NetNews Usenet Archive 1992 #31/NN_1992_31.iso/spool/rec/puzzles/8094

Internet Message Format | 1992-12-22 | 1.7 KB

Path: sparky!uunet!spool.mu.edu!uwm.edu!csd4.csd.uwm.edu!radcliff
From: [e-mail] (David G Radcliffe)
Newsgroups: rec.puzzles
Subject: Re: simple number puzzle
Date: 22 Dec 1992 16:48:27 GMT
Organization: UW-Milwaukee Math Department
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Given a positive integer n, in how many ways can it be expressed
as the sum of (one or more) consecutive positive integers?  The
surprising answer is that the number of such representations is
equal to the number of odd divisors of n.

First we find the number of ways of representing n as the sum of
consecutive integers.  To write n as the sum of an odd number of
consecutive integers, we need to solve (a-b) + . . . + (a+b) = n.
That is, a(2b+1)=n.  The number of solutions of this equation is
equal to the number of odd divisors of n.  Similarly, to write n
as the sum of an even number of consecutive integers, we need to
solve (a-b+1) + . . . + (a+b) = n; that is, (2a+1)b=n.  Again,
the number of solutions of this equation is equal to the number of
odd divisors of n.  The total number of representations is thus
twice the number of odd divisors of n.

If n=a+...+b then n=(1-a)+...+b.  Define an equivalence relation
on the set of all solutions by declaring that {a,...,b} ~ {1-a,...,b}.
Each equivalence class has two elements, and exactly one of these
elements contains only positive integers.  Therefore, half of
the solutions to n=a+...+b are positive.  So, the number of solutions
to n=a+...+b in positive integers is equal to the number of odd
divisors of n.

--
David Radcliffe                              [e-mail]