All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Just Math
Remainder With Square And Cube (Posted on 2008-10-05) Difficulty: 2 of 5
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.)
re: Solution ------- ?? Comment 3 of 3 |
(In reply to Solution by Dej Mar)

I found something different (Dej Mar made a mistake considering the two numbers (square and cube) separately.)

Joining the two numbers, we have these factors:

(2^3 x 251) x ( 7^2 x 41)

So, to have a square and a cube, we need M = 2^3 x 7^4 x 251^3 x 41^2 =  510,588,495,274,648.

Edited to say that is the number Charlie found, too. I missed it.

Edited on October 6, 2008, 12:03 pm
  Posted by pcbouhid on 2008-10-06 10:53:38

Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (2)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2021 by Animus Pactum Consulting. All rights reserved. Privacy Information