2005 AMC 12A Problems/Problem 12

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 $111 \equiv 1 (\mod 11)$, there are $\boxed{8 \textbf{(D)}}$ coordinates for which this is true. ~elpianista227

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png