DS question a-to-the-power-n
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Hi --atishree wrote:Please tell why the answer is (2) only
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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I THINK THE ANSWER IS A. QUESTION ASK IS 8!/ A^N INTEGERatishree wrote:Please tell why the answer is (2) only
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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Why so many A's?[email protected] wrote:I THINK THE ANSWER IS A. QUESTION ASK IS 8!/ A^N INTEGERatishree wrote:Please tell why the answer is (2) only
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
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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SORRY I READ QUESTION WRONGLY.[email protected] wrote:I THINK THE ANSWER IS A. QUESTION ASK IS 8!/ A^N INTEGERatishree wrote:Please tell why the answer is (2) only
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