1993 AHSME Problems/Problem 19

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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

Multiply both sides by $mn$ to clear the denominator. Moving all the terms to the right hand side, the equation becomes $0 = mn-4n-2m$. Adding 8 to both sides allows us to factor the equation as follows: $(m-4)(n-2) = 8$. Since the problem only wants integer pairs $(m,n)$, the pairs are given by the factors of 8, which are 1,2,4, and 8. This results in four pairs, $(5,10), (6,6), (8,4),$ and $(12,3)$, which is answer $\fbox{D}$.

See also

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