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 Power Remainder III (Posted on 2013-04-22)
(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)

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 One thing I was missing Comment 4 of 4 |
I guess I was a little glib in my solution.

I stated that since 2013 and 2014 have no factors in common,
the N's for parts A and B are
(2014)^2 * (2013)^3 and
(2013)^2 * (2014)^3, respectively.

Those are correct, but it is not because they have no factors in common.  It is because neither 2013 nor 2014 has a squared factor.

For instance, if the numbers had been 2012 and 2013, then there are still no factors in common, but now 2012 has 2^2 as a factor. The N's would be
(2013)^2 * (2012)^3 / 2^6 and
(2012)^2 * (2013)^3, respectively.

Also, for instance, if the numbers had been 2010 and 2013, then there is a factor of 3 in common, but N's still
(2013)^2 * (2010)^3 and
(2010)^2 * (2013)^3, respectively.
These N's each have 3^5 as a factor, but there is no division because no factor is ^6 or higher.

Of course, it doesn't matter if all we care about is N mod 7, because (mod 7) m^6 = 1  if m is not a multiple of 7.

Edited on April 23, 2013, 8:37 am
 Posted by Steve Herman on 2013-04-22 22:42:07

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