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B
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Difficulty:
55%
(hard)
Question Stats:
58% (01:27) correct
42%
(01:33)
wrong
based on 570
sessions
History
Date
Time
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Not Attempted Yet
Is the positive integer n equal to the square of an integer?
(1) For every prime number p, if p is a divisor of n, then so is p^2
(2) \(\sqrt{n}\) is an integer
(1) For every prime number p, if p is a divisor of n, then so is p^2
(2) \(\sqrt{n}\) is an integer
22 Apr 2012, 01:36
Kudos
Bookmarks
shikhar
Is the positive integer n equal to the square of an integer?
(1) For every prime number p, if p is a divisor of n, then so is p2.
(2) is an integer.
(1) For every prime number p, if p is a divisor of n, then so is p2.
(2) is an integer.
Is the positive integer n equal to the square of an integer?
Question: is \(n=integer^2\)? So, basically we are asked whether \(n\) is a perfect square (a perfect square, is an integer that can be written as the square of some other integer. For example 16=4^2, is a perfect square.).
(1) For every prime number p, if p is a divisor of n, then so is p^2 --> if \(n=2^2\) then the answer is YES but if \(n=2^3\) then the answer is NO (notice that in both case prime number 2 as well as 2^2 are divisors of n, so our condition is satisfied). Not sufficient.
(2) \(\sqrt{n}\) is an integer --> \(\sqrt{n}=integer\) --> \(n=integer^2\). Sufficient.
Answer: B.
General Discussion
05 May 2013, 00:40
Kudos
Bookmarks
Bunuel
shikhar
Is the positive integer n equal to the square of an integer?
(1) For every prime number p, if p is a divisor of n, then so is p2.
(2) is an integer.
(1) For every prime number p, if p is a divisor of n, then so is p2.
(2) is an integer.
Is the positive integer n equal to the square of an integer?
Question: is \(n=integer^2\)? So, basically we are asked whether \(n\) is a perfect square (a perfect square, is an integer that can be written as the square of some other integer. For example 16=4^2, is a perfect square.).
(1) For every prime number p, if p is a divisor of n, then so is p^2 --> if \(n=2^2\) then the answer is YES but if \(n=2^3\) then the answer is NO (notice that in both case prime number 2 as well as 2^2 are divisors of n, so our condition is satisfied). Not sufficient.
(2) \(\sqrt{n}\) is an integer --> \(\sqrt{n}=integer\) --> \(n=integer^2\). Sufficient.
Answer: B.
ST 1-isnt this telling you all the prime factors of n are raised to even powers which makes n a square number-i got wrong can you please re-explain.
