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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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Explanatory Answer

Method of solving this CAT Question from Number Theory

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]"

Hence, the answer is 3

 

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