1. 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). After 12 games, Pasha is 2023 rubles richer than he was initially. How many of these games did he win?

(L. S. Korechkova, A. A. Tessler)
Answer: 9 (all except 4, 8, and 1024).

Solution. We need to place the signs in the equation $\pm 1 \pm 2 \pm 2^{2} \pm 2^{3} \pm \ldots \pm 2^{9} \pm 2^{10} \pm 2^{11}=2023$. If we choose all plus signs, the sum will be $2^{0}+\ldots+2^{11}=2^{12}-1=4095$, so we need to replace plus signs with minus signs before the numbers whose sum is $\frac{4095-2023}{2}=1036$. There is only one such set of numbers (due to the uniqueness of the binary representation of a number): $1036=1024+8+4$.

Criteria. 1 point - it is stated that the total sum of money is 4095;

another 1 point - it is found that Pasha won 3059;

2 points - the correct answer with an example;

-2 points for the lack of proof of the uniqueness of the representation of 1036.