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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.

  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.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
Bravo!Steve Herman2011-04-02 14:02:58
Some Thoughtsre: Can you really. ...... yes we can...Ady TZIDON2011-04-02 02:27:07
QuestionCan you really....Charlie2011-04-01 23:22:11
SolutionEqual numbersMath Man2011-04-01 21:16:09
Hints/Tipsre: Now I remember the D1 way.Ady TZIDON2011-04-01 18:19:23
SolutionNow I remember the D1 way.Jer2011-04-01 16:51:05
I must be doing something wrong.Jer2011-04-01 16:37:48
Some ThoughtsA more positive answer (spoiler)Steve Herman2011-04-01 16:14:17
Hints/Tipsre: April Fool's answer R E A D T H I SAdy TZIDON2011-04-01 13:39:56
April Fool's answerSteve Herman2011-04-01 12:27:43
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