Divisibility by 8Date: 04/14/97 at 10:41:29 From: Marianne Unruh Subject: Number theory Show that, if n is a positive integer, then 5^n+2*3^(n-1) + 1 is divisible by 8. Date: 04/14/97 at 22:47:53 From: Doctor Steven Subject: Re: Number theory Say we pick an even number for n. Then we have 5^(2k) + 2*3^(2k-1) + 1 = 25^k + 6*3^(2k-2) + 1 = 25^k + 6*9^(k-1) + 1. Note that 25 leaves a remainder of 1 when divided by 8, so 25^k leaves a remainder of one when divided by 8. Nine also leaves a remainder of one when divided by 8, so 9^(k-1) leaves a remainder of one when divided by 8 (since k >= to 1). And 1 of course leaves a remainder of 1 when divided by 8. So we have 1 + 6*1 + 1. When we look at the remainder of this number when divided by 8, 1 + 6*1 + 1 = 8 says the remainder of this number when divided by 8 is 8, so really it has no remainder. Therefore if n is even, it's divisible by 8. Say we pick an odd number for n. Then n = 2k + 1, with k >= 0. So we have 5^(2k + 1) + 2*3^(2k) + 1 = 5*5^(2k) + 2*9^(k) + 1 = 5*25^k + 2*9^k + 1. 25 leaves a remainder of 1 when divided by 8, and so does 9, so when we look at the remainder of this equation when divided by 8 we get: 5*1 + 2*1 + 1 = 8 This says the number is divisible by 8 for n odd, so the number is divisible by eight in any case. Hope this helps. -Doctor Steven, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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