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

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==Solution 2==
 
==Solution 2==
  
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Note that all base <math>7</math> numbers with <math>5</math> or more digits are in fact greater than <math>2019</math>. Since the first answer that is possible using a <math>4</math> digit number is <math>23</math>, we start with the smallest base <math>7</math> number that whose digits sum to <math>23</math>, namely <math>5666_7</math>. But this is greater than <math>2019_10</math>, so we continue by trying <math>4666_7</math>, which is less than 2019. So the answer is <math>\boxed{\textbf{(C) }22}</math>
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Note that all base <math>7</math> numbers with <math>5</math> or more digits are in fact greater than <math>2019</math>. Since the first answer that is possible using a <math>4</math> digit number is <math>23</math>, we start with the smallest base <math>7</math> number that whose digits sum to <math>23</math>, namely <math>5666_7</math>. But this is greater than <math>2019_10</math>, so we continue by trying <math>4666_7</math>, which is less than 2019. So the answer is <math>\boxed{\textbf{(C) }22}</math>.
  
 
==See Also==
 
==See Also==

Revision as of 21:30, 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 $4+6+6+6 = \boxed{\textbf{(C) }22}$.

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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