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

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== Problem ==
 
== Problem ==
  
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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.)
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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.)
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<math> \textrm{(A)}\ 2\qquad\textrm{(B)}\ 4\qquad\textrm{(C)}\ 6\qquad\textrm{(D)}\ 16\qquad\textrm{(E)}\ 46 </math>
 
  
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<math> \textbf{(A)}\ 2\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 16\qquad\textbf{(E)}\ 46 </math>
  
 
== 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. Hence the answer is <math>\boxed{B}</math>.
  
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== See also ==
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== See Also ==
 
{{AHSME box|year=1989|num-b=15|num-a=17}}   
 
{{AHSME box|year=1989|num-b=15|num-a=17}}   
  
 
[[Category: Introductory Number Theory Problems]]
 
[[Category: Introductory Number Theory Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 02:22, 11 July 2020

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.)

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

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. Hence the answer is $\boxed{B}$.

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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