277. Find a two-digit number that is equal to the sum of the cube of its tens digit and the square of its units digit.
277. From the equality $10 x+y=x^{3}+y^{2}$ we find $x(10-$ $\left.-x^{2}\right)=y(y-1)$, hence $0<x<\sqrt{10}$; therefore, $x=1$, 2,3 ; but $y(y-1)$ is an even number, so $x$ is also even; $x=2$, then $y=4$.

Answer. 24.