SOLUTION: if 3^(8n+3) divided by 5 what will be the remainder,where n is a integer

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Question 814632: if 3^(8n+3) divided by 5 what will be the remainder,where n is a integer

Answer by Edwin McCravy(20056) About Me  (Show Source):
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if 3^(8n+3) divided by 5 what will be the remainder,where n is a integer
If n=1, we have 3%5E%288%2A1%2B3%29 = 3%5E11 = 177147, when divided by 5, has remainder 2.

That is 3%5E11+=+5%2A35429+%2B+2

We will try to prove, by induction, that the remainder will always be 2.

That is, we will try to prove 3%5E%288n%2B3%29+=+5%2Ap+%2B+2 for some integer p

Assume that for all n+%3C=+k, 

(1)      3%5E%288n%2B3%29+=+5%2Ap+%2B+2 for some integer p       <--assumed

we want to use (1) to prove: 

         3%5E%288%28n%2B1%29%2B3%29+=+5%2Aq+%2B+2 for some integer q   <--unproved 

or

         3%5E%288n%2B8%2B3%29+=+5%2Aq+%2B+2 for some integer q     <--unproved

or

(2)     3%5E%288n%2B11%29+=+5%2Aq+%2B+2 for some integer q       <--unproved

We want to show that the assumption of (1) imples (2)

Multiply both sides of (1) by 3^8

        3%5E8%2A3%5E%288n%2B3%29+=+3%5E8%2A5p+%2B+3%5E8%2A2

        3%5E%288n%2B11%29+=+5%283%5E8p%29+%2B+13122

        3%5E%288n%2B11%29+=+5%283%5E8p%29+%2B+13120+%2B+2

        3%5E%288n%2B11%29+=+5%283%5E8p%29+%2B+5%2A2624+%2B+2 

        3%5E%288n%2B11%29+=+5%283%5E8p+%2B+2624%29+%2B+2

So we take integer q+=+3%5E8p+%2B+2624 and we have proved (2)

(2)     3%5E%288n%2B11%29+=+5%2Aq+%2B+2 for some integer q

Therefore the remainder is 2 for all n

Edwin