Olympiad Maths Prep

Track / Stage 5 / 223 of 400 #823 of 2000

Problem 823

AIME late
Number theory Difficulty 5.5 Find the answer

24. (HUN 5) IMO6{ }^{\mathrm{IMO6}} 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.

Official solution

24. Let the kk th child receive xkx_{k} coins. By the condition of the problem, the number of coins that remain after him was 6(xk−k)6\left(x_{k}-k\right). This gives us a recurrence relation
xk+1=k+1+6(xk−k)−k−17=67xk+67, x_{k+1}=k+1+\frac{6\left(x_{k}-k\right)-k-1}{7}=\frac{6}{7} x_{k}+\frac{6}{7},
which, together with the condition x1=1+(m−1)/7x_{1}=1+(m-1) / 7, yields
xk=6k−17k(m−36)+6 for 1≤k≤n.  x_{k}=\frac{6^{k-1}}{7^{k}}(m-36)+6 \text { for } 1 \leq k \leq n \text {. }

Since we are given xn=nx_{n}=n, we obtain 6n−1(m−36)=7n(n−6)6^{n-1}(m-36)=7^{n}(n-6). It follows that 6n−1∣n−66^{n-1} \mid n-6, which is possible only for n=6n=6. Hence, n=6n=6 and m=36m=36

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.