2019 AMC 10B Problems/Problem 12

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What is the greatest possible sum of the digits in the base-seven representation of a positive integer less than $2019$?

$\textbf{(A) } 11 \qquad\textbf{(B) } 14 \qquad\textbf{(C) } 22 \qquad\textbf{(D) } 23 \qquad\textbf{(E) } 27$

Solution

Convert 2019 to base 7. This will get you 5613, which will be the upper bound. To maximize the sum of the digits, we want as many 6s as possible (which is the highest value in base 7), and this would be the number "4666". Thus, the answer is $4+6+6+6 = \boxed{C) 22}$

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