Answer should be C

Question states that product of first 8 positive integers i.e. 1*2*3*4*5*6*7*8 is a multiple of a^n. What is a = ?

1. a^n = 64 ==> 2^6 or 4^3 or 8^2. So, a can be equal to 2,4 or 8. So, insufficient.

2. n=6. This alone is insufficient to answer the question.

Together, the only possiblity of n=6 is when a=2. Also, 2^6 is a factor of the product of the first 8 positive integers.

Hope this helps.

rossmj wrote:If the integers a & n are >1 and the product of the first 8 positive integers is a multiple of a^n, what is the value of a?
1) a^n=64
2) n=6

Qtn decomposes to

2^7*3^2*5*7 is a mutiple of a^n. a=?

1. First statment similar evalauation done above

2. n=6

if you consider a =3, 2^7*3^2*5*7 cannot be multiple of 3^6


( How can something containing 3^2 be a multiple of 3^6 given the conditions in this problem)

similarly, if yon consider any damn value for "a" other than 2, 2^7*3^2*5*7 will not be mutiple of a^6


Hint: Look at 2^7 as highest power in the multiple and a^6

so it is clear that "a" has to be 2 to meet the condition 2^7*3^2*5*7 is a mutiple of a^n

Sufficient

IMO B

great expln by imcste.
My vote for B.
what is OA?

OA B

cubicle_bound_misfit wrote:great expln by imcste.
My vote for B.
what is OA?

Why not D?

Stmt 1 - a^n = 2^6 -- thus a = 2 and n = 6?

@xunil56

Vemuri wrote:
1. a^n = 64 ==> 2^6 or 4^3 or 8^2. So, a can be equal to 2,4 or 8. So, insufficient.

.

This is why St 1 is not suff.

can someone please explain this problem again, i still dont understand how statement 2 is the correct answer. Please do explain. Thanks