Difference between revisions of "2005 AMC 12A Problems/Problem 12"
(→Solution 2) |
m (→Solution 2) |
||
Line 10: | Line 10: | ||
== Solution 2 == | == Solution 2 == | ||
− | The slope of the line is | + | The slope of the line is<math> |
\frac{1000-1}{100-1}=\frac{111}{11}, | \frac{1000-1}{100-1}=\frac{111}{11}, | ||
− | + | </math>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 | ||
Latest revision as of 19:18, 15 July 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 isso 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.