# math plllls heelp

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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 -

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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