perplexus.info :: Just Math : Almost equal

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.

Comments: ( You must be logged in to post comments.)

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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