1. How many five-digit numbers have the first digit equal to the last, the digit in the thousands place equal to the digit in the tens place, and the product of all digits equal to the square of a natural number?
IV/1. The product of the digits of a five-digit number of the form $\overline{a b c b a}$ is $a^{2} \cdot b^{2} \cdot c$. The digit $c$ must therefore be a perfect square, so it can be 1, 4, or 9. The other digits are any non-zero digits, so there are 9 possibilities for $a$ and 9 possibilities for $b$. In total, there are thus $9 \cdot 9 \cdot 3=243$ such numbers.

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