Sharing Marbles

Let there be $N$ boys and let boys $1, 2, ... N$ and girls $1, 2 ... N+16$ start with $b_1, b_2, ... b_N, g_1, g_2, ... g_{N+16}$ marbles respectively. Let $g_{total}$ be the total number of girl marbles.

(Boy 1 gives $\frac{b_1}{N+17}$ to himself and the $N+16$ girls etc.)

Boys $i$ and $j$ end up with $\frac{b_i}{N+17} + \frac{g_{total}}{N+1}$ and $\frac{b_j}{N+17} + \frac{g_{total}}{N+1}$ . Hence, $b_i = b_j = b, \forall i,j$ and some $b$ . Similarly for the girls all starting with $g$ each.

We can also note that $(N+17) | b$ and $(N+1) | g$ . So we can let $b = (N+17)B$ and $g = (N+1)G$ . ( $B$ is the number of marbles giving from a boy to a girl etc.)

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Equating the final amounts of the boys and girls, we get:

Fred's final amount, $F = B + (N+16)G = NB + G$ , so $(N - 1)B = ( (N - 1) + 16 )G$ .

$kB = k(1 + \frac{16}{N - 1})G$ , where k is the greatest divisor of $N$ that doesn't divide $16$ (so \frac{16k}{N - 1} is an integer coprime to $k$ ).

To minimise $F$ we must minimise $B$ and $G$ , which have minimum values $k(1 + \frac{16}{N - 1})$ and $k$ respectively. ( $F = k [ N(1 + \frac{16}{N - 1}) + 1]$ ).

Assuming constant $k$ , we can minimise $F$ w.r.t. $N$ by differentiation: $\frac{dF}{dN} = k [1 + \frac{16}{N - 1} - \frac{16N}{(N - 1)^2} ] = 0 \implies (N - 1)^2 + 16(N - 1) - 16N = (N - 1)^2 - 16 = 0 \implies N = 5$ .

(We can note that $N = 5 \implies k = 1$ , thus minimising it, hence this is a true minimum. If $k \neq 1$ , there may be another $N$ nearby with a smaller $k$ and a smaller $F$ ).

For $N = 5, B = 5, G = 1$ , and $F = 26$ .

Note, in this case, the 5 boys start with $110$ marbles and the 21 girls start with $6$ . Totalling $676 = 26^2$ .