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

Home > Just Math
Almost equal (Posted on 2011-04-01) Difficulty: 1 of 5
Integers equal to each other or differing by 1 will be called "almost equal " within the contents of this problem.

In how many ways can 2011 be expressed as a sum of almost equal addends?
The order of the addends in the expression is immaterial.

See The Solution Submitted by Ady TZIDON    
Rating: 3.7500 (4 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Hints/Tips re: Now I remember the D1 way. | Comment 6 of 10 |
(In reply to Now I remember the D1 way. by Jer)

BRAVO!

You have solved it correctly, just  edit your example :  2011= 402+402+402+402+403,
not the way you wrote.

Edited on April 1, 2011, 6:29 pm
  Posted by Ady TZIDON on 2011-04-01 18:19:23

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 (10)
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