Question

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?

Details and assumptions
The consecutive integers are allowed to be a mix of negative integers, 0 and positive integers (as long as they are consecutive).

Answer 1

Don't klnow the answer yet, but Try to put it in equations, maybe will be easyer to visualize
\(n+(n+1)+\cdots + (n+25)=N\)
\(m+(m+1)+\cdots + (m+12)=N\)
other way:
\(26n+(1+2+\cdots +25)=N\)
\(13m+(1+2+\cdots +12)=N\)

Answer 2

Lets call the 1+2+... 25 =A, and 1+2+...+12=B. Then, I think, we will need to solve this inequaity:
\(1 \le 26n+A \le 1000\)
which is:
\(\frac{1-a}{26}\le n \le \frac{1000-A}{26}\)
Still thinking obout the rest...

Answer 3

Claim: Let n be a positive odd integer. Then nk may be written as the sum of n integers.

Proof: Let h = nk, with h, k, n, integers, n > 0

Let \(\large m = \frac{n-1}{2}\in\mathbb{Z}\)

Then \(\large h =(k-m)+(k+1-m)+... +[k+(m-1)-m]+k\\\large \quad\ \ + (k+1) + (k+2) + ...+(k+(m-1))+(k+m)\)

Thus, h is the sum of n consecutive integers.

Answer 4

Sorry, had to assume in the claim that k is an integer.

Answer 5

I claimed this because... isn't it true that any integer that is the sum of 26 consecutive integers is also a sum of 13 consecutive integers?

Answer 6

So... to show that... suppose we have a sum S of 26 consecutive integers... then there exists an integer z such that
\[\Large S = \sum_{i=1}^{26}(z+i)=26z + \frac{26\times27}{2}= 13(2z +27)\]

Answer 7

Means any such sum would be divisible by 13, and further meaning any such sum of 26 consecutive integers is itself a sum of 13 consecutive integers.

Answer 8

So it's enough to consider how many integers from 1 to 1000 may be written as a sum of 26 consecutive integers.

Answer 9

i don't see how 13(2z+27) impies that it is also a sum of 13 consecutive ! integers