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

Home > Just Math
Geometric Integers II (Posted on 2011-08-02) Difficulty: 3 of 5
Consider three positive integers x, y and z in geometric sequence with x < y < z < 2011 and, x+y=z+1 and determine all possible triplets (x,y,z) that satisfy the given conditions.

No Solution Yet Submitted by K Sengupta    
Rating: 5.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution computer solution | Comment 1 of 3

DEFDBL A-Z
FOR x = 1 TO 2011
 FOR y = x + 1 TO 2011
   z = x + y - 1
   IF z < 2011 THEN
      IF x * z = y * y THEN PRINT x; y; z
   END IF
 NEXT
NEXT x

 4  6  9
 25  40  64
 169  273  441


  Posted by Charlie on 2011-08-03 18:30:22
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