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

Home > Numbers
A FIRST and SECOND Problem (Posted on 2006-08-22) Difficulty: 2 of 5
There are n retailers and each of them have copies of two novels left, titled “FIRST” and “SECOND”. Each retailer has a different number of copies of "FIRST", which is always more than the number of copies of "SECOND" stocked by that retailer.

It is observed that the sum of squares of the number of copies of FIRST and the number of copies of SECOND for each of the retailers is always equal to:
(22+32)(32+42)(42+52).

For example, if the respective number of copies of FIRST and SECOND titles possessed by any given retailer i are xi and yi, then :
xi2+yi2 = (22+32)(32+42)(42+52); for all i = 1,2,....,n

How many retailers are there and how many copies of FIRST novel in total are available from all these (n) retailers?

See The Solution Submitted by K Sengupta    
Rating: 4.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Some Thoughts re: An idea, with no computer Comment 5 of 5 |
(In reply to An idea, with no computer by Old Original Oskar!)

Since 3^2+4^2=0^2+5^2, you could also try c=0 and d=5, I think. It's not clear to me, however, whether you'd get all possible solutions using these formulas.
  Posted by Federico Kereki on 2006-08-25 08:58:41

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 (7)
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