Maclaurin 2016 Q5
5. The number of my hotel room is a three-digit integer. I thought that the same number could be obtained by multiplying together all of:
Prove that I was mistaken.
Maclaurin Olympiad 2016, Question 5. © UK Mathematics Trust. Original paper on ukmt.org.uk
Suppose that my hotel room number is ‘\(abc\)’, that is, \(100a + 10b + c\).
If my belief is true, then
which we may rewrite as
Now \((b + 1)\,c\) is at most 90 because \(b\) and \(c\) are digits. Also, \(a\) is at least 1 because the room number is a three-digit integer. Hence \(\big(100 - (b + 1)\,c\big)\,a\) is at least 10; in particular, it is positive.
Furthermore, \(10 - c\) is at least 1 and \(b\) is at least zero, so \((10 - c)\,b\) is at least zero.
It follows that the left-hand side of equation (∗) is positive, which is not possible.
Hence my belief is mistaken.
Solution © UK Mathematics Trust, from the Maclaurin Olympiad 2016 solutions.