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

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 $4+6+6+6 = \boxed{C) 22}$

iron

Note: the number can also be "5566" which will also give the answer of 22.

2019 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

2019 AMC 10B Problems/Problem 12

Problem

Solution

See Also