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

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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>, our answer is <math>16-3=\boxed{\textbf{(C) }13\qquad}</math>
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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>, our answer is <math>16-3=\boxed{\textbf{(C) }13}</math>

Revision as of 21:20, 7 February 2013

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 digits $b$ 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 $3$, our answer is $16-3=\boxed{\textbf{(C) }13}$

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