2019 AMC 10B Problem 12

First, $2019=5613_7$. Any number with leading digit at most $4$ has digit sum at most $4+6+6+6=22$, and $4666_7<5613_7$ attains that sum. If the leading digit is $5$, then a second digit at most $5$ gives digit sum at most $5+5+6+6=22$. If the first two digits are $56$, comparison with $5613_7$ forces the last two digits to contribute at most $1+2$, giving an even smaller sum. Therefore, the largest digit sum is $22$. Thus, the answer is C .