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Re: Is the positive integer N a perfect square? (1) The number [#permalink]
27 Dec 2013, 08:20

mbaMission wrote:

Is the positive integer N a perfect square?

(1) The number of distinct factors of N is even. (2) The sum of all distinct factors of N is even.

Remember two properties 'bout perfect squares

The number of distinct factors of N is even, of course one will always need pairs and will always have the factor 1 remaining hence always odd, so the answer is NO, N is not a perfect squares

The sum of all distinct factors of N is even, of course, same reason, all the pairs will add up to an even number +1 = odd

Re: a perfect square [#permalink]
06 Feb 2014, 23:13

Bunuel wrote:

tingle15 wrote:

I have a doubt...

Consider N=18, Its factors are: 1, 2, 3, 6, 9, 18. The sum of factors is 39 which is odd... Am i missing something?

Tips about the perfect square: 1. The number of distinct factors of a perfect square is ALWAYS ODD. The reverse is also true: if a number has the odd number of distinct factors then it's a perfect square;

2. The sum of distinct factors of a perfect square is ALWAYS ODD. The reverse is NOT always true: a number may have the odd sum of its distinct factors and not be a perfect square. For example: 2, 8, 18 or 50;

3. A perfect square ALWAYS has an ODD number of Odd-factors, and EVEN number of Even-factors. The reverse is also true: if a number has an ODD number of Odd-factors, and EVEN number of Even-factors then it's a perfect square. For example: odd factors of 36 are 1, 3 and 9 (3 odd factor) and even factors are 2, 4, 6, 12, 18 and 36 (6 even factors);

4. Perfect square always has even powers of its prime factors. The reverse is also true: if a number has even powers of its prime factors then it's a perfect square. For example: \(36=2^2*3^2\), powers of prime factors 2 and 3 are even.

NEXT: There is a formula for Finding the Number of Factors of an Integer:

First make prime factorization of an integer \(n=a^p*b^q*c^r\), where \(a\), \(b\), and \(c\) are prime factors of \(n\) and \(p\), \(q\), and \(r\) are their powers.

The number of factors of \(n\) will be expressed by the formula \((p+1)(q+1)(r+1)\). NOTE: this will include 1 and n itself.

Example: Finding the number of all factors of 450: \(450=2^1*3^2*5^2\)

Total number of factors of 450 including 1 and 450 itself is \((1+1)*(2+1)*(2+1)=2*3*3=18\) factors.

Back to the original question:

Is the positive integer N a perfect square?

(1) The number of distinct factors of N is even --> let's say \(n=a^p*b^q*c^r\), given that the number of factors of \(n\) is even --> \((p+1)(q+1)(r+1)=even\). But as we concluded if \(n\) is a perfect square then powers of its primes \(p\), \(q\), and \(r\) must be even, and in this case number of factors would be \((p+1)(q+1)(r+1)=(even+1)(even+1)(even+1)=odd*odd*odd=odd\neq{even}\). Hence \(n\) can not be a perfect square. Sufficient.

(2) The sum of all distinct factors of N is even --> if \(n\) is a perfect square then (according to 3) sum of odd factors would be odd and sum of even factors would be even, so sum of all factors of perfect square would be \(odd+even=odd\neq{even}\). Hence \(n\) can not be a perfect square. Sufficient.

Answer: D.

Hope it helps.

Hello Bunuel

What if n=1 ? Question says n is a Positive Integer. is 1 considered a perfect square ? Please clarify.

Re: a perfect square [#permalink]
07 Feb 2014, 04:21

Expert's post

niyantg wrote:

Bunuel wrote:

tingle15 wrote:

I have a doubt...

Consider N=18, Its factors are: 1, 2, 3, 6, 9, 18. The sum of factors is 39 which is odd... Am i missing something?

Tips about the perfect square: 1. The number of distinct factors of a perfect square is ALWAYS ODD. The reverse is also true: if a number has the odd number of distinct factors then it's a perfect square;

2. The sum of distinct factors of a perfect square is ALWAYS ODD. The reverse is NOT always true: a number may have the odd sum of its distinct factors and not be a perfect square. For example: 2, 8, 18 or 50;

3. A perfect square ALWAYS has an ODD number of Odd-factors, and EVEN number of Even-factors. The reverse is also true: if a number has an ODD number of Odd-factors, and EVEN number of Even-factors then it's a perfect square. For example: odd factors of 36 are 1, 3 and 9 (3 odd factor) and even factors are 2, 4, 6, 12, 18 and 36 (6 even factors);

4. Perfect square always has even powers of its prime factors. The reverse is also true: if a number has even powers of its prime factors then it's a perfect square. For example: \(36=2^2*3^2\), powers of prime factors 2 and 3 are even.

