GMATPrep - DS

Question

I cant comprehend this question. Makes me feel like a moron! 
Can somebody rephrase this question ?

LM · Answer

a & n are = { 2,3,4,5,6,7,8....}

Product of first 8 positive integers is:

(2^7)*(3^2)*(5)*(7)


from (i) a^n = 64

2^6 = 64

4^3 = 64

from (ii)

n=6

we can have it possible only in case a^n = 2^6 which can divide the product of first 8 positive integers.

Therefore answer should be "B"

GMATBLACKBELT · Answer

should be B,

b/c we know a is greater than 1 so it must be +

pmenon · Answer

i dont get this.

product of the first 8 positive integers is 8! isnt it ?

gmatnub · Answer

wow, I thought that the gmat does not test intelligence, I stand corrected.

I would not have known where to begin. 

LM, you are the man/woman.


Let me try to rephrase LM's work for my own practice and reference.

a and n are positive integers having possible values greater 1. 

So a could be any number from 2,3,4,5,6,7,8,9....

So n could also be 2,3,4,5,6,7,8,9....

The phrase "the product of the first 8 positive integers" seems vague. How can you have 8 individual integers when a and n are just two integers? It would be more understandable if they add the word "possible" to the phrase to make it reads "the product of the first 8 'possible' positive integers". (but really, who would write a question like that an expect a real world answer?)

So the product of the first 8 possible integers:

2x3x4x5x6x7x8x9 = 2x3x2x2x5x2x3x7x2x2x2x3x3 = (2^7)*(3^4)*(5)*(7) 


from (i) a^n = 64 

possibilities are 

2^6 = 64 

4^3 = 64 

8^2 = 64

a could be 2, 4, or 8 ---> Insuff

from (ii) 

n=6 ---> SUFF

a can only be 2 because no other number can have the exponent that goes to 6 (although without doing math, i would NOT know for sure that 3^6 is no divisible by (2^7)*(3^4)*(5)*(7) )

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