Difference between revisions of "2019 AMC 10B Problems/Problem 12"
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==Solution== | ==Solution== | ||
| − | Convert 2019 to base 7. This will get you | + | 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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| + | Note: the number can also be <math>5566_7</math>, which will also give the answer of <math>22</math>. | ||
iron | iron | ||
| − | + | Edited by greersc | |
| − | |||
==See Also== | ==See Also== | ||
Revision as of 17:15, 14 February 2019
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 
Note: the number can also be 
, which will also give the answer of 
.
iron Edited by greersc
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. 
