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

Home > Numbers
Some Products Differ by 94 (Posted on 2009-09-03) Difficulty: 2 of 5
Determine the total number of quadruplets (p, q, r, s) of positive integers with p < q < r < s < 100, such that: p+s = q+r and, q*r - p*s = 94.

See The Solution Submitted by K Sengupta    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Analytical solution (spoiler) Comment 2 of 2 |
let q = p + a
     r = p + b

then s = (q + r) - p = p + a +b

94 = (q*r) - (p*s)
    = (p^2 +ap +bp +ab) - (p^2 +ap +bp)
    = ab

so (a,b) = (1,94) or (2,47)

if (a,b) = (1,94) then the minimum value of s occurs when p = 1. In this case, s = p + 1+94 = 96.  Maximum s value = 99, so there are 4 solutions where a = 1.

if (a,b) = (2,47) then the minimum value of s occurs when p = 1. In this case, s = p + 2 + 47 = 50.  Maximum s value = 99, so there are 50 solutions where a = 2.

Total number of solutions = 4 + 50 = 54

 

  Posted by Steve Herman on 2009-09-04 03:12:11
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 (6)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2024 by Animus Pactum Consulting. All rights reserved. Privacy Information