CAT 2020 Question Paper | Quants Slot 3

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Question 26 : How many pairs (a,b) of positive integers are there such that a ≤ b and ab = 4²⁰¹⁷?

Video Explanation

Explanatory Answer

4²⁰¹⁷ = 2^2×2017
Number of factors of 2^2×2017 = 2×2017 + 1= 4035
Number of pairs of a,b such that a×b = 2^2×2017 is equal to half the Number of factors of 2^2×2017
Number of pairs of a,b such that a×b = 2^2×2017
= $\frac{4035}{2}$
You observe that $\frac{4035}{2}$ is not an integer, because 2^2×2017 is a perfect square and one of the pairs of (a,b) is (2017 , 2017).
So, the number of such pairs is the highest integer roundoff of $\frac{4035}{2}$ = 2018.
Among all these 2018 pairs, one of the integer is less than or equal to the other. (Equal in the case of (2017,2017))
We assume that a is the least one of the two...
Hence there are 2018 pairs that satify this condition.

Alternate Method:
a × b = 4²⁰¹⁷
a × b = 2^2×2017

a and b are of the form 2^x and 2^y respectfully...
Since a ≤ b;
2^x ≤ 2^y
x ≤ y
Also, a × b = 2^2×2017
2^x + 2^y = 2^2×2017
x + y = 2×2017

We have 2 conditions to deal with...
x ≤ y and x + y = 2×2017
Let's start with x = y:
Here, x = y = 2017

From here on, we decrement x and increment y to maintain the conditions x ≤ y and x + y = 2×2017
We can keep doing this until x = 0, because if x is negative, a which is 2^x will not remain an integer.

Hence x can range from 2017 to 0; and thereby x can take 2018 values.
Therefore, there can be 2018 pairs of (a,b) that sitisfy a ≤ b and ab = 4²⁰¹⁷

The question is "How many pairs (a,b) of positive integers are there such that a ≤ b and ab = 4²⁰¹⁷?"

Hence, the answer is, "2018"

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