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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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 I must be doing something wrong. | Comment 4 of 10 |
I was thinking this was D1 because the answer would be 2011 when this was in the queue.  When I though of how this could be today, things got messy.

2011 is prime so the only way to do it with equal addends is 2011 1's.

all 1's and 2's can be done in 1005 ways. (from 1 1 and 1005 2's down to 2009 1's and 1 2)
All 2's and 3's can be done in 335 ways. (from  2 2's and 669 3's down to 1004 2's and 1 3, but only about 1/3 of the 2's or about 1/2 of the 3's)
All 3's and 4's can be done in 167 ways (from 1 3 and 502 4's down to
1 4 and 669 3's)
All 4's and 5's can be done in 100 ways (from 4 4's and 399 5's down to 499 4's and 3 5's)
All 5's and 6's can be done in 67 ways
All 6's and 7's can be done in 47 ways
All 7's and 8's can be done in 35 ways
The sequence continues 27, 22, 18, 15, 12, 11, 9, 8, 7, 6, 5, 5, 4, 4, 3*, 3, 3, 3, 2, 2, 2, 2, 2, 2, 1...
(the 1's continue from 32's and 33's to 44's and 45's)

The grand total is 1946
Or 1947 if you count the all 1's, or even 1948 if you count the single 2011.

*This is going to be an under count.  I had assumed the number of ways was [2011/(n(n+1))], but it is not.
2011/(22*23)=3.974, but there are 4 ways (13,75) (36,53) (59,31) (82,9)
rounding to the nearest whole number may be more accurate but I dont think this will be precise easier.

Interestingly I am 63 away from 2011 in my count.
This is precisely the smallest value of n for which 2011/(n(n+1))<.5
The implication is that I should have rounded UP not down.  This doesn't make sense either since for example 2011/(19*20)=5.292 but there are only 5 solutions (9,92)(29,72)(49,54)(69,35)(89,16)

 Posted by Jer on 2011-04-01 16:37:48

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