## 44. Bag of Balls

Children are dividing a bag of balls among themselves. The first child took one ball and a tenth of the remaining balls, then the second took 2 balls and a tenth of the remaining, then the third took 3 balls and a tenth of the remaining, and so on, until the last child took all that was left.

How many children were there and how many balls did each get, if all the children received the same number of balls?
44. Let $n$ be the number of children, $x$ be the share of balloons taken by each of them, and $N$ be the total number of balloons. Obviously, $N=n x$. The share of balloons taken by the first child is

$$
1+(N-1) / 10=x
$$

The share of the last child is $x$. The share of the second-to-last child can be written as:

$$
(n-1)+x / 9=N / x-1+x / 9=x
$$

since $9 / 10$ of the number of balloons remaining after the second-to-last child took $(n-1)$ balloons is $x$, meaning the additional $1 / 10$ part of the remainder he took, in addition to $(n-1)$ balloons, is $(1/9)x$. From the relation indicating the share of the first child, it follows that

$$
N=10 x-9
$$

then the expression defining the share of the second-to-last child gives

$$
(10 x-9) / x-1+x / 9=x
$$

The last equation can be rewritten as:

$$
8 x^{2}-81 x+81=0
$$

or

$$
(x-9)(8 x-9)=0
$$

Since the number of balloons each child received is an integer, the only possible solution is: $x=9$. Therefore, $N=81$ and $n=9$: there were nine children in total, and each received nine balloons (check this!).