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CAT 2017 Slot 2 — QA Question 28

FactorsEasy

How many different pairs (a, b) of positive integers are there such that a ≤ b and 1a+1b=19?

Answer & solution

Correct answer: 3

Answer: 3

Solution

Easy

Clear denominators and use Simon's Favourite Factoring Trick: turn the equation into a product of two integer factors equal to 8181, then count factor pairs that respect a≤ba\le b.

1

Clear fractions. Combine the left side over abab.

1a+1b=19⇒ a+bab=19⇒ 9a+9b=ab\begin{aligned} &\frac{1}{a}+\frac{1}{b}=\frac{1}{9}\\ &\Rightarrow\ \frac{a+b}{ab}=\frac{1}{9}\\ &\Rightarrow\ 9a+9b=ab \end{aligned}
2

Factor. Rearrange and add 8181 to both sides to factor the left side.

ab−9a−9b=0⇒ ab−9a−9b+81=81(add 81 to both sides)⇒ (a−9)(b−9)=81\begin{aligned} &ab-9a-9b=0\\ &\Rightarrow\ ab-9a-9b+81=81 \quad\text{(add 81 to both sides)}\\ &\Rightarrow\ (a-9)(b-9)=81 \end{aligned}
3

Count factor pairs with a≤ba\le b. Since a,ba,b are positive integers with a≤ba\le b, we need (a−9)≤(b−9)(a-9)\le(b-9) as a factorisation of 81=3481=3^4 into positive factors:

81=1×81 ⇒ (a,b)=(10,90)81=3×27 ⇒ (a,b)=(12,36)81=9×9 ⇒ (a,b)=(18,18)\begin{aligned} &81=1\times 81\ \Rightarrow\ (a,b)=(10,90)\\ &81=3\times 27\ \Rightarrow\ (a,b)=(12,36)\\ &81=9\times 9\ \Rightarrow\ (a,b)=(18,18) \end{aligned}

All three give positive a,ba,b with a≤ba\le b, so there are 33 valid pairs.

Number of pairs=3\text{Number of pairs}=3

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CAT 2017 Slot 2 QA Q28: How many different pairs (a, b) of positive integers are there such that a ≤ b and 1 a + 1 b = 1 9 ? — Solution | TheCATExam