2016 AMC 10B Problem 6

This asks for the smallest possible value, so there are two parts: show that $1$, $2$ and $3$ are impossible, then find an example with $4$.

Part 1: $S$ is at least $300$.

1. The two numbers have different hundreds digits, and neither is $0$. The smallest they can be is $1$ and $2$.
2. So the two numbers add to at least $100 + 200 = 300$. Also $S$ has only three digits, so $S$ is somewhere from $300$ to $999$.

Part 2: digit sums $1$, $2$ and $3$ are impossible.

3. The first digit of $S$ is at least $3$, so the digit sum is at least $3$. That already rules out $1$ and $2$.
4. A digit sum of exactly $3$ would need $S = 300$ (a $3$ in front, zeros after).
5. Can two such numbers add to $300$? Look at the hundreds digits.
- If there is no carry into the hundreds place, the hundreds digits add to $3$, so they are $1$ and $2$. Then the last two digits of both numbers must add to $00$ with no carry, which means all four of those digits are $0$. That repeats the digit $0$.
- If there is a carry of $1$, the hundreds digits add to $2$, so they are both $1$. That repeats the digit $1$.

Either way two digits are the same, so $S = 300$ is impossible.

Part 3: digit sum $4$ works.

1. Aim for $S = 400$. Try $143 + 257 = 400$. The six digits $1, 4, 3, 2, 5, 7$ are all different, and $4 + 0 + 0 = 4$.

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

For a "smallest possible" question, first find a lower limit, then show an example that reaches it. Here the lower limit comes from the hundreds digits, which must be different. To build the example, aim for a round number like $400$ and let the ones and tens columns carry.

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 $\boxed{\textbf{(B)}\ 4}$.