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

 Equality? Yes, but not exactly... (Posted on 2017-03-14)
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.

 See The Solution Submitted by Ady TZIDON No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
 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

 Search: Search body:
Forums (0)