3. Pasha and Igor are flipping a coin. If it lands on heads, Pasha wins; if tails, Igor wins. The first time the loser pays the winner 1 ruble, the second time - 2 rubles, then - 4, and so on (each time the loser pays twice as much as the previous time). At the beginning of the game, Pasha had a single-digit amount of money, and Igor had a four-digit amount. By the end, Igor had a two-digit amount, and Pasha had a three-digit amount. What is the minimum number of games Pasha could have won? The players cannot go into debt.

(L. S. Korechkova, A. A. Tessler)
Solution. Let $n$ be the amount of money Pasha has become richer (and Igor poorer). Note that Pasha won the last game (otherwise, he would have lost more money than he gained in all previous stages). Therefore, the sequence of games can be divided into series, in each of which Pasha won the last game and lost all the others (a series can consist of just one game). If a series started with game number $k$ and ended with game number $m$, then Pasha won $-2^{k}-2^{k+1}-\ldots-2^{m-2}+2^{m-1}=2^{k}$ rubles for it. Thus, the binary representation of the number $n$ uniquely describes the set of games Pasha won (except for the number of the last game): the term $2^{k}$ means that the next series started with game number $k+1$, i.e., Pasha won game number $k$.

According to the problem, $901 \leqslant n \leqslant 998$. But all numbers from 901 to 998 contain $2^{7}+2^{8}+2^{9}$ in their binary representation, so Pasha won the seventh, eighth, and ninth games, as well as the last one (its number is greater than 9, otherwise there would be no term $2^{9}$) - already a minimum of 4 games.

In addition, Pasha must have won at least 3 times in the first 6 games:

1) at least one out of the first four games, since $9-1-2-4-8<0$;
2) at least one out of the next two, since $9 \pm 1 \pm 2 \pm 4 \pm 8-16-32<0$;
3) if only one out of the first four was won, then after them the sum is no more than 10, and the fifth and sixth must definitely be won.

Thus, Pasha won at least 7 games. Here is an example for 7 games: initially, Pasha had 9 rubles, and Igor had 1000 rubles, a total of 10 games were played. Then $n=985=2^{0}+2^{3}+$ $2^{4}+2^{6}+2^{7}+2^{8}+2^{9}=\left(-2^{0}-2^{1}+2^{2}\right)+\left(2^{3}\right)+\left(2^{4}\right)+\left(-2^{5}+2^{6}\right)+\left(2^{7}\right)+\left(2^{8}\right)+\left(2^{9}\right)$, i.e., Pasha won games with numbers $3,4,6,7,8,9,10$, and Igor - games $1,2,5$. In the end, Pasha will have 994 rubles, and Igor - 15 rubles.

Answer: 7 games.

Criteria. 5 points are given for the estimate (1 point for estimating that there were at least 10 games in total), 2 points for the example.