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

Revision as of 22:10, 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 1: / Line 1: @@
-sub2pewds
+==Problem==
+What is the greatest possible sum of the digits in the base-seven representation of a positive integer less than <math>2019</math>?
+<math>\textbf{(A) } 11
+\qquad\textbf{(B) } 14
+\qquad\textbf{(C) } 22
+\qquad\textbf{(D) } 23
+\qquad\textbf{(E) } 27</math>
+==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 = 5+5+6+6 = \boxed{\textbf{(C) }22}</math>.
+~IronicNinja went through this test 100 times
+==Solution 2==
+Note that all base <math>7</math> numbers with <math>5</math> or more digits are in fact greater than <math>2019</math>. Since the first answer that is possible using a <math>4</math> digit number is <math>23</math>, we start with the smallest base <math>7</math> number that whose digits sum to <math>23</math>, namely <math>5666_7</math>. But this is greater than <math>2019_{10}</math>, so we continue by trying <math>4666_7</math>, which is less than 2019. So the answer is <math>\boxed{\textbf{(C) }22}</math>.
+LaTeX code fix by EthanYL
+==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 <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
+==Video Solution==
+https://youtu.be/mXvetCMMzpU
+~IceMatrix
+==See Also==
+{{AMC10 box|year=2019|ab=B|num-b=11|num-a=13}}
+{{MAA Notice}}

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