Posted by **andrew** on Wednesday, May 29, 2013 at 1:21am.

How many integers 1≤N≤1000 can be written both as the sum of 26 consecutive integers and as the sum of 13 consecutive integers?

- math plllls heelp -
**Steve**, Wednesday, May 29, 2013 at 11:43am
The sum of 26 consecutive integers starting at n is

n+(n+1)+...(n+25)

= 26n + 1+...+25

= 26n + 25(26)/2

= 26n + 325

The sum of 13 consecutive integers starting at m is

13m + 78

We want

26n+325 = 13m+78

where n <= 25 and m <= 70

That means that m=2n+19

Starting with n=1,

1+...+26 = 351 = 21+...+33

Now, (1000-351)/26 = 24.9, so our last n will be 1+24 = 25:

25+...+50 = 975 = 69+...+81

Looks like there are 25 such integers which are the sum of positive integers.

If we also allow negative integers, then we can start with n = -12, since

-12+...+13 = 13 = -5+...+7

That will add another 13 numbers to the list, making 38 values for N.

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