g.shankaran 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

OAB

Stat. (1) is obviously insufficient: a^n = 64 allows 2^6, or 4^3, or 6^2, so the value of a could be 2, or 4,or 8, depending on 8.

the real question is why stat. (2) is sufficient: why from the fact that n=6, a is determined. Let's see:
the product of the first 8 positive integers is 8*7*6*5*4*3*2*1. a^n must be a part a factor of this list. the problem is that n is a high power - a cannot be 5, or 7, or 8, for example, because 5^6 will come out of the scope of the list of factors of 1*2*3*4*5*6*7*8.

What could a be?
could a equal 2? Are there 6 powers of 2 in the product?
2 is one power
4 is 2^2
6 adds another 2
8 is 2^3
Total of 2 * 2^2 * 2 * 2^3 = 2^(1+2+1+3) = 2^6.

Could a be greater than 2? Could a be 3?

count the powers of 3:
3 - one power
6 adds another power.
that's it - only 3^2. Since there aren't 6 powers of 3 in the product, a cannot equal 3 if n=6.
the same goes for 4, 5, or any other power: 2 is the only base small enough to go 6 times in 1*2*3*4*5*6*7*8. Thus, a must equal 2, and stat. (2) is sufficient.

8! (2^7*3^2*5*7) is multiple of a^n
a)a^n=64
well 8! is multiple of 64. true.
it can be 2^6,4^3,8^2
Insufficient
b) n=6
a can be 2 only (as a^n is multiple of 8! and in 8! only 2 has power>=6)
thus sufficient.
IMO B

Hi,
8!=1.2.3.4.5.6.7.8 = 2^7.3^2.5.7
From(1): a^n = 64. So, the possibilities are:
a=2, n=6
a=4, n=3
a=8, n=2
Insufficient
From(2): n=6
8! = (2)^6.(2)(3^2)(5)(7) This number can be written uniquely as a multiple of a^6 i.e. 2^6
So, a=2
Sufficient

Hence, B