abhasjha wrote:.if the integers a and n are greater than 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

We have k*a^n = 8!, k is a positive integer constant and the integers a and n are greater than 1. What is a?

(1) For the integers a and n greater than 1, we have more than one way to show that a^n = 64; 2^6 = 64, 4^3 = 64, and 8^2 = 64. Insufficient

(2) If n = 6, a^6 can be a factor of 8! for both rational and irrational a. Insufficient

Combine the two

n = 6 is possible only when a = 2.

[spoiler]C[/spoiler]

abhasjha wrote:.if the integers a and n are greater than 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

Product of first 8 positive integers = 8! = 2*3*4*5*6*7*8 = 2^7*3^2*5*7

is a multiple of a^n

1) a^n =64 seems like a weird choice since 8! is not a multiple of 64
2) n=6 also doesnt help

Please check the question

ajith wrote:

abhasjha wrote:.if the integers a and n are greater than 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

Product of first 8 positive integers = 8! = 2*3*4*5*6*7*8 = 2^7*3^2*5*7

is a multiple of a^n

1) a^n =64 seems like a weird choice since 8! is not a multiple of 64
2) n=6 also doesnt help

Please check the question

But 8! is a multiple of 64 or 2^6.

acc to me Ans is B

@ Abhasjha

PLZ DONT FORGET TO PUT OA ALSO