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: ( You must be logged in to post comments.)
  Subject Author Date
Bravo!Steve Herman2011-04-02 14:02:58
Some Thoughtsre: Can you really. ...... yes we can...Ady TZIDON2011-04-02 02:27:07
QuestionCan you really....Charlie2011-04-01 23:22:11
SolutionEqual numbersMath Man2011-04-01 21:16:09
Hints/Tipsre: Now I remember the D1 way.Ady TZIDON2011-04-01 18:19:23
SolutionNow I remember the D1 way.Jer2011-04-01 16:51:05
I must be doing something wrong.Jer2011-04-01 16:37:48
Some ThoughtsA more positive answer (spoiler)Steve Herman2011-04-01 16:14:17
Hints/Tipsre: April Fool's answer R E A D T H I SAdy TZIDON2011-04-01 13:39:56
April Fool's answerSteve Herman2011-04-01 12:27:43
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 (9)
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