Which of the following CANNOT be a product of two distinct

Choosing a specific value for \(b\) for which there is no \(a\) which fulfills the condition is not a proof. There are infinitely many other possible values for \(b.\) The statement is CANNOT be true, not that it MUST be true for any \(a\) and \(b.\)
You have to prove/justify that there is no pair \((a,b)\) of positive integers for which the equality holds.

Ohk, i didnt understand your point first yes you are right ! i probably didnt phrase my solution correctly but i just wanted to show that product of two +ve integers can never be less that either of them.

First of all, we can eliminate answer choices A, B, and E right away. There is no reason to distinguish between \(a\) and \(b\), so if \(a\) cannot be a product, then \(b\) cannot be either and since we cannot have two correct answers in PS problems, then neither A nor B can be correct. As for E it's clearly cannot be correct answer.

So, we are left with options C and D.

For C, can \(ab=3b+2a\)? --> \(ab-2a=3b\) --> \(a(b-2)=3b\) --> \(a=\frac{3b}{b-2}\). Now, if \(b=3\), then \(a=9\), so in this case \(ab=3b+2a\) is possible.

Only option left is D.

Answer: D.

Else, you can notice that since \(a\) and \(b\) are positive integers, then their product MUST be greater than their difference, so D is not possible.

Hope it's clear.

If we pick b = -2 and a = 2 then for d we get -4 which is same as ab....

a and b need to be positive integers.

That is where the trick in this question lies. a and b are distinct positive integers. So their product must be equal to or greater than the greater number.
Considering b - a, b will be greater than a (so that the result is positive) and b - a will be smaller than b. Hence, (b - a) can never be the product of 'b' and 'a'.

Asked: Which of the following CANNOT be a product of two distinct positive integers a and b?

A. a
B. b
C. 3b + 2a
D. b - a
E. ba

b - a < b
ab > b
Therefore b-a CANNOT be a product of two distinct positive integers a and b

IMO D