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Please help me in understanding this question. I am confused between C and D.

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.

Please help me in understanding this question. I am confused between C and D.

Thanks & Regards Vinni

(A) \(a=ab\) for \(b=1\) and any positive integer \(a>1.\) (B) \(b=ab\) for \(a=1\) and any positive integer \(b>1.\) (C) \(3b+2a=ab\) can be written as \(ab-2a-3b=0\) or \(a(b-2)-3(b-2)=6\) and finally \((a-3)(b-2)=6.\) Then we can have for example \(a-2=1\) and \(b-2=6\) or \(a=3\) and \(b=8.\) (D) \(b-a=ab\) or \(b=a(b+1)\). Since \(a\geq{1}\) and \(b>0,\) \(\, \, b=a(b+1)\geq{1}\cdot{(b+1)}=b+1\), impossible. Or, \(b-a<b\) but \(ab>b,\) because \(a\geq{1}.\) So, this equality CANNOT hold.

(E) \(ab=ab,\) no problem for any pair of positive integers \(a\) and \(b.\)

Answer D
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PhD in Applied Mathematics Love GMAT Quant questions and running.

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'.
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Please help me in understanding this question. I am confused between C and D.

Thanks & Regards Vinni

In these questions it is best to take an example as if something is true for all positive integers than it as to be true for the smallest and the easiest integers to work with take a = 1 and b = 2 and work with the options ab = 2 A) a take a =2, b = 1 B) b take a = 1 b = 2 C) 3b + 2a Seems tricky, lets see other options and then come back to it. D) b - a take b = 1 and a = 2 --> b - a = -1 .. How the hell can product of two positive integers be negative ?? or less than each of them? E) ba Always true

Please help me in understanding this question. I am confused between C and D.

Thanks & Regards Vinni

In these questions it is best to take an example as if something is true for all positive integers than it as to be true for the smallest and the easiest integers to work with take a = 1 and b = 2 and work with the options ab = 2 A) a take a =2, b = 1 B) b take a = 1 b = 2 C) 3b + 2a Seems tricky, lets see other options and then come back to it. D) b - a take b = 1 and a = 2 --> b - a = -1 .. How the hell can product of two positive integers be negative ?? or less than each of them? E) ba Always true

You don't even have to worry what C is !

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.
_________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Please help me in understanding this question. I am confused between C and D.

Thanks & Regards Vinni

In these questions it is best to take an example as if something is true for all positive integers than it as to be true for the smallest and the easiest integers to work with take a = 1 and b = 2 and work with the options ab = 2 A) a take a =2, b = 1 B) b take a = 1 b = 2 C) 3b + 2a Seems tricky, lets see other options and then come back to it. D) b - a take b = 1 and a = 2 --> b - a = -1 .. How the hell can product of two positive integers be negative ?? or less than each of them? E) ba Always true

You don't even have to worry what C is !

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.

Re: Which of the following CANNOT be a product of two distinct [#permalink]

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Re: Which of the following CANNOT be a product of two distinct [#permalink]

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08 Dec 2015, 11:23

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