2019 AMC 10B Problem 12

In base seven the place values are $1, 7, 49, 343, 2401$, and each digit is between $0$ and $6$.

1. How many digits? $7^4 = 2401$ is bigger than $2019$. So every number below $2019$ uses at most four base-seven digits: the $343$s, $49$s, $7$s and $1$s places.
2. Highest sum imaginable. Four digits, each at most $6$, sum to at most $24$. The only number with digit sum $24$ is $6666_7 = 6 \cdot 343 + 6 \cdot 49 + 6 \cdot 7 + 6 = 2400$, too big.
3. Can the sum be $23$? Four digits, each at most $6$, adding to $23$ must be three $6$s and one $5$. The smallest such number puts the $5$ in front:
$$5666_7 = 5 \cdot 343 + 6 \cdot 49 + 6 \cdot 7 + 6 = 1715 + 294 + 42 + 6 = 2057.$$
That is more than $2019$, and the other arrangements are even larger. So $23$ is impossible.
4. Sum $22$ works. Lower the front digit to $4$:
$$4666_7 = 4 \cdot 343 + 294 + 42 + 6 = 1372 + 342 = 1714,$$
which is less than $2019$. Its digit sum is $4 + 6 + 6 + 6 = 22$.

Sum $22$ is possible, and sums $23$ and $24$ are not, so the greatest is $22$.

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

To make a digit sum big, you want big digits, but the number must stay under the limit. So work down from the largest sum you can imagine and, for each candidate sum, test the smallest number that has it. If even the smallest is too big, that sum is out. It helps to know that $2019 = 5613_7$: its leading digit $5$ is followed by small digits, so dropping the leading digit to $4$ frees the rest to be $6$s.