Difference between revisions of "2019 AMC 10B Problems/Problem 12"
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==Solution 1== | ==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 = \boxed{\textbf{(C) }22}</math>. | + | 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 | + | ~IronicNinja |
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==Solution 2== | ==Solution 2== |
Revision as of 01:36, 7 January 2020
Contents
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
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
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.