IMO B.

What is the value of A.
a^n is multiple of 8!.

the 1st statment says that a^n = 64.it could be 2^6 or 4^3.
Insufficient.

the 2nd statment says that n = 6 i.e a^6 is multiple of 8!.
sufficient.

Its a C trap question so be careful.

@abhinav85

How do you determine, under 2 minutes, that a^6 may not result in 2 values. Based on that alone I eliminated option 2 and went with C.

thats because..8! = 2*3*4*5*6*7*8.. get in terms of prime itegers..

8! = 2^7*3^2*5*7.....so to come to a^6 form only 2 can ascertain and none ..hence stmnt alone is ok

sorry rohan I don't understand what you mean

How do the prime factorizations give a clue to statement 2's sufficiency?

Prime factors were just a way to get to the lowest levels of factors..u r right that alone getting into terms of prime integers doesnt suffice..

But if you see 8! , you only 2 is having power >=6 .. any combination of factor which you can thus derive is not even close to 6 in its power..
Either you take 6 , 64 , 8 , , whatever

Answer would have been E if there would there would any possibility of getting any factor whose pwoer coluld have been of the form >=6

rohan_vus wrote:Prime factors were just a way to get to the lowest levels of factors..u r right that alone getting into terms of prime integers doesnt suffice..

But if you see 8! , you only 2 is having power >=6 .. any combination of factor which you can thus derive is not even close to 6 in its power..
Either you take 6 , 64 , 8 , , whatever

Answer would have been E if there would there would any possibility of getting any factor whose pwoer coluld have been of the form >=6

OA IS B

However I have the same doubt as obgeni and not cleared, could u try again...

Prime factors were just a way to get to the lowest levels of factors..u r right that alone getting into terms of prime integers doesnt suffice..

But if you see 8! , you only 2 is having power >=6 .. any combination of factor which you can thus derive is not even close to 6 in its power..
Either you take 6 , 64 , 8 , , whatever

Answer would have been E if there would there would any possibility of getting any factor whose pwoer coluld have been of the form >=6

I agree with rohan over here.

Ok.. let me put it this way..

When u get any number in terms of prime factors.. you are at the rootest levels of factors , which in other words is nothing but lowest levels of factors.. I guess atleast you will agree with me on this theory

Later , any other non prime factor which you get is derived by multiplying prime factors itslef.. for e.g .. 20 = is written as 2^2*5.. how many different factors it can have.. it can have 2 , 4 , 5 , 10 ... 4 u get by 2^2 ..and 10 u get by 2 *5 ...

Now our focus was the powers of the factors..Any non prime factor when its bulit ( like 10 and 4) from prime factors will always have least of the powers among the powers of individual prime factors of which its made of .

I dont know i make sense here

Thanks for explaining. It is clearer now