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

Home > Just Math
Q^2 + R = 1977 (Posted on 2011-02-08) Difficulty: 3 of 5
Each of X and Y is a positive integer with X ≤ Y. The quotient and the remainder obtained upon dividing X2 + Y2 by X+Y are respectively denoted by Q and R.

Determine all possible pairs (X, Y) such that Q2 + R = 1977

Supplementary questions:

This problem has been out of circulation for quite some time. Why? When is it likely to come back into favour?

No Solution Yet Submitted by K Sengupta    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
re(2): Supplementary questions | Comment 6 of 7 |
(In reply to re: Supplementary questions by Charlie)

As usual, Charlie, I'm probably wrong, but 44 + 41/44 is less than 45, so that Q^2=1936. 1936+41=1977. Then taking your example, 44+74/44 is 45+30/74, and Q^2 is 2025. Your 'remainder' is larger than 1, so increases the quotient by 1.

If remainders that were improper fractions were allowed, there would be no reason not to use 2010 in place of 1977, as you rightly point out.


  Posted by broll on 2011-02-09 15:56:25
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 (10)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information