2019 AMC 10B Problems/Problem 12

Problem

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 1

Observe that $2019_{10} = 5613_7$ . To maximize the sum of the digits, we want as many $6$ s as possible (since $6$ is the highest value in base $7$ ), and this will occur with either of the numbers $4666_7$ or $5566_7$ . Thus, the answer is $4+6+6+6 = \boxed{\textbf{(C) }22}$ .

Solution 2

Note that all base $7$ numbers with $5$ or more digits are in fact greater than $2019$ . Since the first answer that is possible using a $4$ digit number is $23$ , we start with the smallest base $7$ number that whose digits sum to $23$ , namely $5666_7$ . But this is greater than $2019_10$ , so we continue by trying $4666_7$ , which is less than 2019. So the answer is $\boxed{\textbf{(C) }22}$ .

2019 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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2019 AMC 10B Problems/Problem 12

Contents

Problem

Solution 1

Solution 2

See Also