#
Problem
of the Week

Problem
C and Solution

Divisors
and Number

## Problem

Your friend Cael always likes challenging you. One challenge is
called “*Divisors and Number*”. Cael will tell you certain facts
about the divisors of a number and then challenge you to find the
number. Here is Cael’s challenge.

“I am looking for a positive integer with exactly eight positive
divisors, two of which are \(21\) and
\(33\).”

Determine Cael’s number.

## Solution

Let \(n\) represent the number we
are looking for.

We know that four of the positive divisors of \(n\) are \(1\), \(21\), \(33\), and \(n\). In our solution we will first find the
remaining four positive divisors and then determine \(n\).

Since \(21\) is a divisor of \(n\) and \(21 =
3\times 7\), then \(3\) and
\(7\) must also be divisors of \(n\).

Since \(33\) is a divisor of \(n\) and \(33 =
3\times 11\), then \(11\) must
also be a divisor of \(n\).

Since \(7\) is a divisor of \(n\) and \(11\) is a divisor of \(n\), and since \(7\) and \(11\) have no common divisors, then \(7\times 11 = 77\) must also be a divisor of
\(n\).

We have found all eight of the positive divisors of the unknown
number. The positive divisors are \(1\), \(3\), \(7\), \(11\), \(21\), \(33\), \(77\), and \(n\). We now need to determine \(n\).

From the list of divisors, we can see that the prime factors of \(n\) are \(3\), \(7\), and \(11\). It follows that \(n = 3\times 7\times 11 = 231\).

Therefore, Cael’s number is \(231\).