Difference between revisions of "2005 AMC 12A Problems/Problem 12"
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</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. | </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 | ||
| − | ==Solution 3== | + | ==Solution 3 (modular arithmetic)== |
| − | We can re-write the equation in slope-intercept form (where y is on the left side). We know that the slope is <math>\frac{1000 - 1}{100 - 1} = \frac{999}{99} = \frac{111}{11}</math>. Then, we have <math>y = \frac{111}{11}x - \frac{100}{11}</math> which reduces to <math>y = \frac{111x - 100}{11}</math>. Now, it remains to look for values of <math>x</math> such that <math>111x \ | + | We can re-write the equation in slope-intercept form (where y is on the left side). We know that the slope is <math>\frac{1000 - 1}{100 - 1} = \frac{999}{99} = \frac{111}{11}</math>. Then, we have <math>y = \frac{111}{11}x - \frac{100}{11}</math> which reduces to <math>y = \frac{111x - 100}{11}</math>. Now, it remains to look for values of <math>x</math> such that <math>111x \equiv 1 (\mod 11)</math>. Since <math>111 \equiv 1 (\mod 11)</math>, there are <math>\boxed{8 \textbf{(D)}}</math> coordinates for which this is true. |
~elpianista227 | ~elpianista227 | ||
| − | + | ||
== See also == | == See also == | ||
{{AMC12 box|year=2005|num-b=11|num-a=13|ab=A}} | {{AMC12 box|year=2005|num-b=11|num-a=13|ab=A}} | ||
Latest revision as of 08:31, 16 December 2024
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
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
Solution 3 (modular arithmetic)
We can re-write the equation in slope-intercept form (where y is on the left side). We know that the slope is 
. Then, we have 
which reduces to 
. Now, it remains to look for values of such that 
. Since 
, there are 
coordinates for which this is true.
~elpianista227
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. 
