23(1151). Does there exist a two-digit number that, when divided by the sum of the squares of its digits, gives a quotient of 2 and a remainder of 6, and when divided by the product of its digits, gives a quotient of 4 and a remainder of 6?
Solution. Let $\overline{x y}=10 x+y$ be the desired two-digit number, then from the condition of the problem we have the system of equations:

$$
\left\{\begin{array}{l}
10 x+y=\left(x^{2}+y^{2}\right) \cdot 2+6 \\
10 x+y=x y \cdot 4+6
\end{array}\right.
$$

from which $\left(x^{2}+y^{2}\right) \cdot 2=4 x y, x^{2}+y^{2}=2 x y$, i.e., $(x-y)^{2}=0$, from which $x=y$.

Substituting the value $x=y$ into one of the equations of the system (1), we get the equation $4 x^{2}-11 x+6=0$, from which $x=2\left(x=\frac{3}{4}\right.$ does not satisfy the condition of the problem). We obtained the number $x y=22$, which does not satisfy the second condition of the problem.

Remark. Having obtained $x=y$, the desired two-digit number can be found in another way: choose from the numbers $11,22,33, \ldots, 99$ those that satisfy the condition of the problem. There are no such numbers. Answer: does not exist.

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