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.
The key is not to look at the specific addends but how many there are.
For each number of addends from 1 to 2011 there is exactly one way to do it, so the answer is 2011.
1 addend = 2011
2 addends = 1005+1006
3 addends = 670+670+671
4 addends = 502+503+503+503
This always works. For d addends just look at 2011 = q*d + r
Then (dr)*q + r*(q+1) = 2011
For example with d=5
2011 = 402*5 + 1
(51)*403 + 1(402+1) = 4*403 + 1*402
= 402+403+403+403+403 = 2011

Posted by Jer
on 20110401 16:51:05 