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

Home > Just Math
Sum (Pair Product) - Sum = 21 (Posted on 2010-06-07) Difficulty: 2 of 5
Determine all possible triplet(s) (a, b, c) of positive integers, with a ≤ b ≤ c, that satisfy this equation.

a*b+b*c+c*a - (a+b+c) = 21

No Solution Yet Submitted by K Sengupta    
Rating: 2.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Analytical solution (spoiler) | Comment 3 of 4 |
1) A cannot be greater than 2.
    Let f(a,b,c) = a*b+b*c+c*a - (a+b+c)

    f(3,3,3) = 18, so (3,3,3) is not a solution,
    and
    f(3,3,4) = 23, which is already too big.  Since the products increase faster than the sum, a must be 1 or 2

2) If a = 1, then solving for c in terms of b gives c = 22/b

    The only solutions where a = 1 are (1,1,22) and (1,2,11)

3) If a = 2,  then solving for c in terms of b 
    gives c =  (24/(b+1)) - 1.

    The only solutions where a = 2 are (2,2,7) and (2,3,5).

(Edited to fix copy-and-paste error.  Thanks Dej Mar!)

Edited on June 8, 2010, 9:36 am
  Posted by Steve Herman on 2010-06-07 20:35:34

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 (11)
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