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Revision as of 11:07, 4 July 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 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 $16-3=\boxed{\textbf{(C) }13}$

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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