Bunuel wrote:

If x and y are both integers greater than 1, is xy > 100?

(1) x has exactly 7 unique factors.

(2) y has exactly 9 unique factors.

Given situation:- \(x>1, y>1\) (x and y are integers)

Stem:- Is \(xy>100\)

St1:- x has exactly 7 unique factors

Or, we can write in the prime factorization form,\(x=a^p\), where p+1=7 or, p=6 (I have taken x=a^p not x=a^p*b^q..., because we are given 7 factors ,if we consider more than one prime exponent, we can't get a multiplication result of 7 by multiplying more than one integer values(where none of the integers are 1)) or \({(p+1)(q+1)...}\neq7\)

So,\(x=a^6\)

Now \(minimum(x)=min(a^6)=2^6=64\) (we can't consider a=1 since it yields x=1^6=1)

Therefore, \(x*y=64*2=128>100\) (The lowest integer value of y greater than 1 is 2)

For all other positive integer values of a>2,we have x > 100. Subsequently \(xy>100\).

Sufficient.

St2:-y has exactly 9 unique factors

With the same reasoning as stated in st1, we have \(y=a^p*b^q\), where (p+1)(q+1)=9=3*3.

So, p+1=3 and q+1=3

Or, p=2 and q=2

So, \(min(y)=min(a^2*b^2)=2^2*3^2=36\)

Therefore, \(xy=2*36=72<100\)----------------(a) (when x=2)

and \(xy=3*36=108>100\)-----------------------(b) (when x=3)

From (a) and (b), st2 is insufficient since st2 is inconsistent with the question stem.

Ans. (A)

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Regards,

PKN

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