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

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Is the positive integer N a perfect square? (1) The number [#permalink]  02 Jun 2009, 04:18
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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.
[Reveal] Spoiler: OA

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Last edited by Bunuel on 30 Jul 2012, 03:36, edited 1 time in total.
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Re: a perfect square [#permalink]  02 Jun 2009, 06:49
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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.
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Re: a perfect square [#permalink]  02 Jun 2009, 06:55
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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.

Note: A square never has even number of distinct factors. Also the sum of distinct factors of a square is never even.

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

Suppose N = 4. It has 3 distinct factors: 1, 2 and 4.
Suppose N = 9. It has 3 distinct factors: 1, 3 and 9.
Suppose N = 16. It has 5 distinct factors: 1, 2, 4, 8, and 16.
Suppose N = 64. It has 7 distinct factors: 1, 2, 4, 8, 16, 32, and 64.
But that not the case. In fact, the case is opposite. So it is sufficient because N is not a square.

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

If you follow the above pattern, you see 1 is always there. The sum of all distinct factors except 1 of N is even. If you add 1 on the even sum, that odd. So N is not a square.
But that not the case. In fact, the case is opposite. So it is sufficient because N is again not a square.

So D.

To validate the above premises, lets assume if N = 10. Its factors are 1, 2, 5 and 10 and their sum = 18. So 10 has 4 distinct factors and 18, which is even, sum.

If N = 15. Its factors are 1, 3, 5 and 15 and their sum = 24. So 15 has 4 distinct factors and even sum of its distinct factors.

If N = 30. Its factors are 1, 2, 3, 5, 6, 10, 15 and 30 and their sum = 40. So 15 has 8 distinct factors and even sum of its distinct factors. HTH.

Note: It would be better if you do not post OA or spoiler before any response from other members.

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Re: a perfect square [#permalink]  03 Jun 2009, 23:03
Nice funda from goldern eagle..

i tried with examples and got the same deduction..
let me write it once agian..\

A perfect number always has odd number of factors whose sum is odd...

A perfect number alwyas has odd number of odd factors and even number of even factors...
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Re: a perfect square [#permalink]  30 Sep 2010, 08:56
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?
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Re: a perfect square [#permalink]  30 Sep 2010, 09:08
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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?

The question only states that the sum of N's factors is even. Since ANY perfect square has factors which add up to an ODD number, N cannot be a perfect square. It doesn't matter that there are other, non-perfect-square numbers which have that same quality.
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Re: a perfect square [#permalink]  30 Sep 2010, 09:33
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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.
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Re: a perfect square [#permalink]  30 Sep 2010, 12:18
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.

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Re: a perfect square [#permalink]  30 Sep 2010, 12:21
TehJay 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?

The question only states that the sum of N's factors is even. Since ANY perfect square has factors which add up to an ODD number, N cannot be a perfect square. It doesn't matter that there are other, non-perfect-square numbers which have that same quality.

Don't know what was I thinking when I posted this... B clearly says that I should be considering only those numbers whose factors add up to give an even number... Thanks
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Re: a perfect square [#permalink]  30 Sep 2010, 12:28
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If you don't know the tricks about perfect squares, it will take you a little time to demonstrate :

(1) SUFFICIENT: The factors of any number N can be sorted into pairs that multiply to give N. (For instance, the factors of 24 can be paired as follows: 1 and 24; 2 and 12; 3 and 8; 4 and 6.) However, if N is a perfect square, one of these ‘pairs’ will consist of just one number: the square root of N. (For example, if N were 49, it would have the factor pair 7 × 7) Since all of the other factors can be paired off, it follows that if N is a perfect square, then N has an odd number of factors. (If N is not a perfect square, then all of its factors can be paired off, so it will have an even number of factors.) This statement then implies that N is not a perfect square.

(2) SUFFICIENT: Let N be a perfect square. If N is odd, then all factors of N are odd. Therefore, by the above reasoning, N has an odd number of odd factors, so their sum must be odd. If N is even, let M be the product of all the odd prime factors (as many times as they appear in N – not distinct) of N, which is also a perfect square. (For instance, if N = 100, then M =5 × 5 = 25.) Then the sum of factors of M is odd, by the above reasoning. Furthermore, all other factors of N (i.e., that don’t also divide M) are even. The sum total is thus odd + even = odd.
Therefore, the sum of the factors of any perfect square is odd, so this statement implies that N is not a perfect square.
This statement can also be investigated by trying several cases of even perfect squares (4, 16, 36, 64, 100), noting that in each case the sums of the factors are odd, and generalizing.

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Re: a perfect square [#permalink]  16 Oct 2010, 06:48
good question...learned some facts abt perfect squares...
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Re: a perfect square [#permalink]  08 Nov 2010, 09:32
Expert's post
Fijisurf wrote:
Barkatis wrote:
(2) SUFFICIENT: Let N be a perfect square. If N is odd, then all factors of N are odd. Therefore, by the above reasoning, N has an odd number of odd factors, so their sum must be odd. If N is even, let M be the product of all the odd prime factors (as many times as they appear in N – not distinct) of N, which is also a perfect square. (For instance, if N = 100, then M =5 × 5 = 25.) Then the sum of factors of M is odd, by the above reasoning. Furthermore, all other factors of N (i.e., that don’t also divide M) are even. The sum total is thus odd + even = odd.
Therefore, the sum of the factors of any perfect square is odd, so this statement implies that N is not a perfect square.
This statement can also be investigated by trying several cases of even perfect squares (4, 16, 36, 64, 100), noting that in each case the sums of the factors are odd, and generalizing.

