Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Re: Sum of the distinct factors of square root of a number [#permalink]
26 Nov 2010, 20:32

2

This post received KUDOS

Thanks!

(1) The only squares that have 3 distinct positive factors are 4, 25 and 49. The factors are 1,x,\(\sqrt{x}\). Since \(\sqrt{x}\) for the above three numbers needs to be considered, hence there will be three different values for the sums

Hence (1) is insufficient

(2) \(x^2-1 = 3k\) where k =odd integer

Hence, \(x^2 =3k+1\) is even

There can be multiple values where x^2 is even

Hence (2) is insufficient

(1)+(2) - x^2 should be even and should have 3 distinct positive factors The only value that satisfies both conditions is 4

Re: Sum of the distinct factors of square root of a number [#permalink]
27 Nov 2010, 11:16

4

This post received KUDOS

Expert's post

2

This post was BOOKMARKED

nades09 wrote:

Thanks!

(1) The only squares that have 3 distinct positive factors are 4, 25 and 49. The factors are 1,x,\(\sqrt{x}\). Since \(\sqrt{x}\) for the above three numbers needs to be considered, hence there will be three different values for the sums

Hence (1) is insufficient

(2) \(x^2-1 = 3k\) where k =odd integer

Hence, \(x^2 =3k+1\) is even

There can be multiple values where x^2 is even

Hence (2) is insufficient

(1)+(2) - x^2 should be even and should have 3 distinct positive factors The only value that satisfies both conditions is 4

\(\sqrt{x}\) is 2, hence we can find the sum

Ans: C

You got most of it. If you go through factors theory, it will help you understand that only a square of a prime number can have 3 factors. e.g. 4 or 25 or 49 or 121 or 169.. 1 , 2, 4 are factors of 4 1, 5, 25 are factors of 25 1, 7, 49 are factors of 49 1, 11, 121 are factors of 121 etc

So if \(x\) has 3 factors, \(\sqrt{x}\) must be prime.

Also from statement 2, \(x^2 = 3k + 1\) where k is odd. So 3k is odd and 3k + 1 is even. So \(x^2\) is even. Now, if \(x^2\) is even, x has to be even too (It is not possible that a power of an odd number becomes even. If x is odd, \(x^2, x^3\) etc all will be odd. If x is even, \(x^2, x^3\)etc all will be even.). Then \(\sqrt{x}\) must also be even.

The only number that is even and prime is 2. So \(\sqrt{x}\) must be 2. _________________

Re: If x is an integer, what is the sum of all distinct positive [#permalink]
12 Sep 2013, 08:26

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

If x is an integer, what is the sum of all distinct positive [#permalink]
25 Jul 2014, 05:52

VeritasPrepKarishma wrote:

nades09 wrote:

Thanks!

(1) The only squares that have 3 distinct positive factors are 4, 25 and 49. The factors are 1,x,\(\sqrt{x}\). Since \(\sqrt{x}\) for the above three numbers needs to be considered, hence there will be three different values for the sums

Hence (1) is insufficient

(2) \(x^2-1 = 3k\) where k =odd integer

Hence, \(x^2 =3k+1\) is even

There can be multiple values where x^2 is even

Hence (2) is insufficient

(1)+(2) - x^2 should be even and should have 3 distinct positive factors The only value that satisfies both conditions is 4

\(\sqrt{x}\) is 2, hence we can find the sum

Ans: C

You got most of it. If you go through factors theory, it will help you understand that only a square of a prime number can have 3 factors. e.g. 4 or 25 or 49 or 121 or 169.. 1 , 2, 4 are factors of 4 1, 5, 25 are factors of 25 1, 7, 49 are factors of 49 1, 11, 121 are factors of 121 etc

So if \(x\) has 3 factors, \(\sqrt{x}\) must be prime.

Also from statement 2, \(x^2 = 3k + 1\) where k is odd. So 3k is odd and 3k + 1 is even. So \(x^2\) is even. Now, if \(x^2\) is even, x has to be even too (It is not possible that a power of an odd number becomes even. If x is odd, \(x^2, x^3\) etc all will be odd. If x is even, \(x^2, x^3\)etc all will be even.). Then \(\sqrt{x}\) must also be even.

The only number that is even and prime is 2. So \(\sqrt{x}\) must be 2.

