PK 2019 Q25

Let the 4-digit integer be ‘\(abcd\)’. When it is divided by ‘\(abc\)’, we get ‘\(abcd\)’ = 10 × ‘\(abc\)’ + \(d\). Since ‘\(abc\)’ is a factor of ‘\(abcd\)’, we must have \(d = 0\), and the integer is ‘\(abc\)0’. Similarly, when ‘\(abc\)0’ is divided by ‘\(ab\)0’, we get ‘\(abc\)0’ = 10 × ‘\(ab\)0’ + ‘\(c\)0’. But ‘\(c\)0’ can only be divisible by ‘\(ab\)0’ if \(c = 0\). Thus the integer is ‘\(ab\)00’. This is divisible by ‘\(b\)00’, so we can’t have \(b = 0\). When we divide ‘\(ab\)00’ by ‘\(a\)00’, we get ‘\(ab\)00’ = 10 × ‘\(a\)00’ + ‘\(b\)00’, hence ‘\(b\)00’ must be a multiple of ‘\(a\)00’, and therefore \(b\) is a multiple of \(a\).

Since ‘\(b\)00’ is a factor of ‘\(ab\)00’ and ‘\(ab\)00’ = ‘\(a\)000’ + ‘\(b\)00’, it must be that ‘\(b\)00’ divides ‘\(a\)000’, hence ‘\(a\)0’ is a multiple of \(b\). For \(b\) to be a multiple of \(a\) and a factor of \(10 \times a\), \(b\) must be \(a\), \(2a\), \(5a\) or \(10a\). But \(10a\) is more than one digit long. When \(b = a\) we get ‘\(ab\)00’ is 1100, 2200, 3300, 4400, 5500, 6600, 7700, 8800, 9900. When \(b = 2a\), we get 1200, 2400, 3600, 4800. When \(b = 5a\), we get 1500. This is 9 + 5 = 14 possibilities.