4. Determine the number of all triples of two-digit natural numbers $a, b, c$, whose product $abc$ has a representation in which all digits are the same. Triples differing only in the order of the numbers are considered the same, i.e., they are counted only once.
SOLUTION. For two-digit numbers $a, b, c$, the product $abc$ is a four-digit, five-digit, or six-digit number. If all the digits of the number $abc$ are equal to the same digit $k$, one of the equalities $abc = k \cdot 1111, abc = k \cdot 11111$, or $abc = k \cdot 111111, k \in \{1,2, \ldots, 9\}$ holds.

The numbers $1111 = 11 \cdot 101$ and $11 \cdot 111 = 41 \cdot 271$ have three-digit prime numbers in their factorization, so they cannot be the product of two-digit numbers. Therefore, the only remaining possibility is:

$$
abc = k \cdot 111111 = k \cdot 3 \cdot 7 \cdot 11 \cdot 13 \cdot 37
$$

Let's see how the prime numbers $3, 7, 11, 13, 37$ can be distributed among the individual factors $a, b, c$. Since the products $37 \cdot 3$ and $3 \cdot 7 \cdot 11$ are greater than 100, the prime number 37 must be a factor on its own, and the remaining four prime numbers $3, 7, 11, 13$ must be divided into pairs. Since the product $11 \cdot 13$ is also greater than 100, the only possible distributions are the factors $3 \cdot 11, 7 \cdot 13$, and 37, or the factors $3 \cdot 13, 7 \cdot 11$, and 37. We then add the possible factors from the factorization of the digit $k$ and get two types of solutions:

$$
\begin{array}{ll}
a = 33k_1, b = 91, c = 37k_2, & \text{where } k_1 \in \{1,2,3\}, k_2 \in \{1,2\}, \\
a = 39k_1, b = 77, c = 37k_2, & \text{where } k_1 \in \{1,2\}, k_2 \in \{1,2\},
\end{array}
$$

The number of triples of numbers $a, b, c$ we are looking for is therefore $3 \cdot 2 + 2 \cdot 2 = 10$.

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