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.
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 \(231\).