IMC 2004 Q14

We note first that \(y = 5\) since that is the only non-zero digit which, when it is multiplied by 3, has itself as the units digit. So there is a carry of 1 into the tens column. We note also that \(a = 1\) or \(a = 2\) as “\(fly\)” < 1000 and therefore 3 × “\(fly\)” < 3000. We now need 3 × \(l\) + 1 to end in either 1 or 2 and the only possibility is \(l = 7\), giving \(a = 2\) with a carry of 2 into the hundreds column. As \(a = 2\), \(f\) must be at least 6. However, if \(f = 6\) then \(w = 0\) which is not allowed. Also the letters represent different digits, so \(f \neq 7\) and we can also deduce that \(f \neq 9\) since \(f = 9\) would make \(w = 9\).
Hence \(f = 8\), making \(w = 6\) and the letters represent 875 × 3 = 2625.