Solution to May 2002 Problem

We define the fractional part of r to be

F(r) = r - (the largest integer that divides evenly into r). Some examples: F(12.34) = 0.34, F( 11/2) = 1/2, F( 8/3) = 2/3. Problem for May: Find a positive number r such that F(r) + F( 1/r) = 1.

The source of this month's problem is the fifth annual Team Competition organized by the North Central Section of the Math Association of America. It was held November 10, 2001.



Four correct solutions were submitted. Our solvers are

John Campbell (Edmonton, Alberta)
Normand Laliberté (Kitchener, Ontario)
Juan Mir Pieras (Spain)
Alexander Potapenko (Russia) All answered the more general question of finding all positive real numbers r that satisfy the desired condition. We present a combination of the four solutions.

We first observe that r is not 1 (because r = 1 implies that F(r) = F( 1/r) = 1). We can therefore label our desired pair of numbers so that r > 1 > 1/r > 0. This means that F(r) = r -  (where denotes the greatest integer less than or equal to r), while F( 1/r) = 1/r. Our assumption that F(r) + F( 1/r) = 1 now implies that (r - ) + 1/r = 1, or

r + 1/r = n for some n >= 2 Since r( 1/r) = 1 and r + ( 1/r) = n, r and 1/r must be the roots of x 2 - nx + 1 = 0, namely

Conversely, for any integer n >= 3


Since the product of these two numbers is 1, we deduce that for all n ³ 3. Thus and we conclude that equation (*) provides a solution to our problem for any integer n >= 3. We must exclude n = 2 because n = 2 implies that r = 1/r = 1. Thus, the smallest two examples are n = 3:

so that


n = 4:

so that

Comments on the golden section.

Although all correspondents sent in a solution that addressed the general question (which is certainly more interesting and more challenging), the original problem requested just one example of a number r for which F(r) + F( 1/r) = 1. Anybody with an understanding of the golden section would have been able to devise a solution immediately, after only trivial calculations. The golden section, often denoted by the Greek letter , is generally defined to be the positive solution to x 2 = x&nnbsp;+ 1, namely

This equation, namely , tells us how to compute all powers of just by addition! For example . More relevant, divide both sides of the equation by to get and once more to get Thus and provide a solution to the original problem: One simply checks that while Indeed, is the case n = 3 in our featured solution.