Hi Chetan / Bhaskar,
According to me
Concider 3^C (Numerator), where c is some random fraction ( We are not given (a) is a integer or not ) which could yield any positive number in numerator ( both even & odd ), which when divided by 2^b may yield a prime or may not.
My logic is certainly 3^ some fraction would yield 4, 6 ..etc. If its 4 , and (b) is one, the answer would be yes and for all other + integers for (b) answer would be no.
Please let me know if this approach has any flaw.
(27^a) / (6^b) can be re-written as ( 3^3a ) / ( 2^b * 3^b )
Now Statement 1 :- 3a = 1 + b
Substitute the value of 3a in the re-written prompt. i.e ( 3^1 * 3^b ) / ( 2^b * 3^b ) = 3 / (2^b).
The only unknown value in the equation is b.
Case 1 :- If (b) is zero, than the answer would be yes.
Case 2 :- If (b) is any positive integer, that the result would certainly be a fraction and hence not a prime number. So the answer would be No.
Two different solution, Hence Insufficiant.
Satement 2 - Definitely Not sufficiant.
Combining the 2 satements, (b) is a positive integer ( statement 2 ) and 3 / 2^b from first statement.
If (b) is a positive integer, the resultant would be a fraction, not a prime no.
Hence the Ans is C.
Yes you are right! I understand what you are saying. For example