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

Revision as of 21:10, 26 October 2022

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/mXvetCMMzpU

~IceMatrix

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.

@@ Line 23: / Line 23: @@
 ==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 7.  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>.
+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

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