Let $\underline{A}\ \underline{B}\ \underline{C}$ represent a three-digit base 10 number whose digits are $A$, $B$, and $C$ with $A \ge 1$. Compute the minimum value of$$ \underline{A}\ \underline{B}\ \underline{C} - (A^2 + B^2 + C^2). $$

$\underline{A}\ \underline{B}\ \underline{C}$ represent a three-digit base 10 number whose digits are $A$, $B$, and $C$ with $A \ge 1$. The minimum value of$$ \underline{A}\ \underline{B}\ \underline{C} - (A^2 + B^2 + C^2) is :

$a=1$, $b=0$ and $c=9$.

Explanation:

Simplified the expression given above as $100a-a^2+10b-b^2+c-c^2$.

This expression contains three quadratic expressions,namely $100a-a^2$,$10b-b^2$,$c-c^2$.

You want to minimize the expression $$100a-a^2+10b-b^2+c-c^2,$$ which is the sum of three quadratics.

It is minimal when the three quadratics $$100a-a^2,\qquad 10b-b^2,\qquad c-c^2,$$ are minimal.

These are parabolas with their maxima at $50$, $5$ and $\tfrac{1}{2}$, respectively.

They are minimal when we are as far away from the maximum as possible. This gives us $a=1$, $b=0$ and $c=9$.