2016 AMC 10B Problem 6

The two hundreds digits are nonzero and different from each other, so they add to at least $1 + 2 = 3$. Hence $S \ge 300$, and $S$ is a three-digit number, so $300 \le S \le 999$.

A three-digit number with digit sum $1$ is only $100$, and with digit sum $2$ only $101, 110, 200$; all are below $300$. Digit sum $3$ with $S \ge 300$ forces $S = 300$. But $S = 300$ needs the last two digits of the addends to sum to $00$ (no carry, hundreds digits $1$ and $2$, then all four remaining digits are $0$) or to $100$ with hundreds digits $1$ and $1$; either way a digit repeats. So digit sum $3$ is impossible.

Digit sum $4$ is achievable: $143 + 257 = 400$, and the digits $1, 4, 3, 2, 5, 7$ are all different.

The answer is $\boxed{\textbf{(B)}\ 4}$.

"Smallest possible" problems are two-part: a lower bound from structure (distinct hundreds digits push $S$ to at least $300$, and small digit sums require very special numbers), then an explicit construction that meets the bound. Building the example is easiest by aiming at a round number like $400$ and choosing the tens and units digits to carry.

• Answering $1$ by imagining $S = 100$, forgetting that two three-digit numbers with distinct hundreds digits already add to at least $300$.
• Accepting an example like $105 + 195 = 300$ without checking that all six digits are different (here $1$ and $5$ repeat).