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Almost equal (
Posted on 20110401
)
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:
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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: (
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Subject
Author
Date
Bravo!
Steve Herman
20110402 14:02:58
re: Can you really. ...... yes we can...
Ady TZIDON
20110402 02:27:07
Can you really....
Charlie
20110401 23:22:11
Equal numbers
Math Man
20110401 21:16:09
re: Now I remember the D1 way.
Ady TZIDON
20110401 18:19:23
Now I remember the D1 way.
Jer
20110401 16:51:05
I must be doing something wrong.
Jer
20110401 16:37:48
A more positive answer (spoiler)
Steve Herman
20110401 16:14:17
re: April Fool's answer R E A D T H I S
Ady TZIDON
20110401 13:39:56
April Fool's answer
Steve Herman
20110401 12:27:43
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