I am having trouble understanding this reasoning. If we are interested in sum of factors, why are you talking about product of odd prime factors?
Thanks.

Take an example N = $$2^4 * 3^2 *5^4$$
If we write down its factors, we get: 1, 2, 3, 4, 5, 6, 8, 10, 12..... total 75 factors.
Lets just consider the odd factors i.e. factors made by $$3^2 *5^4$$ (including 1)
These will be a total of 3*5 = 15 odd factors.
When you add an odd number of odd factors, result will be an odd number.
The rest of the factors of N will be even i.e. they will include at least one 2. their sum will definitely be even because no matter how many even numbers you add, you always get an even result.
Adding an odd number to an even number, you will get an odd number. Hence, sum of all factors of a perfect square will always be odd.
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Re: a perfect square [#permalink]  08 Nov 2010, 10:17
VeritasPrepKarishma wrote:
Fijisurf wrote:
Barkatis wrote:
(2) SUFFICIENT: Let N be a perfect square. If N is odd, then all factors of N are odd. Therefore, by the above reasoning, N has an odd number of odd factors, so their sum must be odd. If N is even, let M be the product of all the odd prime factors (as many times as they appear in N – not distinct) of N, which is also a perfect square. (For instance, if N = 100, then M =5 × 5 = 25.) Then the sum of factors of M is odd, by the above reasoning. Furthermore, all other factors of N (i.e., that don’t also divide M) are even. The sum total is thus odd + even = odd.
Therefore, the sum of the factors of any perfect square is odd, so this statement implies that N is not a perfect square.
This statement can also be investigated by trying several cases of even perfect squares (4, 16, 36, 64, 100), noting that in each case the sums of the factors are odd, and generalizing.

I am having trouble understanding this reasoning. If we are interested in sum of factors, why are you talking about product of odd prime factors?
Thanks.

Take an example N = $$2^4 * 3^2 *5^4$$
If we write down its factors, we get: 1, 2, 3, 4, 5, 6, 8, 10, 12..... total 75 factors.
Lets just consider the odd factors i.e. factors made by $$3^2 *5^4$$ (including 1)
These will be a total of 3*5 = 15 odd factors.
When you add an odd number of odd factors, result will be an odd number.
The rest of the factors of N will be even i.e. they will include at least one 2. their sum will definitely be even because no matter how many even numbers you add, you always get an even result.
Adding an odd number to an even number, you will get an odd number. Hence, sum of all factors of a perfect square will always be odd.

I do understand now.
However still not sure why perfect squares have "EVEN number of Even-factors".
I understand that power of the only even prime factor "2" determines the number of even factors. So, if a number is a perfect square then all the powers will be even. Then by adding 1 to all powers we get odd numbers, which we multiply. Then the number of even factors will be odd.
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Re: a perfect square [#permalink]  08 Nov 2010, 10:30
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Fijisurf wrote:

I do understand now.
However still not sure why perfect squares have "EVEN number of Even-factors".
I understand that power of the only even prime factor "2" determines the number of even factors. So, if a number is a perfect square then all the powers will be even. Then by adding 1 to all powers we get odd numbers, which we multiply. Then the number of even factors will be odd.

OK. If you understand the theory above, will you agree that total number of factors of a perfect square will be odd (e.g. 75 above)? Also, all factors will be either even or odd. Will you also agree that total number of odd factors of perfect square will be odd? See the e.g. above $$3^2.5^4$$ will give 3*5 = 15 i.e. odd number of factors because powers are always even.
Then, the number of even factors should be Odd (75) - Odd(15) = Even(60).
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Re: a perfect square [#permalink]  08 Nov 2010, 11:19
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If you want the proper method:
We will obtain an even factor by multiplying all the 15 odd factors we obtained above by $$2 or 2^2 or 2^3 or 2^4$$.
Therefore, we will obtain 15*4 = 60 even factors. (We can take 2 in four ways)
If the number is a perfect square, power of 2 will always be even. (We do not use 4 + 1 here because to make it an even number, there has to be at least one 2.)
Therefore, even factors of a perfect square will always be even.
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Re: a perfect square [#permalink]  08 Nov 2010, 12:36
VeritasPrepKarishma wrote:
(We do not use 4 + 1 here because to make it an even number, there has to be at least one 2.)

I got it. The above phrase was the key: we do not add "1" to the power of 2, because we do not want to count factors with no twos.

Great. Thanks.
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Re: a perfect square [#permalink]  28 Jun 2011, 02:16
Bunuel your explanation is so comprehensive and amazing Thank you so much
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Re: a perfect square [#permalink]  19 Apr 2012, 12:08
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;

I think we should mention that the unwritten assumption is that we are only talking about positive factors. If we include -ve factors, the number of factors of every integer is always even.
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Re: a perfect square [#permalink]  19 Apr 2012, 20:26
Expert's post
glores1970 wrote:

I think we should mention that the unwritten assumption is that we are only talking about positive factors. If we include -ve factors, the number of factors of every integer is always even.

Factors are positive numbers only. On the exam if you have 'a has 4 factors', it means it has 4 positive factors.
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Re: Is the positive integer N a perfect square? (1) The number [#permalink]  02 Sep 2012, 21:45
Expert's post
Responding to a pm:

The concepts of both the statements are discussed in detail in this post: http://www.veritasprep.com/blog/2010/12 ... t-squares/

Get back if you still have doubts.
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Re: Is the positive integer N a perfect square? (1) The number   [#permalink] 02 Sep 2012, 21:45

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