Difference between revisions of "1989 AHSME Problems/Problem 16"

m
(Problem)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
  
A lattice point is a point in the plane with integer coordinates. How many lattice points are on the line segment whose endpoints are (3,17) and (48,281)? (Include both endpoints of the segment in your count.)
+
A lattice point is a point in the plane with integer coordinates. How many lattice points are on the line segment whose endpoints are <math>(3,17)</math> and <math>(48,281)</math>? (Include both endpoints of the segment in your count.)
  
 
<math> \textrm{(A)}\ 2\qquad\textrm{(B)}\ 4\qquad\textrm{(C)}\ 6\qquad\textrm{(D)}\ 16\qquad\textrm{(E)}\ 46 </math>
 
<math> \textrm{(A)}\ 2\qquad\textrm{(B)}\ 4\qquad\textrm{(C)}\ 6\qquad\textrm{(D)}\ 16\qquad\textrm{(E)}\ 46 </math>
 
  
 
== Solution ==
 
== Solution ==

Revision as of 17:50, 14 January 2016

Problem

A lattice point is a point in the plane with integer coordinates. How many lattice points are on the line segment whose endpoints are $(3,17)$ and $(48,281)$? (Include both endpoints of the segment in your count.)

$\textrm{(A)}\ 2\qquad\textrm{(B)}\ 4\qquad\textrm{(C)}\ 6\qquad\textrm{(D)}\ 16\qquad\textrm{(E)}\ 46$

Solution

Since the endpoints are (3,17) and (48,281), the line that passes through these 2 points has slope $m=\frac{281-17}{48-3}=\frac{264}{45}=\frac{88}{15}$. The equation of the line passing through these points can then be given by $y=17+\frac{88}{15}(x-3)$. Since $\frac{88}{15}$ is reduced to lowest terms, in order for $y$ to be integral we must have that $15|x-3$. Hence $x$ is 3 more than a multiple of 15. Note that $x=3$ corresponds to the endpoint $(3,17)$. Then we have $x=18$, $x=33$, and $x=48$ where $x=48$ corresponds to the endpoint $(48,281)$. Hence there are 4 in all.

See also

1989 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
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 26 27 28 29 30
All AHSME Problems and Solutions

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