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# Is the positive integer N a perfect square? (1) The number

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Re: Is the positive integer N a perfect square? (1) The number [#permalink]  07 Jul 2013, 23:53
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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

Hence, D is the correct answer choice

Hope it helps

Cheers!
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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?

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.

Hope it helps.

Hello Bunuel

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

Thankyou
Math Expert
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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?

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.

Hope it helps.

Hello Bunuel

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

Thankyou

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

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.

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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Re: Is the positive integer N a perfect square? (1) The number [#permalink]  24 Aug 2014, 20:49
Expert's post

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

Here is a post that explains this: http://www.veritasprep.com/blog/2010/12 ... t-squares/
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Manager Joined: 22 Aug 2014 Posts: 204 Followers: 0 Kudos [?]: 7 [0], given: 49 Re: Is the positive integer N a perfect square? (1) The number [#permalink] 17 Feb 2015, 07:58 Hi @veritasprepkarishma/@Bunuel, For 2nd statement if we take 1)4-perfect square-sum of distinct factors is 2 or 4(2*2 or 4*1) Condition satisfied 2)8-not a perfect square-sum of distinct factors is 2 or 8(2*2*2 or 8*1) Condition satisfied still not perfect square Then how can D be the answer? Math Expert Joined: 02 Sep 2009 Posts: 31303 Followers: 5363 Kudos [?]: 62508 [0], given: 9457 Re: Is the positive integer N a perfect square? (1) The number [#permalink] 17 Feb 2015, 08:39 Expert's post ssriva2 wrote: Hi @veritasprepkarishma/@Bunuel, For 2nd statement if we take 1)4-perfect square-sum of distinct factors is 2 or 4(2*2 or 4*1) Condition satisfied 2)8-not a perfect square-sum of distinct factors is 2 or 8(2*2*2 or 8*1) Condition satisfied still not perfect square Then how can D be the answer? (2) says that the sum of all distinct factors of N is even. If N = 4, then its factors are 1, 2, and 4 --> the sum = 1 + 2 + 4 = 7 = odd. If N = 8, then its factors are 1, 2, 4 and 8 --> the sum = 1 + 2 + 4 +8 = 15 = even. _________________ Intern Joined: 14 Jan 2015 Posts: 7 Followers: 0 Kudos [?]: 0 [0], given: 0 Re: Is the positive integer N a perfect square? (1) The number [#permalink] 18 Feb 2015, 10:21 goldeneagle94 wrote: Interesting Question !!! A few facts to review: A perfect sqaure ALWAYS has an ODD number of factors, whose sum is ALWAYS ODD. A perfect sqaure ALWAYS has an ODD number of Odd-factors, and EVEN number of Even-factors. Using the above facts, you can conclude that both statements are sufficient to answer the question. Perfect Square 36: (6 x 6) (3 X 3 X 2 X 2) 4 total factors, 2 distinct factors, and sum is even…? EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 5618 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: 340 Q170 V170 Followers: 227 Kudos [?]: 1600 [0], given: 154 Re: Is the positive integer N a perfect square? (1) The number [#permalink] 18 Feb 2015, 20:33 Expert's post Hi Kitzrow, You have to note the difference between "factors" and "prime factors" 36 has the following FACTORS: 1 and 36 2 and 18 3 and 12 4 and 9 6 So, there are 9 factors and the sum of those factors is 91. This example matches the prior statements - 36 has an ODD number of factors and the sum of those factors is ODD. GMAT assassins aren't born, they're made, Rich _________________ # Rich Cohen Co-Founder & GMAT Assassin # Special Offer: Save$75 + GMAT Club Tests

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Re: Is the positive integer N a perfect square? (1) The number [#permalink]  05 Aug 2015, 00:31
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?

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.

Hope it helps.

Bunuel Do you have 5-10 questions to practice on Perfect Square.

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Re: Is the positive integer N a perfect square? (1) The number [#permalink]  16 Aug 2015, 10:22
Expert's post
honchos 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?

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.

Hope it helps.

Bunuel Do you have 5-10 questions to practice on Perfect Square.

Check questions from the following list: search.php?st=0&sk=t&sd=d&search_tags=any&author_id=73391
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Re: Is the positive integer N a perfect square? (1) The number   [#permalink] 16 Aug 2015, 10:22

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