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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.

 Submitted by Ady TZIDON Rating: 3.7500 (4 votes) Solution: (Hide) For each number of addends from 1 to 2011 there is exactly one way to do it, so the answer is 2011. If we diregard the trivial case of 1 addend i.e. 2011=2011 , then the answer is 2010.

 Subject Author Date Bravo! Steve Herman 2011-04-02 14:02:58 re: Can you really. ...... yes we can... Ady TZIDON 2011-04-02 02:27:07 Can you really.... Charlie 2011-04-01 23:22:11 Equal numbers Math Man 2011-04-01 21:16:09 re: Now I remember the D1 way. Ady TZIDON 2011-04-01 18:19:23 Now I remember the D1 way. Jer 2011-04-01 16:51:05 I must be doing something wrong. Jer 2011-04-01 16:37:48 A more positive answer (spoiler) Steve Herman 2011-04-01 16:14:17 re: April Fool's answer R E A D T H I S Ady TZIDON 2011-04-01 13:39:56 April Fool's answer Steve Herman 2011-04-01 12:27:43

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