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.)
Some Thoughts re: Can you really. ...... yes we can... | Comment 9 of 10 |
(In reply to Can you really.... by Charlie)

IMHO, grammatically no and mathematically yes.

Mathematicians  usually extend meanings beyond the original definitions. like raising a number to a zero(or negative , or fractional)   power, extracting a square root from a negative number , building a 7-dimensional "cube" etc.

Such extensions are only required not to inroduce contradictions in any previously defined operations.
Following this reasoning a sum of a single addend is the addend itself and a sum of zero  addends is zero.

However, Charlie's remark should be addressed in the solution, so the proper answer should be either  2011, including the trivial case of one addend, or 2010, discounting this trivial case.

 I trust this resolution satisfies everyone.


  Posted by Ady TZIDON on 2011-04-02 02:27:07
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 (4)
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