Difference between revisions of "2013 AMC 10A Problems/Problem 19"
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<math> \textbf{(A)}\ 6\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 13\qquad\textbf{(D)}\ 16\qquad\textbf{(E)}\ 18 </math> | <math> \textbf{(A)}\ 6\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 13\qquad\textbf{(D)}\ 16\qquad\textbf{(E)}\ 18 </math> | ||
| − | ==Solution== | + | ==Solution 1== |
| − | 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> | + | 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> |
==Video Solution by SpreadTheMathLove== | ==Video Solution by SpreadTheMathLove== | ||
Revision as of 23:51, 3 November 2025
Contents
Problem
In base 
, the number 
ends in the digit 
. In base 
, on the other hand, the same number is written as 
and ends in the digit 
. For how many positive integers 
does the base--representation of end in the digit ?
Solution 1
We want the integers such that 
is a factor of 
. Since 
, it has 
factors. Since cannot equal 
or , as these cannot have the digit in their base representations, our answer is 
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
