atishree wrote:Please tell why the answer is (2) only

Hi --
a,n> 1

to find the value of a.

product of first 8 positive integers -- 1*2*3*4*5*6*7*8 .

Using STATEMENT 1-- we get (a^n = 64)

8^2 = 64.

2^6 = 64

4^3 = 64

clearly, we are getting 3 different values of a here. HENCE, INSUFFICIENT


Using option 2. we have n = 6.

given, the product of first 8 positive numbers is a multiple of a^n. Put n= 6 here.

we get. the product of first 8 positive numbers is a multiple of a^6

or we can rewrite this statement as a^6 is a factor of the product of first 8 positive numbers.

the important thing to note here is that the product is nothing but 8! [or I am so stupid that I just looked at it ]

Anyways, 8! can be broken into -- 2^7*3^2*5*7 -- the only possibility that a^6 is a factor of this number is when a=2. Other numbers don't have the power equal to 6 or more...

HENCE, STATEMENT 2 is SUFFICIENT>

oh cool, I understood the 8! thing while i was trying but forgot to break it up for the power of factors. thanks.[/u]

atishree wrote:Please tell why the answer is (2) only

I THINK THE ANSWER IS A. QUESTION ASK IS 8!/ A^N INTEGER
FROM 1: A^N = 64 . BOTH A AND N ARE INTEGERS WHICH ARE GREATER THAN 1
THEREFOR A COULD BE = 2,4,8, AND N COULD BE = 6, 3, 2. IN EVERY CONDITION WE GET AN INTEGER VALUE. THUS SUFFICIENT
FROM 2: N IS = 6. BUT A CAN HAVE ANY VALUE EG 2,3,4,5,6,7,ETC THUS INSUFFICIENT.

THEREFORE ANSWER IS AAAAA. 100% SURE

[email protected] wrote:

atishree wrote:Please tell why the answer is (2) only

I THINK THE ANSWER IS A. QUESTION ASK IS 8!/ A^N INTEGER
FROM 1: A^N = 64 . BOTH A AND N ARE INTEGERS WHICH ARE GREATER THAN 1
THEREFOR A COULD BE = 2,4,8, AND N COULD BE = 6, 3, 2. IN EVERY CONDITION WE GET AN INTEGER VALUE. THUS SUFFICIENT
FROM 2: N IS = 6. BUT A CAN HAVE ANY VALUE EG 2,3,4,5,6,7,ETC THUS INSUFFICIENT.

THEREFORE ANSWER IS AAAAA. 100% SURE

Why so many A's?
Anyways, the basic rule in DS questions is to find a unique value. Can you do that using statement 1 only?

[email protected] wrote:

atishree wrote:Please tell why the answer is (2) only

I THINK THE ANSWER IS A. QUESTION ASK IS 8!/ A^N INTEGER
FROM 1: A^N = 64 . BOTH A AND N ARE INTEGERS WHICH ARE GREATER THAN 1
THEREFOR A COULD BE = 2,4,8, AND N COULD BE = 6, 3, 2. IN EVERY CONDITION WE GET AN INTEGER VALUE. THUS SUFFICIENT
FROM 2: N IS = 6. BUT A CAN HAVE ANY VALUE EG 2,3,4,5,6,7,ETC THUS INSUFFICIENT.

THEREFORE ANSWER IS AAAAA. 100% SURE

SORRY I READ QUESTION WRONGLY.

Using the information in 2, that n = 6...

can someone explain to me why a couldn't equal 11, subsequently a^6 = 11^6 and a multiple of this being (8!)*11^6 ?