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

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Observe that <math>2019_{10} = 5613_7</math>. To maximize the sum of the digits, we want as many <math>6</math>s as possible (since <math>6</math> is the highest value in base <math>7</math>), and this will occur with either of the numbers <math>4666_7</math> or <math>5566_7</math>. Thus, the answer is <math>4+6+6+6 = \boxed{\textbf{(C) }22}</math>.
 
Observe that <math>2019_{10} = 5613_7</math>. To maximize the sum of the digits, we want as many <math>6</math>s as possible (since <math>6</math> is the highest value in base <math>7</math>), and this will occur with either of the numbers <math>4666_7</math> or <math>5566_7</math>. Thus, the answer is <math>4+6+6+6 = \boxed{\textbf{(C) }22}</math>.
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~IronicNinja
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Note: the number can also be <math>5566_7</math>, which will also give the answer of <math>22</math>.
  
 
==Solution 2==
 
==Solution 2==

Revision as of 23:02, 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 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 $4+6+6+6 = \boxed{\textbf{(C) }22}$.

~IronicNinja

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 $5$ 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 $\boxed{\textbf{(C) }22}$.

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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