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June 27th, 2018, 09:46 PM  #1 
Member Joined: Aug 2017 From: India Posts: 42 Thanks: 1  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 $ Last edited by skipjack; June 28th, 2018 at 03:55 AM. 
June 27th, 2018, 09:56 PM  #2 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,184 Thanks: 481 Math Focus: Calculus/ODEs 
I think I would write: $\displaystyle 27^{999}=(281)^{999}=1+28k$ And so what will the remainder be? 
June 27th, 2018, 11:30 PM  #3 
Member Joined: Aug 2017 From: India Posts: 42 Thanks: 1 
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 03:57 AM. 
June 27th, 2018, 11:42 PM  #4 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,184 Thanks: 481 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.

June 28th, 2018, 04:27 AM  #5 
Global Moderator Joined: Dec 2006 Posts: 19,191 Thanks: 1649  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. 