16*. Find a two-digit number that has the property that the cube of the sum of its digits is equal to the square of the number itself.
16. Solution. Let \(x\) be the sum of the digits of a two-digit number \(y\). According to the condition, \(x^{3}=y^{2}\). This equality in natural numbers is possible only when \(x=z^{2}\) and \(y=z^{3}\), where \(z \in N\). Since the sum of the digits of a two-digit number is no more than 18, then \(z^{2} \leqslant 18\), and therefore \(z \leqslant 4\). By checking, we find that there is only one number that satisfies the conditions of the problem: 27.