In Kocourkov, coins are used with only two values, expressed in Kocourkov crowns as positive integers. With a sufficient amount of such coins, it is possible to pay any integer amount greater than 53 Kocourkov crowns exactly and without change. However, the amount of 53 Kocourkov crowns cannot be paid exactly without change.

Determine which values could be on the Kocourkov coins. Find at least two solutions.

(A. Bohiniková)

Hint. Could one of the Kocourkov coins be a one-crown coin, or a two-crown coin, ...?
Let's consider the value of one set of cat coins and the value of another set to meet the requirements of the problem.

1) If one of the coins had a value of 1, then any amount could be paid.
2) Suppose one set of coins has a value of 2. The other coins cannot have an even value, because with such coins only even amounts could be paid.

The other coins cannot have an odd value less than 55, because with such coins an amount of 53 could be paid (e.g., $2 \cdot 2 + 49, 2 + 51$). If the other coins had a value of 55, then any even amount could be paid using a sufficient number of 2s, and any odd amount greater than 53 could be paid using 55 and a sufficient number of 2s:

$$
\begin{array}{lll}
54 = 27 \cdot 2 & 56 = 28 \cdot 2 & 58 = 29 \cdot 2 \\
55 = 1 \cdot 55 & 57 = 2 + 1 \cdot 55 & 59 = 2 \cdot 2 + 1 \cdot 55
\end{array}
$$

3) Suppose one set of coins has a value of 3. The other coins cannot have a value divisible by three, because with such coins only amounts divisible by three could be paid.

The other coins cannot have a value of $5, 8, 11, \ldots, 53$, because with such coins an amount of 53 could be paid (e.g., $16 \cdot 3 + 5, 15 \cdot 3 + 8, \ldots, 3 + 50$). In this group, we considered values that give a remainder of 2 when divided by 3. Other values with the same remainders are 56, 59, etc. These options also do not work, because with such coins an amount of 55 (which is smaller and gives a remainder of 1 when divided by 3) could not be paid.

The other coins cannot have a value of $4, 7, 10, \ldots, 25$, because with such coins an amount of 53 could also be paid (e.g., $15 \cdot 3 + 2 \cdot 4, 13 \cdot 3 + 2 \cdot 7, \ldots, 3 + 2 \cdot 25$). In this group, we considered values that give a remainder of 1 when divided by 3. If the other coins had a value of 28, then any amounts greater than 53 could be paid, for example, as follows:

$$
\begin{aligned}
& 54 = 18 \cdot 3 \quad 57 = 19 \cdot 3 \quad 60 = 20 \cdot 3 \\
& 55 = 9 \cdot 3 + 1 \cdot 28 \quad 58 = 10 \cdot 3 + 1 \cdot 28 \quad 61 = 11 \cdot 3 + 1 \cdot 28 \\
& 56 = 2 \cdot 28 \quad 59 = 1 \cdot 3 + 2 \cdot 28 \quad 62 = 2 \cdot 3 + 2 \cdot 28
\end{aligned}
$$

So far, we have two solutions: the values on the cat coins could be 2, 55 or 3, 28. In a similar (but somewhat more laborious) way, we can also discover the possible values 4, 19 and 7, 10.

Note. The problem we are solving in this task is known as the Frobenius problem: the suitable pairs of coin values $a, b$ must satisfy $a b - a - b = 53$ or

$$
(a-1)(b-1) = 54
$$

With this (non-trivial) insight, we can indirectly verify that the four cases mentioned are the only possible ones.