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

Revision as of 08:27, 16 December 2024

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.

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 $(1+11t, 1+111t)$ for some value of $t$ (the rise is 111 and the run is 11). Such a point has integer coordinates if and only if $t$ is an integer, and the point is strictly between $A$ and $B$ if and only if $0<t<9$ . Thus, there are $\boxed{8}$ 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 $\frac{1000 - 1}{100 - 1} = \frac{999}{99} = \frac{111}{11}$ . Then, we have $y = \frac{111}{11}x - \frac{100}{11}$ which reduces to $y = \frac{111x - 100}{11}$ . Now, it remains to look for values of $x$ such that $111x \equiv 1 (\mod 11)$ . Since $111 \equiv 1 (\mod 11)$ , the only values that work are $x = 12, 23, 34, 45, 56, 67, 78, 89$ . Therefore, there are $\boxed{8 \textbf{(D)}}$ coordinates for which this is true. ~elpianista227

@@ Line 14: / Line 14: @@
 </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
-==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 \cong 1 (mod 11)</math>. Since <math>111 \cong 1 (mod 11)</math>, the only values that work are <math>x = 12, 23, 34, 45, 56, 67, 78, 89</math>. Therefore, there are <math>\boxed{8 \textbf{(D)}}</math> coordinates for which this is true.
+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>, the only values that work are <math>x = 12, 23, 34, 45, 56, 67, 78, 89</math>. Therefore, there are <math>\boxed{8 \textbf{(D)}}</math> coordinates for which this is true.
 ~elpianista227

2005 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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