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Difference between revisions of "2013 AMC 10A Problems/Problem 19"

(Created page with "==Problem== In base <math>10</math>, the number <math>2013</math> ends in the digit <math>3</math>. In base <math>9</math>, on the other hand, the same number is written as <ma...")
 
 
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==Problem==
 
==Problem==
  
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In base <math>10</math>, the number <math>2013</math> ends in the digit <math>3</math>.  In base <math>9</math>, on the other hand, the same number is written as <math>(2676)_{9}</math> and ends in the digit <math>6</math>.  For how many digits <math>b</math> does the base-<math>b</math>-representation of <math>2013</math> end in the digit <math>3</math>?
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In base <math>10</math>, the number <math>2013</math> ends in the digit <math>3</math>.  In base <math>9</math>, on the other hand, the same number is written as <math>(2676)_{9}</math> and ends in the digit <math>6</math>.  For how many positive integers <math>b</math> does the base-<math>b</math>-representation of <math>2013</math> end in the digit <math>3</math>?
  
  
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==Solution==
 
==Solution==
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We want the integers <math>b</math> such that <math> 2013\equiv 3\pmod{b} \Rightarrow b </math> is a factor of <math>2010</math>. Since <math>2010=2 \cdot 3 \cdot 5 \cdot 67</math>, it has <math>(1+1)(1+1)(1+1)(1+1)=16</math> factors. Since <math>b</math> cannot equal <math>1, 2, </math> or <math>3</math>, as these cannot have the digit <math>3</math> in their base representations, our answer is <math>16-3=\boxed{\textbf{(C) }13}</math>
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==Video Solution by SpreadTheMathLove==
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https://www.youtube.com/watch?v=msGdQB7_-50
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== Video Solution by OmegaLearn ==
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https://youtu.be/ZhAZ1oPe5Ds?t=4366
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 +
~ pi_is_3.14
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==See Also==
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{{AMC10 box|year=2013|ab=A|num-b=18|num-a=20}}
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[[Category:Introductory Number Theory Problems]]
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{{MAA Notice}}

Latest revision as of 12:22, 2 April 2023

Problem

In base $10$, the number $2013$ ends in the digit $3$. In base $9$, on the other hand, the same number is written as $(2676)_{9}$ and ends in the digit $6$. For how many positive integers $b$ does the base-$b$-representation of $2013$ end in the digit $3$?


$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 13\qquad\textbf{(D)}\ 16\qquad\textbf{(E)}\ 18$

Solution

We want the integers $b$ such that $2013\equiv 3\pmod{b} \Rightarrow b$ is a factor of $2010$. Since $2010=2 \cdot 3 \cdot 5 \cdot 67$, it has $(1+1)(1+1)(1+1)(1+1)=16$ factors. Since $b$ cannot equal $1, 2,$ or $3$, as these cannot have the digit $3$ in their base representations, our answer is $16-3=\boxed{\textbf{(C) }13}$

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=msGdQB7_-50

Video Solution by OmegaLearn

https://youtu.be/ZhAZ1oPe5Ds?t=4366

~ pi_is_3.14

See Also

2013 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
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

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