1993 AHSME Problems/Problem 19

Problem

How many ordered pairs $(m,n)$ of positive integers are solutions to \[\frac{4}{m}+\frac{2}{n}=1?\]

$\text{(A) } 1\quad \text{(B) } 2\quad \text{(C) } 3\quad \text{(D) } 4\quad \text{(E) } \text{more than }6$

Solution

$\fbox{D}$

See also

1993 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 18
Followed by
Problem 20
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