A number's cube consists of five digits, of which the first is 3, the fifth is 4. Which is this number?
The number $N$, whose cube is a five-digit number, consists of two digits. The cube number

$$
30000<N^{3}<40000
$$

and thus

$$
30<N<40
$$

The last digit of $N^{3}$ is 4, so the second digit of $N$ is even, and it can only be 4. The sought number is 34 and $34^{3}=39304$.

[Indeed, $2^{3}=8,4^{3}=64,6^{3}=216,8^{3}=512$. If the second digit of the two-digit number is $2,6,8$, then in the cube number, the digit in the units place is $8,6,2$].

Holló György (Dobó István g. VI. o. Eger)

Note: Even if we phrase the problem more generally: The first digit of the cube of a two-digit number is 3, and the last digit is 4. Which is this number?