Question

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$$ \underline{A}\ \underline{B}\ \underline{C} - (A^2 + B^2 + C^2). $$

Answer 1

To find the minimum value for the expression «AC« - (A^2 + B^2 + C^2), set A=1, B=0, C=0 to get the smallest three-digit number 100. Then, 100 -(1^2 + 0^2 + 0^2) equals 99, which is the minimum value of the expression.

Step-by-step explanation:

To compute the minimum value of the expression «AC« - (A^2 + B^2 + C^2), we look for the smallest possible three-digit number «AC«, where A is the hundred's digit, B is the ten's digit, and C is the unit's digit. Given that A ≥ 1 and we want the smallest such number, we set A=1. Additionally, to minimize the expression, we should aim to minimize B and C as well, setting them to 0, while still keeping the result positive if possible.

Thus, our number would be 100 (with A=1, B=0, C=0), which is the smallest three-digit number. Now, calculate the expression:

100 - (1^2 + 0^2 + 0^2) = 100 - (1 + 0 + 0) = 99.

The minimum value of the expression is therefore 99.