If two integers are chosen at random out of first 5 positive integers,

TL;DR

#s: 1, 2, 3, 4, 5

Total Outcome: 5C2 = 5!/3!2! = 10
Favourable Cases:

n1.n2 = a^2 - b^2 = (a+b)(a-b)

If n1=2 (3.5-1.5) and n2=5 (3.5+1.5), then a=3.5 b=1.5 => Not possible since a & b are + int

For a & b to be int, avg needs to be int, which is possible only when both chosen numbers are even or odd
#even: 2 => Fav case = 1
#odd: 3 => Fav case = 3C2 = 6
Total Fav case: 7 => ANSWER = 7/10

Official Solution

Solution: This might look like a probability question but isn’t. Questions like these are the reason we ask you to go through basics of every topic including probability. If you do not know probability at all, you may skip this question even though it needs very basic knowledge of probability.

Probability will tell you that

Required probability = Favorable cases/Total cases

Total cases are very easy to find: 5C2 = 10 or 5*4/2 = 10 whatever you prefer. This is the number of ways in which you select any 2 distinct numbers out of the given 5 distinct numbers.

Number of favorable cases is the challenge here. That is why it is a number properties question and not so much a probability question. Let’s focus on the main part of the question:

First five positive integers: 1, 2, 3, 4, 5

We need to select two integers such that their product is of the form a^2 – b^2. What does a^2 – b^2 remind you of? It reminds me of (a + b)(a – b). So the product needs to be of the form (a + b)(a – b). So is it necessary that of the two numbers we pick, one must be of the form (a + b) and the other must be (a – b)? No. Note that we should be able to write the product in this form. It is not necessary that the numbers must be of this form only.

But first let’s focus on numbers which are already of the form (a + b) and (a – b).

Say you pick two numbers, 2 and 5. Are they of the form (a + b) and (a – b) such that a and b are integers? No.

5 = 3.5 + 1.5

2 = 3.5 – 1.5

So a = 3.5, b = 1.5.

a and b are not integers.

What about numbers such as 3 and 5? Are they of the form (a + b) and (a – b) such that a and b are integers? Yes.

5 = 4 + 1

3 = 4 – 1

Note that whenever the average of the numbers will be an integer, we will be able to write them as a+b and a – b because one number will be some number more than the average and the other will be the same number less than average. So a will be the average and the amount more or less will be b.

When will the average of two numbers (Number1 + Number2)/2 be an integer? When the sum of the two numbers is even! When is the sum of two numbers even? It is when both the numbers are even or when both are odd. So then does the question boil down to “favorable cases are when we select both numbers even or both numbers odd?” Yes and No. When we select both even numbers or both odd numbers, the product can be written as a^2 – b^2. But are those the only cases when the product can be written as a^2 – b^2?

The question is not so much as whether both the numbers are even or both are odd as whether the product of the numbers can be written as product of two even numbers or two odd numbers. We need to be able to write the product (whatever we obtain) as product of two even or two odd numbers.

To explain this, let’s say we pick two numbers 4 and 5

4*5 = 20

Can we write 20 as product of two even numbers? Yes 2*10.

So even though, 4 is even and 5 is odd, their product can be written as product of two even numbers. So in which all cases will this happen?

- Whenever you have at least 4 in the product, you can write it as product of two even numbers: give one 2 to one number and the other 2 to the other number to make both even.

If the product is even but not a multiple of 4, it cannot be written as product of two even numbers or product of two odd numbers. It can only be written as product of one even and one odd number.

If the product is odd, it can always be written as product of two odd numbers.

Let’s go back to our question:

We have 5 numbers: 1, 2, 3, 4, 5

Our favorable cases constitute those in which either both numbers are odd or the product has 4 as a factor.

3 Odd numbers: 1, 3, 5

2 Even numbers: 2, 4

Number of cases when both numbers are odd = 3C2 = 3 (select 2 of the 3 odd numbers)

Number of cases when 4 is a factor of the product = Number of cases such that we select 4 and any other number = 1*4C1 = 4

Total number of favorable cases = 3 + 4 = 7

Note that this includes the case where we take both even numbers. Had there been more even numbers such as 6, we would have included more cases where we pick both even numbers such as 2 and 6 since their product would have 4 as a factor.

Required Probability = 7/10

Answer (C)

Takeaway:

When can we write a number as difference of squares?

- When the number is odd

or

- When the number has 4 as a factor