TITLE: A riddle involving series.
QUESTION [8 upvotes]: 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 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 then find the number of children and gold coins.
I tried to write $x_k$ as some function of $k$ (where $x_k$ is the number of coins taken by the $k_{th}$ child) but failed. All I could write is $x_k= k + \frac{1}{7}(n-k- S_{k-1})$ where $S_k$ denotes the sum of first $k$ terms then $S_k=\frac{n}{7} + \frac{6}{7}(S_{k-1} +k)$ but I cannot proceed further, please help.

REPLY [2 votes]: If $n>0$ is the number of children and we denote by $c_k$ the number of coins left after the first $k$ children have taken their share, we have
$$ c_{k+1} = \frac{6}{7} \bigl(c_k - (k+1)\bigr) ~~~\text{ for } k=0,\ldots,n-1 \enspace, $$
with $c_n = 0$.  Applying generating functions,
$$ C(x) \biggl(\frac{1}{x} - \frac{6}{7} \biggr) = \frac{c_0}{x} - \frac{6}{7}\biggl(\frac{nx^n}{x-1} - \frac{x^n-1}{(x-1)^2}\biggr) \enspace. $$
Taking $x=\frac{7}{6}$, the left-hand side vanishes and we can solve for $c_0$:
$$ c_0 = 36 + 6 (n-6)\biggl(\frac{7}{6}\biggr)^{\!n} \enspace. $$
The only positive value of $n$ for which $c_0$ is an integer is $6$.  Hence $c_0=36$.