Solution #41: A Bag Of Marbles

Let n be the number of children, x each child's share of marbles, and N the total number of marbles. We have N = nx.

First child's share: 1 + (N-1)/10 = x.

Last child's share: x.

Penultimate child's share: n - 1 + x/9 = (N/x) - 1 + x/9 = x.

Consider the expression for the first child's share: N = 10x - 9.

Substituting into the equation for the penultimate child's share gives us

(10x - 9)/x - 1 + x/9 = x
which can be written
8x2 - 81x + 81 = 0 or (x - 9)(8x - 9) = 0
The number of marbles taken by each child being a whole number, the only possible solution is x = 9;

hence N = 81 and n = 9.

There were nine children. Each got nine marbles.

Source: Berrondom, Marie
Categories: Number sense, Modular arithmetic, Algebra
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