Difference between revisions of "2005 AMC 12A Problems/Problem 12"
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| − | The slope of the line is | + | The slope of the line is\[ |
| − | \frac{1000-1}{100-1}=\frac{111}{11} | + | \frac{1000-1}{100-1}=\frac{111}{11}, |
\]so all points on the line have the form <math>(1+11t, 1+111t)</math> for some value of <math>t</math> (the rise is 111 and the run is 11). Such a point has integer coordinates if and only if <math>t</math> is an integer, and the point is strictly between <math>A</math> and <math>B</math> if and only if <math>0<t<9</math>. Thus, there are <math>\boxed{8}</math> points with the required property. | \]so all points on the line have the form <math>(1+11t, 1+111t)</math> for some value of <math>t</math> (the rise is 111 and the run is 11). Such a point has integer coordinates if and only if <math>t</math> is an integer, and the point is strictly between <math>A</math> and <math>B</math> if and only if <math>0<t<9</math>. Thus, there are <math>\boxed{8}</math> points with the required property. | ||
-Paixiao | -Paixiao | ||
Revision as of 10:43, 28 May 2021
Contents
Problem
A line passes through 
and 
. How many other points with integer coordinates are on the line and strictly between 
and 
?
Solution
For convenience’s sake, we can transform to the origin and to 
(this does not change the problem). The line 
has the equation 
. The coordinates are integers if 
, so the values of 
are 
, with a total of 
coordinates.
Solution 2
The slope of the line is\[
\frac{1000-1}{100-1}=\frac{111}{11},
\]so all points on the line have the form 
for some value of 
(the rise is 111 and the run is 11). Such a point has integer coordinates if and only if is an integer, and the point is strictly between and if and only if 
. Thus, there are 
points with the required property.
-Paixiao
See also
| 2005 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 11 |
Followed by Problem 13 |
| 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 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
