RMO 1998 Questions, Solutions, Discussions

Since \(a_1a_2, a_1a_2a_3a_4, a_1a_2a_3a_4a_5a_6\) are divisible by \(2\), it follows that \(a_2, a_4, a_6\) are even, and hence they are equal to \(2, 4, 6\) in some order. Using that \(a_1a_2a_3a_4a_5\) is divisible by \(5\), we get that \(a_5 = 5\). So \(a_1, a_3\) are equal to \(1, 3\) in some order. Using that \(a_1a_2a_3\) is a multiple of \(3\), we obtain \[ a_1+ a_2+ a_3\equiv 0 \bmod 3, \] which yields \[ a_2 \equiv 2 \bmod 3, \] and hence, \(a_2 = 2\) holds. Note that \(1234, 3214\) are not divisible by \(4\). This shows that \(a_4 = 6\), and hence \(a_6 = 4\). It follows that \(a_1a_2a_3a_4a_5a_6\) is equal to \(321654\), or \(123654\). Note that the integers \(321654, 123654\) satisfy the required conditions too. This proves that \(321654, 123654\) are precisely all the \(6\)-digit numbers satisfying the given condition.