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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)

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Solution Now I remember the D1 way. | Comment 5 of 10 |
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 (d-r)*q + r*(q+1) = 2011

For example with d=5
2011 = 402*5 + 1
(5-1)*403 + 1(402+1) = 4*403 + 1*402
= 402+403+403+403+403 = 2011
 

  Posted by Jer on 2011-04-01 16:51:05
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