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

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==Solution==
 
==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 <math>4+6+6+6 = \boxed{C) 22}</math>
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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{C) 22}</math>
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Note: the number can also be <math>5566_7</math>, which will also give the answer of <math>22</math>.
  
 
iron
 
iron
 
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Edited by greersc
Note: the number can also be "5566" which will also give the answer of 22.
 
  
 
==See Also==
 
==See Also==

Revision as of 18:15, 14 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 $4+6+6+6 = \boxed{C) 22}$

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

iron Edited by greersc

See Also

2019 AMC 10B (ProblemsAnswer KeyResources)
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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