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

Home > Numbers
Equality? Yes, but not exactly... (Posted on 2017-03-14) Difficulty: 3 of 5
Let's call two integers approximately equal if they differ by at most 1.
How many different ways are there to write 2017 as a sum of two or more positive integers which are all approximately equal to each other?

The order of terms does not matter: two ways which only differ in the order of terms are counted as one.

No Solution Yet Submitted by Ady TZIDON    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Solution Comment 1 of 1
Assume a sum consists of N summands.  Then for each integer N there is exactly one sum.  Let q=floor(2017/N) and r=2017 mod N.  q and q+1 are the approximately equal integers.  Then the sum consisting of N summands consists of q occurring N-r times and q+1 occurring r times.

There are 2016 positive integers greater than 1 and not exceeding 2017, therefore there are 2016 different ways to write 2017 as a sum of two or more positive integers which are all approximately equal.

  Posted by Brian Smith on 2017-03-14 11:03:31
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 (0)
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