AMC10 2019 B

Answer (C): One can convert 2018 to base seven by repeatedly dividing by 7; the successive remainders are the digits in the base-seven representation, from right to left. Thus $2018 = 5612_{\text{seven}}$. It follows that the base-seven representations of positive integers less than 2019 have at most four digits, each digit is at most 6, and the leftmost digit is at most 5. If the leftmost digit is 4, then the remaining digits can all be 6 for a sum of $4+6+6+6=22$. If the leftmost digit is 5, then the remaining digits cannot all be 6. Therefore the required sum of digits cannot exceed $5+5+6+6=22$. Because $5566_{\text{seven}}<5612_{\text{seven}}<2019$ (and $4666_{\text{seven}}<5612_{\text{seven}}<2019$), the requested maximum sum is 22.