272. Find a three-digit number that is equal to the sum of the tens digit, the square of the hundreds digit, and the cube of the units digit.

$273 *$. Find the number $\overline{a b c d}$, which is a perfect square, if $\overline{a b}$ and $\bar{c} \bar{d}$ are consecutive numbers, with $\overline{a b}>\overline{c d}$.
272. $100 x+10 y+z=y+x^{2}+z^{3}$, from here $9 y=(z-$ $-1)(z+1) z+x^{2}-x-99 x, \quad(z-1) z(z+1)+x(x-1)$ must be divisible by 9; since $z(z-1)(z+1)$ is divisible by $3, x(x-1)$ is also divisible by 3, therefore, $x=1$, $3,4,6,7,9$.

When $x=3 y=\frac{z^{3}-z-291}{9}, z^{3}>z+291$, hence $z=7$, 8,9, but only when $z=7$ the number $\frac{z^{3}-z-291}{9}$ will be an integer; therefore $z=7, y=5$. The desired number is 357.

Verification. $357=5+3^{2}+7^{3}$.

Let $x=4$, then $y=\frac{z^{3}-z-384}{9},$ therefore, $z^{3}>z+384$, hence $z=8,9$, but for these values of $z$ we do not get an integer for $y$. Considering the values $x=6,7,9,1$, we will not find valid values for $y$ and $z$.

Answer. 357.