Difference between revisions of "2019 AMC 10B Problems/Problem 12"
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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{\textbf{(C) }22}</math> | 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{\textbf{(C) }22}</math> | ||
| − | + | ~IronicNinja, edited by some people | |
Note: the number can also be <math>5566_7</math>, which will also give the answer of <math>22</math>. | Note: the number can also be <math>5566_7</math>, which will also give the answer of <math>22</math>. | ||
Revision as of 20:03, 17 February 2019
Contents
Problem
What is the greatest possible sum of the digits in the base-seven representation of a positive integer less than 
?
Solution
Convert to base 
. This will get you 
, which will be the upper bound. To maximize the sum of the digits, we want as many 
s as possible (which is the highest value in base ), and this would be the number 
. Thus, the answer is 
~IronicNinja, edited by some people
Note: the number can also be 
, which will also give the answer of 
.
Solution 2
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 2019, so we continue with trying 4666, which is less than 2019. So the answer is 
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. 
