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Algebra Difficulty 4.8 AIME Find the answer

How many pairs of positive integers (a,b)(a, b) with aba \leq b satisfy 1a+1b=16\frac{1}{a} + \frac{1}{b} = \frac{1}{6}?

A number or a short expression. Spacing and $ signs are ignored.

Solution

1a+1b=16a+bab=16ab=6a+6bab6a6b=0\frac{1}{a} + \frac{1}{b} = \frac{1}{6} \Rightarrow \frac{a+b}{ab} = \frac{1}{6} \Rightarrow ab = 6a + 6b \Rightarrow ab - 6a - 6b = 0. Factoring yields (a6)(b6)=36(a-6)(b-6) = 36. Because aa and bb are positive integers, only the factor pairs of 36 are possible values of a6a-6 and b6b-6. The possible pairs are: a6=1,b6=36a6=2,b6=18a6=3,b6=12a6=4,b6=9a6=6,b6=6\begin{aligned} & a-6=1, b-6=36 \\ & a-6=2, b-6=18 \\ & a-6=3, b-6=12 \\ & a-6=4, b-6=9 \\ & a-6=6, b-6=6 \end{aligned} Because aba \leq b, the symmetric cases, such as a6=12,b6=3a-6=12, b-6=3 are not applicable. Then there are 5 possible pairs.

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