NEXT: There is a formula for Finding the Number of Factors of an Integer:

First make prime factorization of an integer \(n=a^p*b^q*c^r\), where \(a\), \(b\), and \(c\) are prime factors of \(n\) and \(p\), \(q\), and \(r\) are their powers.

The number of factors of \(n\) will be expressed by the formula \((p+1)(q+1)(r+1)\). NOTE: this will include 1 and n itself.

Example: Finding the number of all factors of 450: \(450=2^1*3^2*5^2\)

Total number of factors of 450 including 1 and 450 itself is \((1+1)*(2+1)*(2+1)=2*3*3=18\) factors.

Back to the original question:

Is the positive integer N a perfect square?

(1) The number of distinct factors of N is even --> let's say \(n=a^p*b^q*c^r\), given that the number of factors of \(n\) is even --> \((p+1)(q+1)(r+1)=even\). But as we concluded if \(n\) is a perfect square then powers of its primes \(p\), \(q\), and \(r\) must be even, and in this case number of factors would be \((p+1)(q+1)(r+1)=(even+1)(even+1)(even+1)=odd*odd*odd=odd\neq{even}\). Hence \(n\) can not be a perfect square. Sufficient.

(2) The sum of all distinct factors of N is even --> if \(n\) is a perfect square then (according to 3) sum of odd factors would be odd and sum of even factors would be even, so sum of all factors of perfect square would be \(odd+even=odd\neq{even}\). Hence \(n\) can not be a perfect square. Sufficient.

Answer: D.

Hope it helps.

Hello Bunuel

What if n=1 ? Question says n is a Positive Integer. is 1 considered a perfect square ? Please clarify.

Thankyou

Yes, 1 is a perfect square: 1 = 1^1. _________________

Re: Is the positive integer N a perfect square? (1) The number [#permalink]
21 Aug 2014, 00:26

Bunuel wrote:

tingle15 wrote:

I have a doubt...

Consider N=18, Its factors are: 1, 2, 3, 6, 9, 18. The sum of factors is 39 which is odd... Am i missing something?

Tips about the perfect square: 1. The number of distinct factors of a perfect square is ALWAYS ODD. The reverse is also true: if a number has the odd number of distinct factors then it's a perfect square;

2. The sum of distinct factors of a perfect square is ALWAYS ODD. The reverse is NOT always true: a number may have the odd sum of its distinct factors and not be a perfect square. For example: 2, 8, 18 or 50;

3. A perfect square ALWAYS has an ODD number of Odd-factors, and EVEN number of Even-factors. The reverse is also true: if a number has an ODD number of Odd-factors, and EVEN number of Even-factors then it's a perfect square. For example: odd factors of 36 are 1, 3 and 9 (3 odd factor) and even factors are 2, 4, 6, 12, 18 and 36 (6 even factors);

4. Perfect square always has even powers of its prime factors. The reverse is also true: if a number has even powers of its prime factors then it's a perfect square. For example: \(36=2^2*3^2\), powers of prime factors 2 and 3 are even.

NEXT: There is a formula for Finding the Number of Factors of an Integer:

First make prime factorization of an integer \(n=a^p*b^q*c^r\), where \(a\), \(b\), and \(c\) are prime factors of \(n\) and \(p\), \(q\), and \(r\) are their powers.

The number of factors of \(n\) will be expressed by the formula \((p+1)(q+1)(r+1)\). NOTE: this will include 1 and n itself.

Example: Finding the number of all factors of 450: \(450=2^1*3^2*5^2\)

Total number of factors of 450 including 1 and 450 itself is \((1+1)*(2+1)*(2+1)=2*3*3=18\) factors.

Back to the original question:

Is the positive integer N a perfect square?

(1) The number of distinct factors of N is even --> let's say \(n=a^p*b^q*c^r\), given that the number of factors of \(n\) is even --> \((p+1)(q+1)(r+1)=even\). But as we concluded if \(n\) is a perfect square then powers of its primes \(p\), \(q\), and \(r\) must be even, and in this case number of factors would be \((p+1)(q+1)(r+1)=(even+1)(even+1)(even+1)=odd*odd*odd=odd\neq{even}\). Hence \(n\) can not be a perfect square. Sufficient.

(2) The sum of all distinct factors of N is even --> if \(n\) is a perfect square then (according to 3) sum of odd factors would be odd and sum of even factors would be even, so sum of all factors of perfect square would be \(odd+even=odd\neq{even}\). Hence \(n\) can not be a perfect square. Sufficient.

Answer: D.

Hope it helps.

Hi, could you explain why " A perfect square ALWAYS has an ODD number of Odd-factors, and EVEN number of Even-factors" is true? Thanks _________________

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