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 Almost equal (Posted on 2011-04-01)
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.

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 A more positive answer (spoiler) | Comment 3 of 10 |
Looks like a spreadsheet or computer application, but here's my pencil and paper start:

2011 is a prime, so there are only two ways to do it with equal addends:  2011 1's or 1 2011.

With unequal "almost equal" addends, the ways include:

1's and 2's: 1005 ways, from 1 1 and 1005 2's to 2009 1's and 1 2
2's and 3's: 335 ways, from 2 2's and 669 3's to 1004 2's and 1 3
3's and 4's: 168 ways, from 1 3 and 502 4's to 669 3's and 1 4
4's and 5's: 100 ways, from 4 4's and 399 5's to 499 4's and 3 5's
.
.
.
502's and 503's: 1 way, 1 502 and 3 503's
670's and 671's: 1 way, 2 670's and 1 671
1005 and 1006: 1 way

Note: if the addends are greater than the square root of 2011,
i.e greater than 44.7, then there is at most one way to do it.  This is because if the addends are n and n+1, then the sum can only be constant if you use (n+1) more n's and n fewer (n+1)'s, where n(n+1) must obviously be less than 2011 (unless we use a negative number of addends, which contradicts the definition of addends).

Edited on April 1, 2011, 4:15 pm
 Posted by Steve Herman on 2011-04-01 16:14:17

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