IMO 1967 LL HUN24

Origin: HUN

Problem

Father has left to his children several identical gold coins. According to his will, the oldest child receives one coin and one-seventh of the remaining coins, the next child receives two coins and one-seventh of the remaining coins, the third child receives three coins and one-seventh of the remaining coins, and so on through the youngest child. If every child inherits an integer number of coins, find the number of children and the number of coins.

Solution

Let the kth child receive xk coins. By the condition of the problem, the number of coins that remain after him was 6(xk −k). This gives us a recurrence relation xk+1 = k + 1 + 6(xk −k) −k −1 = 6 7xk + 6 7,

which, together with the condition x1 = 1 + (m −1)/7, yields xk = 6k−1 7k (m −36) + 6 for 1 \leqk \leqn. Since we are given xn = n, we obtain 6n−1(m−36) = 7n(n−6). It follows that 6n−1 | n −6, which is possible only for n = 6. Hence, n = 6 and m = 36.