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

Home > Just Math
No. of Distinct Solutions(1) (Posted on 2008-07-05) Difficulty: 3 of 5
Given that a and b are non-negative integers and f(x,y)=(x+y-a)(x+y-b), then if f(x,y)=0 has n distinct non-negative integer solutions for (x,y), find how many different polynomials f(x,y) can take.

Note: For example, (1,0) and (0,1) are not distinct solutions.

No Solution Yet Submitted by Praneeth    
Rating: 3.5000 (2 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Just counting (spoiler) Comment 1 of 1
Well, let's first solve an easier problem:  Given that a is a non-negative integer and (x+y-a) = 0 has n distinct non-negative integer solutions for (x,y), find a.

Well, if n = 1, then a = 0 or 1, because x + y = 0 implies (x,y) = (0,0) and x + y = 1 implies one distinct non-negative integral solutions, (0,1).

If n = 2 then a = 2 or 3, because x + y = 2 implies two distinct non-negative integral solutions, (0,2), and  (1,1) and x + y = 3 implies two distinct solutions, (0,3) and (1,2).

If n = 3 then a = 4 or 5, because x + y = 4 implies three distinct solutions, (0,4), (1,3), (2,2) and  x + y = 5 implies (x,y) is, (0,5), (1,4), or (2,3).

In general, a = (2n - 2) or (2n - 1).

*****************************************

In the original problem, the number of distinct polynomials is just the number of distinct values of (a,b).  Note that (1,0) and (0,1) yield the same polynomial, and are not distinct.

If n = 1, then the distinct (a,b) must be (0,0) or (1,1).  If (a,b) = (0,0) then (x,y) = (0,0).  If (a,b) = (1,1) then (x,y) must equal (0,1).

If n = 2,  then the distinct (a,b) must be equal to (2,2) or (3,3) or (also) (0,1).  If (a,b) = (0,1) then (x,y) = (0,0) or (0,1).  If (a,b) = (2,2) then (x,y) = (0,2) or (1,1).  If (a,b) = (3,3) then (x,y) = (0,3) or (1,2).  

If n = 3, then the distinct (a,b) must be equal to (4,4) or (5,5) or (0 or 1,2 or 3).  By (0 or 1,2 or 3), I mean (0,2), (0,3), (1,2) or (1,3).  Altogether, there are six distinct polynomials.

If n = 4, then the distinct (a,b) must be equal to (6,6) or (7,7) or (0 or 1,4 or 5) or (also) (2,3).  By (0 or 1,4 or 5), I mean (0,4), (0,5), (1,4) or (1,5).  Altogether, there are 7 distinct polynomials.

If n = 5, then the distinct (a,b) must be equal to (8,8) or (9,9) or (0 or 1,6 or 7) or (2 or 3,4 or 5), Altogether, there are 10 distinct polynomials.

If n = 6, then the distinct (a,b) must be equal to (10,10) or (11,11) or (0 or 1,8 or 9) or (2 or 3,6 or 7), or (also) (4,5).  Altogether, there are 11 distinct polynomials.

In general,
if n is even, there are 2n distinct polynomials
if n is odd,  there are 2n - 1 distinct polynomials 
  Posted by Steve Herman on 2008-07-06 01:35:12
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 (2)
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