How many different pairs (a, b) of positive integers are there such that a ≤ b and 1a+1b=19?
Explanation:
Given 1a+1b=19
b+aab=19
⇒ 9b + 9a = ab ⇒ ab - 9b – 9a = 0 ⇒ ab - 9b – 9a + 81 = 81 [Adding 81 both sides] ⇒ b(a – 9) – 9(a – 9) = 81 ⇒ (a – 9) × (b – 9) = 81
Now 81 as a product of 2 positive integers can be written as 1 × 81, 3 × 21 or 9 × 9
If a - 9 = 1 ⇒ a = 10. Also b - 9 = 81 ⇒ b = 90
Further, if a – 9 = 3, ⇒ a = 12 Also b – 9 = 27 ⇒ b = 36
If a – 9 = 9 ⇒ a = 18. Also b – 9 = 9 ⇒ b = 18
∴ 3 pairs of number i.e., (10, 90), (12, 36) and (18, 18) will satisfy the given equation.
Hence, 3.
» Your doubt will be displayed only after approval.
Help us build a Free and Comprehensive Preparation portal for various competitive exams by providing us your valuable feedback about Apti4All and how it can be improved.