aditiniyer wrote:If the integer a & 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 are given that integers a and n are greater than 1, and the product of the first 8 positive integers is a multiple of a^n. Thus:
8!/a^n = integer
Recall that 8! = 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = (2^7) x (3^2) x 5 x 7.
We must determine the value of a.
Statement One Alone:
a^n = 64
There are multiple possible values of a. For instance, a = 2 and n = 6 (since 2^6 = 64), or a = 4 and n = 3 (since 4^3 = 64). Statement one alone is not sufficient to answer the question. We can eliminate answer choices A and D.
Statement Two Alone:
n = 6
Since n = 6, we know the following:
8!/a^6 = integer
[(2^7) x (3^2) x 5 x 7]/a^6 = integer
The only factor in the prime factorization of 8! that has a power greater than or equal to 6 is 2^7. Thus, the only possible value of a that will allow for 8!/a^6 to be an integer is 2.
Answer:
B
