Manpreet Singh 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


Is the integer n odd?
(1) n is divisible by 3
(2) 2n is divisible by twice as many positive integers as n

Please don't post multiple question in one post. It won't help others to make proper use of the solution or idea posted in your post.

Now lets take your questions.

1.

Product of 1st +ve integers will be 1*2*3*4*5*6*7*8 = 2^7 * 3^2 * 5 * 7.
This is a multiple of a^n, where a,n > 1

a. 2^6 = 64 or 4^2 = 64. hence not sufficient.

b only 2 is the possible value of a as it can only be raised to 6.


Answer is B.

2.
n = odd ?

a. 3 is odd while 6 is even. hence not sufficient.
b. 2n has double the factor as n has.
lets say n is odd and of form a^p * b^q * c^r where a,b and c are all odd prime integer and p,q and r are just +ve number.
so no. of factor of n will (p+1)*(q+1)*(r+1) which will be odd.
so 2n will be of form 2 * a^p * b^q * c^r and have total factor of
(1 + 1)* (p+1)*(q+1)*(r+1).
but if n is even then this statement won't stand.

Hence b is sufficient.

Answer is B.

Manpreet Singh 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


Is the integer n odd?
(1) n is divisible by 3
(2) 2n is divisible by twice as many positive integers as n

Here is my two cents for the second question.
Statement 1: clearly insufficient becasue n can be 3, 12, 6, 15 etc
statement 2: If n is an odd number than 2n has twice the number of factors than that of n.

For Eg n=3 has two factors 1 and 3.
2n= 6 has four factors 1,2,3 and 6.

Thus B alone is sufficient.