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1993 AMC 12 #19 - Ordered Integral Solutions

Source:

January 2, 2012
calculusintegrationSFFTspecial factorizationsAMC

Problem Statement

How many ordered pairs (m,n)(m,n) of positive integers are solutions to 4m+2n=1\frac{4}{m}+\frac{2}{n}=1?
<spanclass=′latex−bold′>(A)</span> 1<spanclass=′latex−bold′>(B)</span> 2<spanclass=′latex−bold′>(C)</span> 3<spanclass=′latex−bold′>(D)</span> 4<spanclass=′latex−bold′>(E)</span> more than 4 <span class='latex-bold'>(A)</span>\ 1 \qquad<span class='latex-bold'>(B)</span>\ 2 \qquad<span class='latex-bold'>(C)</span>\ 3 \qquad<span class='latex-bold'>(D)</span>\ 4 \qquad<span class='latex-bold'>(E)</span>\ \text{more than}\ 4
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