, 2012

Let $(S, G, P)$ be the set of gold, silver and platinum coins of Bill at some moment. The initial set is $(16, 15, 14)$. Note that Bob and Bill have together $30$ gold, $30$ silver and $30$ platinum coins. So, $S \le 30$, $G \le 30$, and $P \le 30$ at any moment. By condition, $S + G + P = 16 + 15 + 14 = 45$ at any time. Therefore, if $G = 0$ at some moment, then $P = 45 - S \ge 45 - 30 = 15$ at the same moment, so $15 \le P \le 30$.

By condition, after any interchange of coins the set $(S, G, P)$ can be one of the following sets
1) $(S+2, G-1, P-1)$,
2) $(S-2, G+1, P+1)$,
3) $(S-1, G+2, P-1)$,
4) $(S+1, G-2, P+1)$,
5) $(S-1, G-1, P+2)$,
6) $(S+1, G+1, P-2)$.
It is easy to see that the difference $G - P$ in the old and in any new set are congruent modulo $3$ in any case. We have $G - P = 15 - 14 = 1$ for the initial set, so $G - P \equiv 1 \pmod 3$ at any time. For $G = 0$ we have $P \equiv 2 \pmod 3$. We see that the numbers $17, 20, 23, 26,$ and $29$ are congruent $2$ modulo $3$ (among the numbers from $15$ to $30$). Therefore, when Bill has $0$ gold coins he can have only one of these five values of platinum coins. On the other hand, Bill can have any of these five numbers of platinum coins.

Indeed, if Bill gives two gold coins to Bill seven times successively, then $(16, 15, 14) \rightarrow (23, 1, 21)$. The first table shows how Bill can get the smallest number of platinum coins ($17$), and the second table shows how Bill can get the greatest number of platinum coins ($29$). We also see that Bill can also get $20, 23, 26$ coins.

| $S$ | 23 | 21 | 22 | 23 | 24 | 25 | 26 | 27 | 28 |
|-----|----|----|----|----|----|----|----|----|----|
| $G$ | 1 | 2 | 0 | 1 | 2 | 0 | 1 | 2 | 0 |
| $P$ | 21 | 22 | 23 | 21 | 19 | 20 | 18 | 16 | 17 |

| $S$ | 23 | 21 | 22 | 20 | 18 | 19 | 17 | 15 | 16 |
|-----|----|----|----|----|----|----|----|----|----|
| $G$ | 1 | 2 | 0 | 1 | 2 | 0 | 1 | 2 | 0 |
| $P$ | 21 | 22 | 23 | 24 | 25 | 26 | 27 | 28 | 29 |