My Math Forum Struck up with this problem

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June 27th, 2018, 10:46 PM   #1
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Struck up with this problem

I am struck up with this problem. Any hints of how to proceed? The maximum I could reach is

$\displaystyle 27^{999}/7=> 3^{2997}/7$
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Last edited by skipjack; June 28th, 2018 at 04:55 AM.

 June 27th, 2018, 10:56 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs I think I would write: $\displaystyle 27^{999}=(28-1)^{999}=-1+28k$ And so what will the remainder be? Thanks from MathsLearner123
 June 28th, 2018, 12:30 AM #3 Member   Joined: Aug 2017 From: India Posts: 50 Thanks: 2 I am not sure whether I have understood completely. But I took an example of $\displaystyle k=1; Value = 27 => Mod(27,7) => 6.$ For any value of k, the remainder is 6. Last edited by skipjack; June 28th, 2018 at 04:57 AM.
 June 28th, 2018, 12:42 AM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs Yes, as 28 is a multiple of 7, we find the expansion is a multiple of 7 less 1, which means the remainder is 6. Thanks from MathsLearner123
June 28th, 2018, 05:27 AM   #5
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Quote:
 Originally Posted by MathsLearner123 For any value of k, the remainder is 6.
If two integers, b and c, differ by an integer multiple of a third integer, n, it follows that equal powers of b and c differ by some integer multiple of n.

For example, 4 and 24 differ by a multiple of 10. It follows that 64 and 13824 (the cubes of 4 and 24) differ by an integer multiple of 10.

As 27 and -1 differ by an integer multiple of 7, it follows that 27^999 and (-1)^999 differ by an integer multiple of 7. Now, one can use the fact that any odd integer power of -1 has the value -1 to conclude that 27^999 - (-1) is an integer multiple of 7. One can make a similar conclusion using 28 instead of 7, but that's not needed for the original question.

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