175. I thought of two two-digit numbers starting with the digit 6, and the other digits of these numbers are not 6. If the digits in each of the thought-of numbers are swapped, the value of the product of the two-digit numbers does not change. What numbers did I think of
Solution. Method I. Let the units digits of the two-digit numbers be $x$ and $y$, with $x \neq 6$ and $y \neq 6$. According to the problem, $\overline{6 x} \cdot \overline{6 y}=\overline{x 6} \cdot \overline{y 6}$, i.e., we have the equation

$$
(6 \cdot 10 + x)(6 \cdot 10 + y) = (10 x + 6)(10 y + 6)
$$

Solving this equation, we get:

$$
36 \cdot 10^{2} + 60 x + 60 y + x y = 10^{2} x y + 60 x + 60 y + 36,
$$

from which $99 x y = 36 \cdot 99, x y = 36$.

Considering that $x$ and $y$ are natural numbers less than 10 (since $x$ and $y$ are digits) and not equal to 6 by the condition, we get $x=9, y=4$ or $x=4, y=9$. The numbers thought of are 69 and 64.

Method II. According to the problem, we have $\overline{6 x} \cdot \overline{6 y}=\overline{x 6} \cdot \overline{y 6}$.

Since the product of 6.6 ends in the digit 6, the product of $x \cdot y$ must also end in the digit 6. Let's list all pairs of single-digit numbers whose product ends in the digit 6:

$(3 ; 2),(4 ; 4),(2 ; 8),(9 ; 4),(6 ; 6),(7 ; 8)$.

Now let's check which of the two-digit numbers satisfy the condition of the problem (we can skip the pair $(6 ; 6)$, as $x \neq 6, y \neq 6$ by the condition):

1) $63 \cdot 62 \neq 36 \cdot 26$
2) $64 \cdot 64 \neq 46 \cdot 46$
3) $62 \cdot 68 \neq 26 \cdot 86$
4) $69 \cdot 64 = 96 \cdot 46$
5) $67 \cdot 68 \neq 76 \cdot 86$

As a result of the check, we established that only the pair $(9 ; 4)$ satisfies the condition of the problem. Therefore, the numbers thought of are 69 and 64.

Answer: 69 and 64.