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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.)
re: What am I missing? (spoiler) | Comment 2 of 4 |
(In reply to What am I missing? (spoiler) by Steve Herman)

I agree Steve. I can't help suspecting that other years having more interesting factorisations may have been used in an earlier version of the problem.

(A) N is the minimum positive integer such that 2013*N is a perfect square and 2014*N is a perfect cube.


Let a^2=2013n, b^3=2014n; a^2/2013=b^3/2014, so b^3 = (2014a^2)/2013
2013=3x11x61; 2014= 2x19x53; neither have powered factors
2013 does not divide 2014, so 2013 divides a^2 and so a; let a=2013c
b^3 = (2014(2013c)^2)/2013
b^3 = 4054182c^2; b= 4054182, c= 4054182, a=8161068366

2013n=(8161068366)^2
2014n=(4054182)^3
n=33,086,456,470,206,612
So the remainder is 4.

In like manner the solution for (B) is 5:

(B) N is the minimum positive integer such that 2013*N is a perfect cube and 2014*N is a perfect square.

Let  a^2=2014n, b^3=2013n; then a^2/2014=b^3/2013, so b^3=(2013a^2)/2014
b^3=(2013(2014c)^2)/2014
b^3 = 4054182c^2; b= 4054182, c= 4054182, a=8165122548

2014n=(8165122548)^2
2013n=(4054182)^3
n=33,102,892,861,895,736
So the remainder is 5.

Edited on April 22, 2013, 3:27 pm
  Posted by broll on 2013-04-22 15:20:32

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