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

Revision as of 05:10, 28 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

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: @@
+==Problem==
 What is the greatest possible sum of the digits in the base-seven representation of a positive integer less than <math>2019</math>?
@@ Line 7: / Line 9: @@
 \qquad\textbf{(E) } 27</math>
-==Solution==
+==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
+==Video Solution==
+https://youtu.be/mXvetCMMzpU
+~IceMatrix
-Convert 2019 to base 7. This will get you 5613, which will be the upper bound. To maximize the sum of the digits, we want as many 6s as possible (which is the highest value in base 7), and this would be the number "4666". Thus, the answer is <math>4+6+6+6 = \boxed{C) 22}</math>
+==See Also==
-iron
+{{AMC10 box|year=2019|ab=B|num-b=11|num-a=13}}
+{{MAA Notice}}

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