Solution. Since $A$ is non-zero and a perfect cube, $A=b^{3}$ with $b \in N^{*}$. Since $A<(100)^{3}$, it follows that $b<100$. $A$ is even, so $b$ is an even number. From $b=2 c$, we get $A=8 \cdot c^{3}$, with $1 \leq c \leq 49 . A: 2=4 \cdot c^{3}=(2 \cdot c)^{2} \cdot c$ being a perfect square, leads to the conclusion that $c$ is a perfect square. We get $c \in\{1,4,9,16,25,36,49\}$ and thus, there are 7 numbers that satisfy the given conditions.
Problem 3. Specify how many non-zero numbers $A$, of at most six digits, satisfy, at the same time, the conditions:

1) $A$ is even
2) $A$ is a perfect cube
3) half of $A$ is a perfect square.