#
Problem
of the Week

Problem
C and Solution

And
the Numbers Are...

## Problem

John and Betty each choose a positive integer that is greater than
\(1\). Betty increases her number by
\(1\). John then takes this new number
and multiplies it by his number. This product is equal to \(260\).

If Betty’s number is larger than John’s number, determine all
possible pairs of integers that John and Betty could have chosen.

## Solution

Let John’s integer be \(j\) and
Betty’s integer be \(b\). We’re given
\(j \times (b+1) = 260\).

In considering the equation \(j\times
(b+1)=260\), we are looking for two integers, each greater than
\(1\), that multiply to \(260\). We also want Betty’s integer \(b\) to be greater than John’s integer \(j\).

We generate the following list of ways to factor \(260\) as a product of two integers:

\(1\times 260\), \(2\times 130\), \(4\times 65\), \(5\times 52\), \(10\times 26\), \(13\times 20\)

We can exclude \(260=1\times 260\)
because both integers must be greater than \(1\).

Since Betty’s integer is larger than John’s integer, we get the
following possibilities:

\(j=2\) and \(b+1=130\). Thus, \(j=2\) and \(b=129\).

\(j=4\) and \(b+1=65\). Thus, \(j=4\) and \(b=64\).

\(j=5\) and \(b+1=52\). Thus, \(j=5\) and \(b=51\).

\(j=10\) and \(b+1=26\). Thus, \(j=10\) and \(b=25\).

\(j=13\) and \(b+1=20\). Thus, \(j=13\) and \(b=19\).

Therefore, there are five pairs of integers that John and Betty could
have chosen. John could have chosen \(2\) and Betty chose \(129\), John could have chosen \(4\) and Betty chose \(64\), John could have chosen \(5\) and Betty chose \(51\), John could have chosen \(10\) and Betty chose \(25\), or John could have chosen \(13\) and Betty chose \(19\).