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

 Q^2 + R = 1977 (Posted on 2011-02-08)
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(3): Supplementary questions Comment 7 of 7 |
(In reply to re(2): Supplementary questions by broll)

"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. "

When the dividend is 31^2+53^2 = 3770 and the divisor is 31 + 53 = 84, making the quotient to be 44 and the remainder to be 74 does not imply that 3770/84 = 44 + 74/44, but rather that 3770/84 = 44 + 74/84; and 74/84 is not larger than 1. Again, the remainder must be smaller than the divisor, not necessarily smaller than the quotient.

 Posted by Charlie on 2011-02-12 13:11:07

 Search: Search body:
Forums (0)