) is divisble by 6 (i.e., it's divisible by both 2 and 3). But I'm stuck on the third one. I know that n^5 - n = (n-1)n(n+1)(n^2+1). How do I show that this is divisible by 2, 3, and 5, and thus by 30?
Date: 09/11/2003 at 12:48:43 From: Doctor Edwin Subject: Re: divisibility Hi, Inari. Ooh, neat puzzle. You've done well so far -- well enough that I don't want to spoil the fun by giving too much away. So here's your hint: Think only about the last digit of n. Write back if you're still stuck and I'll help unstick you. - Doctor Edwin, The Math Forum http://mathforum.org/dr.math/
Date: 09/12/2003 at 01:54:45 From: Inari Subject: divisibility Whatever the last digit of n is, (n^5 - n) will always have 0 as the units digit. This means that (n^5 - n) is divisible by both 2 and 5. Also, n^5 - n = (n-1)n(n+1)(n^2+1) We've already shown that (n-1)n(n+1) is divisible by 3 (from the 2nd part of the question). Therefore, (n^5 - n) is divisible by 2, 3, and 5, and is therefore divisible by 30. Is this correct?
Date: 09/12/2003 at 08:00:29 From: Doctor Edwin Subject: Re: divisibility Nice! And neater than my proof, which was unnecessarily complicated. Good job. - Doctor Edwin, The Math Forum http://mathforum.org/dr.math/
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