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### Sums of Consecutive Integers

```
Date: 01/03/2001 at 01:23:37
From: Roger Wiltfong
Subject: Different ways on breaking numbers (with a twist)

This was a question in a recent math competition. How many different
ways can 2000 be expressed as the sum of two or more consecutive
positive integers?

We only had an hour to complete the test and there were many more
questions. I couldn't see a QUICK way of solving this.

Thanks,
Roger
```

```
Date: 01/03/2001 at 10:07:12
From: Doctor Rob
Subject: Re: Different ways on breaking numbers (with a twist)

Thanks for writing to Ask Dr. Math, Roger.

If the number n of these consecutive integers is odd, then their mean
m is an integer, and m = 2000/n. Thus n must be an odd divisor of
2000, and n >= 2.

If n is even, then m is half an odd integer, and m = 2000/n. Thus the
odd integer 2*m is an odd divisor of 4000, and 1 <= 2*m <= 2000 (since
n >= 2).

Furthermore, the smallest of the numbers will be m - (n-1)/2, and the
largest will be m + (n-1)/2. That implies that:

m - (n-1)/2 >= 1
so
2000/n - (n-1)/2 >= 1
4000 - n^2 + n >= 2*n
0 >= n^2 + n - 4000
63 >= n >= 2

Also
m - ([2000/m]-1)/2 >= 1
2*m^2 - 2000 + m >= 2*m
2*m^2 - m - 2000 >= 0
m >= 32
2000 >= 2*m >= 64

In each case, pick out the numbers that satisfy both the divisor
condition and the corresponding inequality.

- Doctor Rob, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Number Theory

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