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

(Created page with "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 o...")
 
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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.
 
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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{{MAA Notice}}

Revision as of 13:48, 5 July 2013

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

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