vinnik wrote:

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

Answer is

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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