How many different pairs (a,b) of positive integers are there such taht $a \leq b a n d \frac{1}{a} + \frac{1}{b} = \frac{1}{9}$ ? ___

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Solution

We have $\frac{1}{a} + \frac{1}{b} = \frac{1}{9}$
Above equation can also be written as (a - 9)(b - 9) = 81
Now factors of 81 = 1, 3, 9, 27, 81
Out of 5 ways , 2 times ‘a’ will be greater than ‘b’, 2 times less than ‘b’ and once equal to ‘b’
Hence, numbers of ways possible = 3

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