Problem E and Solution

Four Numbers

Norbert has four favourite numbers. Each of these is a three-digit number \(ABC\) with the following two properties:

The digits \(A\), \(B\), and \(C\) are all different.

The product \(A \times B \times C\) is equal to the two-digit number \(BC.\)

For example, one of Norbert’s favourite numbers is \(236\), since \(2 \times 3 \times 6 = 36.\)

Find Norbert’s other three favourite numbers.

Note: It may be helpful to recall that any two-digit number of the form \(BC\) can be represented by the sum \(10B + C.\) For example, \(32 = 10(3) + 2.\)

We know that there are only four answers, so we could attempt a trial and error approach to find the remaining three numbers, or even write a program that could find them using a brute force approach. However here we will present a more systematic approach.

First, since \(A\), \(B\), and \(C\) are digits, they must be positive integers between \(0\) and \(9\), inclusive. Also, since the product \(A \times B \times C\) is a two-digit number, none of \(A\), \(B\), or \(C\) can equal zero.

We want to find all three-digit numbers \(ABC\) such that \(A\times B\times C = 10B + C.\) Since \(B\neq 0\), we can divide by \(B\) and the problem becomes equivalent to finding all integers \(A\), \(B\), \(C\) with \(1\leq A, B, C \leq 9\) and \(A\), \(B\), and \(C\) distinct such that \[\begin{aligned} A\times C = 10 + \frac{C}{B} \end{aligned}\] Since \(A\) and \(C\) are integers, then so is \(A \times C\), so it follows that \(\frac{C}{B}\) must be an integer as well. Therefore, for each possible value of \(C\), we must have that \(B\) divides exactly into \(C.\)

We will break the problem into cases based on the value of \(C\), and then sub-cases based on the possible values of \(B.\)

**Case 1:**\(C=1\).

There are no values of \(B\) where \(\frac{C}{B}\) is an integer and \(B\neq C.\)**Case 2:**\(C=2\).

For \(\frac{C}{B}\) to be an integer and \(B\neq C\), we must have \(B=1.\) If \(B=1\) and \(C=2\), \(A\times B\times C = 10B + C\) becomes \(2A = 12\) and so \(A=6.\) Therefore, one of the three-digit numbers is \(612.\)**Case 3:**\(C=3\).

For \(\frac{C}{B}\) to be an integer and \(B\neq C\), we must have \(B=1.\) If \(B=1\) and \(C=3\), \(A\times B\times C = 10B + C\) becomes \(3A = 13\) and so \(A=\frac{13}{3}.\) Since \(A\) is not an integer, there is no solution in this case.**Case 4:**\(C=4\).

For \(\frac{C}{B}\) to be an integer and \(B\neq C\), we must have \(B=1\) or \(B=2.\)**Case 4a:**\(B=1\).

In this case, \(A\times B\times C = 10B + C\) becomes \(4A = 14\) and so \(A=\frac{7}{2}.\) Since \(A\) is not an integer, there is no solution in this case.**Case 4b:**\(B=2\).

In this case, \(A\times B\times C = 10B + C\) becomes \(8A = 24\) and so \(A=3.\) Therefore, one of the three-digit numbers is \(324.\)

**Case 5:**\(C=5\).

For \(\frac{C}{B}\) to be an integer and \(B\neq C\), we must have \(B=1.\) If \(B=1\) and \(C=5\), \(A\times B\times C = 10B + C\) becomes \(5A = 15\) and so \(A=3.\) Therefore, one of the three-digit numbers is \(315.\)**Case 6:**\(C=6\).

For \(\frac{C}{B}\) to be an integer and \(B\neq C\), we must have \(B=1\), \(B=2\), or \(B=3.\)**Case 6a:**\(B=1\).

In this case, \(A\times B\times C = 10B + C\) becomes \(6A = 16\) and so \(A=\frac{8}{3}.\) Since \(A\) is not an integer, there is no solution in this case.**Case 6b:**\(B=2\).

In this case, \(A\times B\times C = 10B + C\) becomes \(12A = 26\) and so \(A=\frac{13}{6}.\) Since \(A\) is not an integer, there is no solution in this case.**Case 6c:**\(B=3\).

In this case, \(A\times B\times C = 10B + C\) becomes \(18A = 36\) and so \(A=2.\) Therefore, one of the three-digit numbers is \(236.\) This is the number given in the example.

We can actually stop here since we have found \(4\) different three-digit numbers that satisfy the conditions outlined in the problem. If we had not been given the number of possible solutions, we would need to continue by checking cases when \(C=7\), \(C=8\), and \(C=9.\) It turns out that there are no solutions in these cases.

Therefore, Norbert’s four favourite numbers are \(612\), \(324\), \(315\), and \(236.\)