2019 AMC 10B Problem 12

Since $7^4 = 2401 > 2019$, every positive integer below $2019$ has at most four base-seven digits, each at most $6$. So the digit sum is at most $24$, attained only by $6666_7 = 2400$, which is too large.

A digit sum of $23$ would need four digits totaling $23$ with each at most $6$, so the digits must be $6, 6, 6, 5$ in some order. The smallest such number is
$$5666_7 = 5 \cdot 343 + 6 \cdot 49 + 6 \cdot 7 + 6 = 1715 + 294 + 42 + 6 = 2057 > 2019.$$
So $23$ is impossible.

A digit sum of $22$ is achievable:
$$4666_7 = 4 \cdot 343 + 294 + 42 + 6 = 1714 < 2019.$$

The answer is $\boxed{\textbf{(C)}\ 22}$.

Maximizing a digit sum is a trade-off between using large digits and staying under the limit. Bound the digit count from the powers of the base, find the theoretical maximum sum, and then check whether the smallest number achieving each candidate sum fits under the limit. Here $2019 = 5613_7$, so the leading digit $5$ forces the next digits down sharply, while dropping to leading digit $4$ frees all the remaining digits to be $6$.

• Assuming the maximum is 6666 base 7 (sum 24) without converting it, or reading 2019 as if it were already in base 7.
• Insisting on the largest leading digit $5$ and settling for $5566_7 = 2008$ (also sum $22$) without realizing $4666_7$ gives the same sum; both are fine, but forgetting to check $23$ is the real risk.