Difference between revisions of "2005 AMC 8 Problems/Problem 18"
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<math> \textbf{(A)}\ 7\qquad\textbf{(B)}\ 67\qquad\textbf{(C)}\ 69\qquad\textbf{(D)}\ 76\qquad\textbf{(E)}\ 77</math> | <math> \textbf{(A)}\ 7\qquad\textbf{(B)}\ 67\qquad\textbf{(C)}\ 69\qquad\textbf{(D)}\ 76\qquad\textbf{(E)}\ 77</math> | ||
| − | ==Solution== | + | ==Solution 1== |
Let <math>k</math> be any positive integer so that <math>13k</math> is a multiple of <math>13</math>. For the smallest three-digit number, <math>13k>100</math> and <math>k>\frac{100}{13} \approx 7.7</math>. For the greatest three-digit number, <math>13k<999</math> and <math>k<\frac{999}{13} \approx 76.8</math>. The number <math>k</math> can range from <math>8</math> to <math>76</math> so there are <math>\boxed{\textbf{(C)}\ 69}</math> three-digit numbers. | Let <math>k</math> be any positive integer so that <math>13k</math> is a multiple of <math>13</math>. For the smallest three-digit number, <math>13k>100</math> and <math>k>\frac{100}{13} \approx 7.7</math>. For the greatest three-digit number, <math>13k<999</math> and <math>k<\frac{999}{13} \approx 76.8</math>. The number <math>k</math> can range from <math>8</math> to <math>76</math> so there are <math>\boxed{\textbf{(C)}\ 69}</math> three-digit numbers. | ||
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| + | ==Solution 2== | ||
| + | EVERYONE LOVES <math>69</math> and thus that's the answer | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2005|num-b=17|num-a=19}} | {{AMC8 box|year=2005|num-b=17|num-a=19}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 18:39, 22 February 2019
Contents
Problem
How many three-digit numbers are divisible by 13?
Solution 1
Let 
be any positive integer so that 
is a multiple of 
. For the smallest three-digit number, 
and 
. For the greatest three-digit number, 
and 
. The number can range from 
to 
so there are 
three-digit numbers.
Solution 2
EVERYONE LOVES 
and thus that's the answer
See Also
| 2005 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 17 |
Followed by Problem 19 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
