2021 AMC 10B Problem 13

Step 1: turn the numerals into ordinary numbers.

1. In base $n$, the places are worth $n^2$, $n$, $1$. So $\underline{32d}$ in base $n$ means $3n^2 + 2n + d$, and the first fact is
$$3n^2 + 2n + d = 263.$$
2. In the same way, $\underline{324}$ in base $n$ is $3n^2 + 2n + 4$.
3. In base six the places are worth $216$, $36$, $6$, $1$. So $\underline{11d1}$ in base six is $216 + 36 + 6d + 1 = 253 + 6d$. The second fact is
$$3n^2 + 2n + 4 = 253 + 6d.$$

Step 2: subtract to get rid of $n$.

4. Both equations contain the same piece $3n^2 + 2n$, so subtracting one from the other removes it. First minus second:
$$d - 4 = 263 - (253 + 6d) = 10 - 6d.$$
5. So $7d = 14$ and $d = 2$.

Step 3: find $n$.

1. Put $d = 2$ into the first fact: $3n^2 + 2n = 261$.
2. Estimate: $3n^2$ is about $261$, so $n^2$ is a bit under $87$. Try $n = 9$: $3 \cdot 81 + 18 = 243 + 18 = 261$. It works.
3. Check the digits make sense: $3$, $2$, $4$ are all less than $9$, and $d = 2$ is less than $6$, so both numerals are allowed.

So $n + d = 9 + 2 = 11$.

The answer is $\boxed{\textbf{(B)}\ 11}$.

A number written in base $n$ is just a sum of digits times powers of $n$. When two equations share the same messy part, subtracting them cancels it and leaves something simple. Then the remaining equation is small enough to solve by trying a value close to your estimate.

Squeeze $n$ from both sides. From the first fact, $3n^2 \le 263$, so $n^2 \le 87$ and $n \le 9$. From the second fact, $3n^2 + 2n + 4 \ge 253$, and $n = 8$ gives only $192 + 16 + 4 = 212$, so $n \ge 9$. Hence $n = 9$, and then $243 + 18 + d = 263$ gives $d = 2$.