Is 27^a/6^b a prime number?

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 understanding is that " 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 )

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 ( statement 1 ).

If (b) is a positive integer, the resultant would be a fraction, not a prime no.

Hence the Ans is C.

Hi goldfinchmonster,

Yes you are right! I understand what you are saying. For example \(3^x=4\) yields x=\(\frac{Ln(4)}{Ln(3).}\) I think your approach is right.

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem.
Remember equal number of variables and independent equations ensures a solution.

Is 27^a/6^b a prime number?

(1) 3a – b = 1
(2) b is a positive integer

Transforming the original condition and the question, we have 3^3a/(3^b)(2^b) = prime and 3^(3a-b)/2^b=prime.
In case of 1), if 3a-b=1, then b=1 and the answer is no, while the answer is yes if b=0. therefore the condition is not sufficient.
In case of 2), if b=positive integer then 3^(3a-b)/2^b=not prime and therefore the answer is no. Thus the condition is sufficient, and the answer is B.

1) Once we modify the original condition and the question according to the variable approach method 1, we can solve approximately 30% of DS questions.


2) Normally for cases where we need 2 more equations, such as original conditions with 2 variable, or 3 variables and 1 equation, or 4 variables and 2 equations, we have 1 equation each in both 1) and 2). Therefore C has a high chance of being the answer, which is why we attempt to solve the question using 1) and 2) together. Here, there is 70% chance that C is the answer, while E has 25% chance. These two are the key questions. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer according to DS definition, we solve the question assuming C would be our answer hence using ) and 2) together. (It saves us time). Obviously there may be cases where the answer is A, B, D or E.

The answer to the question cannot be B, it's C.

For (2) for some irrational a 27^a/6^b could be a prime. For example, if b = 1, then 27^a/6 could be a prime, say 2, for some irrational a.

Does it mean 3^x will be equal to all values from 1, 1.1, to 1000000 etc and there is some x for it...
say for eg 3^x =2, can the value of x as some int, fraction or decimal be specified by someone here..

Yes. There is specific x which satisfies 3^x = 2: https://www.wolframalpha.com/input/?i=3%5Ex%3D2

Bunuel am I correct in saying that for (2), because no statement is made about a that it can be any real number?

i.e. 6^b = 6 since b can = 1, & 27^a = 42 or 27^x = 48 , the result will be 7 and 8 respectively. (since no statement is made about a, we cannot assume it is an integer and 27^x is a continuous function in the real number space).

While the answer, as others are pointing out for B, is C, is there any OG question that considers the irrational number in the power of a number that results in the integer? (say, a^3 = 5, for some irrational number "a") I have come across a few more questions in the GMAT club but haven't found any in the OG.