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Power Remainder III (Posted on 2013-04-22) Difficulty: 4 of 5
(A) N is the minimum positive integer such that 2013*N is a perfect square and 2014*N is a perfect cube.

Determine (with proof) the remainder when N is divided by 7.

(B) What would have been the answer if 2013*N was a perfect cube and 2014*N was a perfect square?

No Solution Yet Submitted by K Sengupta    
Rating: 2.5000 (2 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
What am I missing? (spoiler) | Comment 1 of 4
Observations
   (mod 7) 2013 = 4 = -3
   (mod 7) 2014 = 5 = -2
   Since 2013 and 2014 differ by 1, they have no factors in common

Part (A)

Since 2013 and 2014 have no factors in common, 
  N = (2014)^2 * (2013)^3
Mod 7, N = (-2)^2 * (-3)^3 = -108 = 4

Part (B)

Since 2013 and 2014 have no factors in common, 
  N = (2013)^2 * (2014)^3
Mod 7, N = (-3)^2 * (-2)^3 = -72 = 5

While I may have made a mistake, I think the problem is difficulty 2 or 3, certainly not 4.

Edited on April 22, 2013, 10:30 pm
  Posted by Steve Herman on 2013-04-22 14:23:29

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