Question

How many pairs \((a, b)\) of positive integers are there such that \(a≤b\) and \(ab=4^{2017}\) ?

• A. 2017
• B. 2019
• C. 2020
• D. 2018

Answer 1

The Correct Option is D

The correct answer is (D): \(2018\)
Given, \(a.b = 4^{2017} = 2^{4034}\)
Since \(a.b\) = \(4^{2017}\), is a perfect square the number of factors of \(2^{4034}\) is odd.
Required answer, the number of values of \(A\) = \(\frac{4034+1+1}{2} = 2018\)

Answer 2

Given, \(ab=4^{2017}= (2^2)^{2017}=2^{4034}\)
So, Total number of factors = \(2 \times2017+1=4035\)
Among these 4035 factors, we can choose two numbers a, b such that a<b in \([\frac{4035}{2}]\) = 2017.
Since the given number is a Perfect Square , we have one set of two equal factors.
The number of such pairs is the highest integer which is when round off =\(\frac{4035}{2}= 2018\)
Therefore, There are 2018 pairs (a,b) of positive integer such that a≤b and \(ab=4^{2017}\)

∴ Correct Answer is Option (D) : 2018