1. In a game, three types of tokens are used, each with a different value expressed in denars. The value of each token is a natural number. Bojan, Ace, and Sasha each have at least one token of each type. Bojan has 4 tokens with a total value of 28 denars, Ace has 5 tokens with a total value of 21 denars, and Sasha has 3 tokens. What is the total value of Sasha's tokens?
Solution. Let $a, b, c$ be the values of the tokens. Then the total value of Bojan's tokens is $2a + b + c = 28$. The value of Ace's tokens can be $3a + b + c, 2a + 2b + c, 2a + b + 2c, a + 2b + 2c, a + b + 3c$ or $a + 3b + c$. Since $3a + b + c = 28 + a > 28, 2a + 2b + c = 28 + b > 28$ and $2a + b + 2c = 28 + c > 28$, we conclude that the following cases are possible: $a + 2b + 2c = 21, a + 3b + c = 21$ or $a + b + 3c = 21$.

If $a + 2b + 2c = 21$, then from $2a + b + c = 28$ it follows that $3(a + b + c) = 49$, which is not possible since 49 is not divisible by 3.

If $a + 3b + c = 21$, then from $2a + b + c = 28$ it follows that $a - 2b = 7$, i.e., $a = 2b + 7$. According to this, the number $a$ is odd and is greater than or equal to 9. Now for $a = 9$, we get $b = 1$ and therefore $c = 9$, which is not possible, since $a, b, c$ are different natural numbers. Thus, $a \geq 11$. Since $b + c \geq 3$, we get $2a = 28 - (b + c) \leq 25$, so $a \leq \frac{25}{2}$ and since $a$ is an odd number and $a \geq 11$, it follows that $a = 11$. Finally, $a = 11, b = 2$ and $c = 4$, so $a + b + c = 17$.

In exactly the same way, it is obtained that from $a + b + 3c = 21$ it follows that $a = 11$, $b = 4$ and $c = 2$, i.e., $a + b + c = 17$.

Finally, the total value of Sasha's tokens is 17 denars.