Difference between revisions of "2005 AMC 12A Problems/Problem 12"

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== Solution ==
 
== Solution ==
For convenience’s sake, we can transform <math>A</math> to the origin and <math>B</math> to <math>(99,999)</math> (this does not change the problem). The line <math>AB</math> has the [[equation]] <math>y = \frac{999}{99}x = \frac{111}{11}x</math>. The coordinates are integers if <math>11|x</math>, so the values of <math>x</math> are <math>11, 22 \ldots 88</math>, with a total of <math>8\ \mathrm{(D)}</math> coordinates.
+
For convenience’s sake, we can transform <math>A</math> to the origin and <math>B</math> to <math>(99,999)</math> (this does not change the problem). The line <math>AB</math> has the [[equation]] <math>y = \frac{999}{99}x = \frac{111}{11}x</math>. The coordinates are integers if <math>11|x</math>, so the values of <math>x</math> are <math>11, 22 \ldots 88</math>, with a total of <math>8\implies \boxed{\mathrm{(D)}}</math> coordinates.
  
 
== See also ==
 
== See also ==

Revision as of 13:58, 7 November 2019

Problem

A line passes through $A\ (1,1)$ and $B\ (100,1000)$. How many other points with integer coordinates are on the line and strictly between $A$ and $B$?

$(\mathrm {A}) \ 0 \qquad (\mathrm {B}) \ 2 \qquad (\mathrm {C})\ 3 \qquad (\mathrm {D}) \ 8 \qquad (\mathrm {E})\ 9$

Solution

For convenience’s sake, we can transform $A$ to the origin and $B$ to $(99,999)$ (this does not change the problem). The line $AB$ has the equation $y = \frac{999}{99}x = \frac{111}{11}x$. The coordinates are integers if $11|x$, so the values of $x$ are $11, 22 \ldots 88$, with a total of $8\implies \boxed{\mathrm{(D)}}$ coordinates.

See also

2005 AMC 12A (ProblemsAnswer KeyResources)
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

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