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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.


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

     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   
Associated Topics:
High School Number Theory

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