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DS question a-to-the-power-n

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Source: — Data Sufficiency |
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by saketk » Tue Sep 06, 2011 10:30 am
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> :)
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by atishree » Tue Sep 06, 2011 10:34 am
oh cool, I understood the 8! thing while i was trying but forgot to break it up for the power of factors. thanks.[/u]
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by [email protected] » Wed Sep 07, 2011 8:19 am
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
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by saketk » Wed Sep 07, 2011 8:52 am
[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?
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by [email protected] » Wed Sep 07, 2011 9:33 am
[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.
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by razorback » Mon Nov 07, 2011 6:00 pm
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 ?
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