Posted by andrew on .
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,
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, (1000351)/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.