I thought statement 2 was sufficient as \(\sqrt{x}\) is supposed to be an integer as only integer can have factors . Based on this, statement 2 says product of 2 consecutive odd integers is odd (x-1)(x+1)= odd

1*3= 3 here x= 2 but \(\sqrt{2}\) is not an integer hence we cannot take x= 2

3*5 = 15 here x= 4 and \(\sqrt{4}\) is an integer hence we can take x= 4

5*7 = cannot take, as this is not in the form 3K

7*9=63 here x = 8 but \(\sqrt{8}\) is not an integer hence we cannot take x= 8

9*11=99 X=10 but \(\sqrt{10}\) is not an integer hence we cannot take x= 10

so from statement 2 after testing various numbers I felt that x = 4 was the only value that qualified hence B was the answer. Are there any values that I missed ? or \(\sqrt{x}\) need not be an integer?

If x is an integer, what is the sum of all distinct positive [#permalink]
25 Jul 2014, 09:40

1

This post received KUDOS

Quote:

I thought statement 2 was sufficient as \(\sqrt{x}\) is supposed to be an integer as only integer can have factors . Based on this, statement 2 says product of 2 consecutive odd integers is odd (x-1)(x+1)= odd

1*3= 3 here x= 2 but \(\sqrt{2}\) is not an integer hence we cannot take x= 2

3*5 = 15 here x= 4 and \(\sqrt{4}\) is an integer hence we can take x= 4

5*7 = cannot take, as this is not in the form 3K

7*9=63 here x = 8 but \(\sqrt{8}\) is not an integer hence we cannot take x= 8

9*11=99 X=10 but \(\sqrt{10}\) is not an integer hence we cannot take x= 10

so from statement 2 after testing various numbers I felt that x = 4 was the only value that qualified hence B was the answer. Are there any values that I missed ? or \(\sqrt{x}\) need not be an integer?

Factors are not essentially Integers otherwise the expressions of "HCF (Highest Common Factor) of Fractions" would not have existed.

In order to ensure that x is a perfect square we will have to combine the information of first statement with second statement. I hope this will give you an insight why answer option C is the correct answer. _________________

If x is an integer, what is the sum of all distinct positive [#permalink]
25 Jul 2014, 15:12

GMATinsight wrote:

Quote:

I thought statement 2 was sufficient as \(\sqrt{x}\) is supposed to be an integer as only integer can have factors . Based on this, statement 2 says product of 2 consecutive odd integers is odd (x-1)(x+1)= odd

1*3= 3 here x= 2 but \(\sqrt{2}\) is not an integer hence we cannot take x= 2

3*5 = 15 here x= 4 and \(\sqrt{4}\) is an integer hence we can take x= 4

5*7 = cannot take, as this is not in the form 3K

7*9=63 here x = 8 but \(\sqrt{8}\) is not an integer hence we cannot take x= 8

9*11=99 X=10 but \(\sqrt{10}\) is not an integer hence we cannot take x= 10

so from statement 2 after testing various numbers I felt that x = 4 was the only value that qualified hence B was the answer. Are there any values that I missed ? or \(\sqrt{x}\) need not be an integer?

Factors are not essentially Integers otherwise the expressions of "HCF (Highest Common Factor) of Fractions" would not have existed.

In order to ensure that x is a perfect square we will have to combine the information of first statement with second statement. I hope this will give you an insight why answer option C is the correct answer.

well if non integers can have factors then the answer is C , Karishma if you can add little more to this that would surely help.

Last edited by qlx on 26 Jul 2014, 01:36, edited 1 time in total.

Re: If x is an integer, what is the sum of all distinct positive [#permalink]
25 Jul 2014, 15:32

Expert's post

qlx wrote:

VeritasPrepKarishma wrote:

nades09 wrote:

Thanks!

(1) The only squares that have 3 distinct positive factors are 4, 25 and 49. The factors are 1,x,\(\sqrt{x}\). Since \(\sqrt{x}\) for the above three numbers needs to be considered, hence there will be three different values for the sums

Hence (1) is insufficient

(2) \(x^2-1 = 3k\) where k =odd integer

Hence, \(x^2 =3k+1\) is even

There can be multiple values where x^2 is even

Hence (2) is insufficient

(1)+(2) - x^2 should be even and should have 3 distinct positive factors The only value that satisfies both conditions is 4

\(\sqrt{x}\) is 2, hence we can find the sum

Ans: C

You got most of it. If you go through factors theory, it will help you understand that only a square of a prime number can have 3 factors. e.g. 4 or 25 or 49 or 121 or 169.. 1 , 2, 4 are factors of 4 1, 5, 25 are factors of 25 1, 7, 49 are factors of 49 1, 11, 121 are factors of 121 etc

So if \(x\) has 3 factors, \(\sqrt{x}\) must be prime.

