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Question 28 : How many different pairs (a, b) of positive integers are there such that a ≤ b and \\frac{1}{a}) + \\frac{1}{b}) = \\frac{1}{9}) ? [TITA]

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Given that \\frac{1}{a}) + \\frac{1}{b}) = \\frac{1}{9})

Let us assume that a ≤ b.

For first case, let a = b, then

\\frac{1}{18}) + \\frac{1}{18}) = \\frac{1}{9})

Hence (18 , 18) is one pair of possibility.

In all the cases, both a and b should be greater than 9 to satisfy the equation.

So, we can take a = 10 , 11 , 12 , ..... , 18 and check for the different possibilities.

For example, let us take a = 15

⟹ \\frac{1}{b}) = \\frac{1}{9}) - \\frac{1}{15})

⟹ \\frac{5 - 3}{45}) = \\frac{2}{45})

Hence a = 15 is not possible.

Similarly when a = 17, 16, 14, 13, and 11 does not work either.

Now let us take a = 12,

⟹ \\frac{1}{b}) = \\frac{1}{9}) - \\frac{1}{12})

⟹ \\frac{4 - 3}{36}) = \\frac{1}{36})

Hence, a = 12 and b = 36 is the other pair of possibility.

When a = 10,

⟹ \\frac{1}{b}) = \\frac{1}{9}) - \\frac{1}{10})

⟹ \\frac{10 - 9}{90}) = \\frac{1}{90})

Hence, a = 10 and b = 90 is the other pair of possibility.

Therefore (18,18) , (12,36) and (9,90) are the three different pairs of positive integers which satisfy the condition

The question is **"How many different pairs (a, b) of positive integers are there such that a ≤ b and \\frac{1}{a}) + \\frac{1}{b}) = \\frac{1}{9}) ? [TITA]" **

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