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

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\qquad\textbf{(E) } 27</math>
 
\qquad\textbf{(E) } 27</math>
  
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==Solution==
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==Solution 1==
  
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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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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 = 5+5+6+6 = \boxed{\textbf{(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>.
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~IronicNinja went through this test 100 times
  
 
==Solution 2==
 
==Solution 2==
  
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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 2017, so we continue with trying 4666, which is less than 2019. So the answer is <math>\boxed{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>.
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LaTeX code fix by EthanYL
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==Solution 3==
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Again note that you want to maximize the number of <math>6</math>s to get the maximum sum.  Note that <math>666_7=342_{10}</math>, so you have room to add a thousands digit base <math>7</math>.  Fix the <math>666</math> in place and try different thousands digits, to get <math>4666_7</math> as the number with the maximum sum of digits.  The answer is <math>\boxed{\textbf{(C)} 22}</math>.
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~mwu2010
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==Video Solution==
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https://youtu.be/jaNRwYiLbxE
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~Education, the Study of Everything
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==Video Solution==
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https://youtu.be/mXvetCMMzpU
  
 
==See Also==
 
==See Also==

Latest revision as of 09:27, 24 June 2023

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 $4+6+6+6 = 5+5+6+6 = \boxed{\textbf{(C) }22}$.

~IronicNinja went through this test 100 times

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

LaTeX code fix by EthanYL

Solution 3

Again note that you want to maximize the number of $6$s to get the maximum sum. Note that $666_7=342_{10}$, so you have room to add a thousands digit base $7$. Fix the $666$ in place and try different thousands digits, to get $4666_7$ as the number with the maximum sum of digits. The answer is $\boxed{\textbf{(C)} 22}$.

~mwu2010

Video Solution

https://youtu.be/jaNRwYiLbxE

~Education, the Study of Everything

Video Solution

https://youtu.be/mXvetCMMzpU

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

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

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