CAT 2017 Question Paper | Quants Slot 2

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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]

Video Explanation

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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