# 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 $$231$$.