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

m (Solution 1: minor edit)
m (Solution 1)
Line 13: Line 13:
 
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>.
 
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 went through this test 100 times
  
 
==Solution 2==
 
==Solution 2==

Revision as of 18:40, 13 January 2020

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

See Also

2019 AMC 10B (ProblemsAnswer KeyResources)
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. AMC logo.png