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 Post subject: Help with Induction problemPosted: Thu, 25 Nov 2004 22:51:17 UTC

Joined: Wed, 24 Nov 2004 19:55:09 UTC
Posts: 5
Location: El Paso, TX,USA
I was wondering if you could help me solve an induction problem... I have to prove that for any non negative integer n, the following is true:

7^(n+2)+8^(2n+1) = 0(mod 57)

*note "=" sign represents congruent to

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 Post subject: Posted: Thu, 25 Nov 2004 23:24:25 UTC
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Joined: Sun, 7 Nov 2004 08:21:37 UTC
Posts: 2987
First start off by considering the case n = 1. I think you can verify it using a calculator.

Then, assume the formula is true for n = k:

7^(k+2) + 8^(2k+1) = 0(mod 57)

Now, try to prove it for n = k+1

7^(k+3) + 8^(2k+3) = 7*(7^(k+2)) + 64*(8^(2k+1))

If you can somehow isolate 7^(k+2) + 8^(2k+1), it should be easy from there.

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 Post subject: How????Posted: Fri, 26 Nov 2004 00:56:20 UTC

Joined: Wed, 24 Nov 2004 19:55:09 UTC
Posts: 5
Location: El Paso, TX,USA
Can anyone show us how to issolate them?!?!?!?!?!?!?!?!?!?

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 Post subject: Posted: Fri, 26 Nov 2004 00:57:00 UTC
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Joined: Wed, 1 Oct 2003 04:45:43 UTC
Posts: 9633
I was wondering if you could help me solve an induction problem... I have to prove that for any non negative integer n, the following is true:

7^(n+2)+8^(2n+1) = 0(mod 57)

*note "=" sign represents congruent to

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 Post subject: Posted: Fri, 26 Nov 2004 04:13:00 UTC
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Joined: Sun, 7 Nov 2004 08:21:37 UTC
Posts: 2987
Matthew wrote:

True, but now he's trying to prove it by induction, so he will need a different method. Unfortunately, I probably don't know enough about modular math to isolate
7^(k+2) + 8^(2k+1) and hence solve the problem.

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 Post subject: Posted: Fri, 26 Nov 2004 06:13:25 UTC
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Joined: Sun, 1 Aug 2004 17:25:34 UTC
Posts: 929

jinydu wrote:
7^(k+3) + 8^(2k+3) = 7*(7^(k+2)) + 64*(8^(2k+1))

and ask yourself what is 64 mod 57.

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 Post subject: Still confusedPosted: Mon, 9 Feb 2009 20:01:16 UTC
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Joined: Mon, 9 Feb 2009 19:59:10 UTC
Posts: 3
I am also trying to work on this problem and I do not understand how the fact that 64 = 7 mod 57 helps to conclude that the whole thing is congruent to 0 mod 57

Last edited by helpplease on Mon, 10 May 2010 16:22:03 UTC, edited 1 time in total.

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 Post subject: Posted: Wed, 11 Feb 2009 06:11:30 UTC
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Joined: Tue, 6 Feb 2007 15:49:02 UTC
Posts: 2594
7*(7^(k+2)) + 64*(8^(2k+1)) = 7*(7^(k+2)) + (7 + 57)*(8^(2k+1)) = 7*(7^(k+2) + 8^(2k+1)) + 57*(8^(2k+1))
Can you finish now?

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 Post subject: Posted: Wed, 11 Feb 2009 17:30:48 UTC
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Joined: Mon, 9 Feb 2009 19:59:10 UTC
Posts: 3
Yes, thanks!

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