Find all three digit numbers which is sum of square of tens digit and cube of its units digit.
My Try:
Let $xyz$ be three digit number so we have to find all integral solutions of the equation
$$xyz=y^2+z^3 \tag{1}$$ such that $0 \lt x \le 9$ and $0\le y,z\le 9$
Case$1.$ Obviously $z$ cannot be $0,1,2$, since $y^2+z^3$ can never become three digit number.
Case $2.$ Let $z=3$ we have $$xy3=y^2+27$$ so $y^2$ should end with digit $6$.
Hence either $y=4$ or $y=6$. but in both cases RHS is not three digit number.
Case $3.$ Let $z=4$ then $$xy4=y^2+64$$ so $y^2$ should end with digit $0$. so $y=0$ for which RHS is not three digit number.
Similarly with above reason $z$ cannot take values $5$ ,$6$ and $9$.
Case $4.$ if $z=7$ we have
$$xy7=y^2+343$$ so $y^2$ should end with digit $4$.so $y$ can be either $2$ or $8$. but in both cases not feasible.
Finally if $z=8$ we have $$xy8=y^2+512$$ so $y$ can be either $4$ or $6$ which also not feasible.
hence there is no Three digit number with the property mentioned in title.