1989 AHSME Problems/Problem 16

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.

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1989 AHSME Problems/Problem 16