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Difference between revisions of "2019 AMC 10B Problems/Problem 12"

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Convert <math>2019</math> to base <math>7</math>. This will get you <math>5613_7</math>, which will be the upper bound. To maximize the sum of the digits, we want as many <math>6</math>s as possible (which is the highest value in base <math>7</math>), and this would be the number <math>4666_7</math>. Thus, the answer is <math>4+6+6+6 = \boxed{\textbf{(C) }22}</math>
 
Convert <math>2019</math> to base <math>7</math>. This will get you <math>5613_7</math>, which will be the upper bound. To maximize the sum of the digits, we want as many <math>6</math>s as possible (which is the highest value in base <math>7</math>), and this would be the number <math>4666_7</math>. Thus, the answer is <math>4+6+6+6 = \boxed{\textbf{(C) }22}</math>
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iron, edited by someone people
  
 
Note: the number can also be <math>5566_7</math>, which will also give the answer of <math>22</math>.
 
Note: the number can also be <math>5566_7</math>, which will also give the answer of <math>22</math>.

Revision as of 20:31, 17 February 2019

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 $5613_7$, which will be the upper bound. To maximize the sum of the digits, we want as many $6$s as possible (which is the highest value in base $7$), and this would be the number $4666_7$. Thus, the answer is $4+6+6+6 = \boxed{\textbf{(C) }22}$

iron, edited by someone people

Note: the number can also be $5566_7$, which will also give the answer of $22$.

Solution 2

Note that all base 7 numbers with 5 digits or more is 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 digits adds up to 23, 5666. 5666 in base 10 is greater than 2019, so we continue with trying 4666, which is less than 2019. So the answer is $\boxed{\textbf{(C) }22}$

See Also

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

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