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

  1. 2019
  2. 2018
  3. 2020
  4. 2017

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Explanatory Answer

42017 = 22×2017
Number of factors of 22×2017 = 2×2017 + 1= 4035
Number of pairs of a,b such that a×b = 22×2017 is equal to half the Number of factors of 22×2017
Number of pairs of a,b such that a×b = 22×2017
= \\frac{4035}{2})
You observe that \\frac{4035}{2}) is not an integer, because 22×2017 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 = 22×2017

a and b are of the form 2x and 2y respectfully...
Since a ≤ b;
2x ≤ 2y
x ≤ y
Also, a × b = 22×2017
2x + 2y = 22×2017
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 2x will not remain an integer.

x y
2017 2017
2016 2018
2015 2019
2 4033
1 4034
0 4035

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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