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

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== Solution ==
 
== Solution ==
  
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Since the endpoints are (3,17) and (48,281), the line that passes through these 2 points has slope <math>m=\frac{281-17}{48-3}=\frac{264}{45}=\frac{88}{15}</math>. The equation of the line passing through these points can then be given by <math>y=17+\frac{88}{15}(x-3)</math>. Since <math>\frac{88}{15}</math> is reduced to lowest terms, in order for <math>y</math> to be integral we must have that <math>15|x-3</math>. Hence <math>x</math> is 3 more than a multiple of 15. Note that <math>x=3</math> corresponds to the endpoint <math>(3,17)</math>. Then we have <math>x=18</math>, <math>x=33</math>, and <math>x=48</math> where <math>x=48</math> corresponds to the endpoint <math>(48,281)</math>. Hence there are 4 in all.
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The difference in the <math>y</math>-coordinates is <math>281 - 17 = 264</math>, and the difference in the <math>x</math>-coordinates is <math>48 - 3 = 45</math>.
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The gcd of 264 and 45 is 3, so the line segment joining <math>(3,17)</math> and <math>(48,281)</math> has slope
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<cmath>\frac{88}{15}.</cmath>
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The points on the line have coordinates
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<cmath>\left(3+t,17+\frac{88}{15}t\right).</cmath>
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If <math>t</math> is an integer, the <math>y</math>-coordinate of this point is an integer if and only if <math>t</math> is a multiple of 15. The points where <math>t</math> is a multiple of 15 on the segment <math>3\leq x\leq 48</math> are <math>3</math>, <math>3+15</math>, <math>3+30</math>, and <math>3+45</math>. There are 4 lattice points on this line.
  
 
== See also ==
 
== See also ==

Revision as of 07:51, 2 April 2018

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

The difference in the $y$-coordinates is $281 - 17 = 264$, and the difference in the $x$-coordinates is $48 - 3 = 45$. The gcd of 264 and 45 is 3, so the line segment joining $(3,17)$ and $(48,281)$ has slope \[\frac{88}{15}.\] The points on the line have coordinates \[\left(3+t,17+\frac{88}{15}t\right).\] If $t$ is an integer, the $y$-coordinate of this point is an integer if and only if $t$ is a multiple of 15. The points where $t$ is a multiple of 15 on the segment $3\leq x\leq 48$ are $3$, $3+15$, $3+30$, and $3+45$. There are 4 lattice points on this line.

See also

1989 AHSME (Problems • Answer Key • Resources)
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

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