Hamilton 2016 Q6

An integer is divisible by 4 when the number formed from the rightmost two digits is a multiple of 4, and not otherwise. But 74 is not a multiple of 4, so \(N\) is not divisible by 4.

However, when two even numbers are multiplied together, the result is a multiple of 4. We conclude that Tony's list of consecutive integers does not include two even numbers.

There are two possibilities: either he has multiplied three consecutive integers ‘odd’, ‘even’, ‘odd’; or he has multiplied two consecutive integers. But when two consecutive integers are multiplied together the last digit is never 4—the only options for the last digits are \(0 \times 1\), \(1 \times 2\). \(2 \times 3\), \(3 \times 4\), \(4 \times 5\), \(5 \times 6\), \(6 \times 7\), \(7 \times 8\), \(8 \times 9\), \(9 \times 0\), so the result ends in 0, 2, 6, 2, 0, 0, 2, 6, 2 or 0.

We are therefore trying to find an odd integer \(n\) such that \(n \times (n + 1) \times (n + 2) =\) ‘47… 74’. We also know that \(n + 1\) is not a multiple of 4.

Now \(81 \times 82 \times 83\) fails because it ends in 6. It is also too big, since it is bigger than \(80 \times 80 \times 80 = 512\,000\).

Also \(73 \times 74 \times 75\) fails because it ends in 0. It is also too small, since it is smaller than \(75 \times 75 \times 75 = 421\,875\).

The only remaining possibility is \(77 \times 78 \times 79 = 474\,474\), which works.