 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Remainder With Square And Cube (Posted on 2008-10-05) M is the smallest positive integer such that 2008*M is a perfect square and 2009*M is a perfect cube.

Analytically determine the remainder when M is divided by 25.

 See The Solution Submitted by K Sengupta Rating: 4.0000 (2 votes) Comments: ( Back to comment list | You must be logged in to post comments.) solution | Comment 1 of 3

2008 = 2^3 * 251
2009 = 7^2 * 41

m needs to have, as factors, an odd power of 2 and an odd power of 251, and they need be multiples of 3, so the power of each needs be 3 to be a minimum.

It also needs to have an even power of 7 such that two more than this power is a multiple of 3; 4 will do. And an even power of 41 such that the power plus 1 will be a multiple of 3; 2 will do.

So m = 2^3 * 251^3 * 7^4 * 41^2.

Mod 25:

2^3 = 8
251^3 = 1
7^4 = (-1)^2 = 1
41^2 = 16^2 = 6

Product is 48 = 23

Verify:

m comes out to be 510588495274648, which is 23 mod 25.

 Posted by Charlie on 2008-10-05 14:19:47 Please log in:
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