Sums of Consecutive IntegersDate: 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/ |
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