2019 AMC 10B Problems/Problem 12: Difference between revisions
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==Problem== | |||
What is the greatest possible sum of the digits in the base-seven representation of a positive integer less than <math>2019</math>? | |||
<math>\textbf{(A) } 11 | |||
\qquad\textbf{(B) } 14 | |||
\qquad\textbf{(C) } 22 | |||
\qquad\textbf{(D) } 23 | |||
\qquad\textbf{(E) } 27</math> | |||
==Solution 1== | |||
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>. | |||
~IronicNinja went through this test 100 times | |||
==Solution 2== | |||
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>. | |||
LaTeX code fix by EthanYL | |||
==Solution 3== | |||
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>. | |||
~mwu2010 | |||
==Video Solution== | |||
https://youtu.be/jaNRwYiLbxE | |||
~Education, the Study of Everything | |||
==Video Solution== | |||
https://youtu.be/mXvetCMMzpU | |||
==See Also== | |||
{{AMC10 box|year=2019|ab=B|num-b=11|num-a=13}} | |||
{{MAA Notice}} | |||
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 
?
Solution 1
Observe that 
. To maximize the sum of the digits, we want as many 
s as possible (since is the highest value in base 
), and this will occur with either of the numbers 
or 
. Thus, the answer is 
.
~IronicNinja went through this test 100 times
Solution 2
Note that all base numbers with 
or more digits are in fact greater than . Since the first answer that is possible using a 
digit number is 
, we start with the smallest base number that whose digits sum to , namely 
. But this is greater than 
, so we continue by trying , which is less than 2019. So the answer is 
.
LaTeX code fix by EthanYL
Solution 3
Again note that you want to maximize the number of s to get the maximum sum. Note that 
, so you have room to add a thousands digit base . Fix the 
in place and try different thousands digits, to get as the number with the maximum sum of digits. The answer is 
.
~mwu2010
Video Solution
~Education, the Study of Everything
Video Solution
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 | ||
These problems are copyrighted © by the Mathematical Association of America. 
