Problem:
How many integers less than can be written as the sum of consecutive positive odd integers for exactly 5 values of ?
Solution:
Recall that the sum of the first positive odd integers is . Thus if is equal to the sum of the th through th positive odd integers, then . Let , and let . Note that , and and have the same parity. Thus is either odd or a multiple of . Conversely, if , where and are positive integers with the same parity and , then , where and , and it follows that is the sum of the th through th odd integers. Thus there is a one-to-one correspondence between the sets of consecutive positive odd integers whose sum is and the ordered pairs of positive integers such that and are of the same parity, , and .
First consider the case where is odd. All the divisors of have the same parity because they are all odd. Since five pairs of positive integers have product must have either or divisors. must therefore have the form , or , where and are distinct odd primes. But cannot have the form or , because that would imply that . If has the form , then because , and there are two possible values of , namely and . If has the form , then must be , or .
In the case where is even, , where and for positive integers and . In this case, has five pairs of divisors of the same parity if and only if has or divisors. Count the number of positive integers less than that are of the previously mentioned forms, except that now or can be . There are no integers less than that are of the form or ; there are four such integers of the form , namely, , , and ; and there are six such integers of the form , namely, , , and .
Thus there are a total of possible values for .
The problems on this page are the property of the MAA's American Mathematics Competitions