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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.

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.

Re: Is the positive integer n equal to the square [#permalink]

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05 May 2013, 00:40

Bunuel wrote:

shikhar wrote:

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.

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.

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.

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.

No, the first statement says that if a prime number p is a factor of n, then so is p^2, which means that the power of p is more than or equal to 2: it could be 2, 3, ... So, n is not necessarily a prefect square. For example, 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).

Re: Is the positive integer n equal to the square of an integer? [#permalink]

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06 May 2013, 13:00

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) root n is an integer

From 1 ) If p=4, than 16 is also a factor. Which can qualify n to be a perfect square.But if p=2 than 4 is also a factor. However we can't say if n is square of an integer or not. Hence Insufficient.

2) If root n is an integer -> N has to be the square of an integer. Sufficient.

Re: Is the positive integer n equal to the square of an integer? [#permalink]

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12 Dec 2014, 07:04

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