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...") |
|||
Line 1: | Line 1: | ||
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. | ||
+ | {{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 . The equation of the line passing through these points can then be given by
. Since
is reduced to lowest terms, in order for
to be integral we must have that
. Hence
is 3 more than a multiple of 15. Note that
corresponds to the endpoint
. Then we have
,
, and
where
corresponds to the endpoint
. Hence there are 4 in all.
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.