Also from statement 2, \(x^2 = 3k + 1\) where k is odd. So 3k is odd and 3k + 1 is even. So \(x^2\) is even. Now, if \(x^2\) is even, x has to be even too (It is not possible that a power of an odd number becomes even. If x is odd, \(x^2, x^3\) etc all will be odd. If x is even, \(x^2, x^3\)etc all will be even.). Then \(\sqrt{x}\) must also be even.

The only number that is even and prime is 2. So \(\sqrt{x}\) must be 2.

I thought statement 2 was sufficient as \(\sqrt{x}\) is supposed to be an integer as only integer can have factors . Based on this, statement 2 says product of 2 consecutive odd integers is odd (x-1)(x+1)= odd

1*3= 3 here x= 2 but \(\sqrt{2}\) is not an integer hence we cannot take x= 2

3*5 = 15 here x= 4 and \(\sqrt{4}\) is an integer hence we can take x= 4

5*7 = cannot take, as this is not in the form 3K

7*9=63 here x = 8 but \(\sqrt{8}\) is not an integer hence we cannot take x= 8

9*11=99 X=10 but \(\sqrt{10}\) is not an integer hence we cannot take x= 10

so from statement 2 after testing various numbers I felt that x = 4 was the only value that qualified hence B was the answer. Are there any values that I missed ? or \(\sqrt{x}\) need not be an integer?

Here is another value: \(x^2 = 3*85 + 1 = 256\) --> \(x = 16\) --> \(\sqrt{x}=4\). _________________

Re: If x is an integer, what is the sum of all distinct positive [#permalink]
25 Jul 2014, 15:34

1

This post received KUDOS

Expert's post

GMATinsight wrote:

Quote:

I thought statement 2 was sufficient as \(\sqrt{x}\) is supposed to be an integer as only integer can have factors . Based on this, statement 2 says product of 2 consecutive odd integers is odd (x-1)(x+1)= odd

1*3= 3 here x= 2 but \(\sqrt{2}\) is not an integer hence we cannot take x= 2

3*5 = 15 here x= 4 and \(\sqrt{4}\) is an integer hence we can take x= 4

5*7 = cannot take, as this is not in the form 3K

7*9=63 here x = 8 but \(\sqrt{8}\) is not an integer hence we cannot take x= 8

9*11=99 X=10 but \(\sqrt{10}\) is not an integer hence we cannot take x= 10

so from statement 2 after testing various numbers I felt that x = 4 was the only value that qualified hence B was the answer. Are there any values that I missed ? or \(\sqrt{x}\) need not be an integer?

Factors are not essentially Integers otherwise the expressions of "HCF (Highest Common Factor) of Fractions" would not have existed.

In order to ensure that x is a perfect square we will have to combine the information of first statement with second statement. I hope this will give you an insight why answer option C is the correct answer.

That's not true for the GMAT. Only positive integers are considered as factors and only integers could have factors on the GMAT. _________________

If x is an integer, what is the sum of all distinct positive [#permalink]
26 Jul 2014, 01:25

Bunuel wrote:

GMATinsight wrote:

Quote:

I thought statement 2 was sufficient as \(\sqrt{x}\) is supposed to be an integer as only integer can have factors . Based on this, statement 2 says product of 2 consecutive odd integers is odd (x-1)(x+1)= odd

1*3= 3 here x= 2 but \(\sqrt{2}\) is not an integer hence we cannot take x= 2

3*5 = 15 here x= 4 and \(\sqrt{4}\) is an integer hence we can take x= 4

5*7 = cannot take, as this is not in the form 3K

7*9=63 here x = 8 but \(\sqrt{8}\) is not an integer hence we cannot take x= 8

9*11=99 X=10 but \(\sqrt{10}\) is not an integer hence we cannot take x= 10

so from statement 2 after testing various numbers I felt that x = 4 was the only value that qualified hence B was the answer. Are there any values that I missed ? or \(\sqrt{x}\) need not be an integer?

Factors are not essentially Integers otherwise the expressions of "HCF (Highest Common Factor) of Fractions" would not have existed.

In order to ensure that x is a perfect square we will have to combine the information of first statement with second statement. I hope this will give you an insight why answer option C is the correct answer.

That's not true for the GMAT. Only positive integers are considered as factors and only integers could have factors on the GMAT.

Thanks for clearing that, should have tested few more values

Few more values that qualify for B 15*17 = 3.K ( for some positive integer K) here x= 16 and \(\sqrt {16} =\) 4 sum of factors =7

63.65= 3.K here x = 64, \(\sqrt {64}\) = 8

99.101= 3.k here X= 100, \(\sqrt {100}\) =10 etc .

If x is an integer, what is the sum of all distinct positive [#permalink]
27 Jul 2014, 20:26

Expert's post

qlx wrote:

I thought statement 2 was sufficient as \(\sqrt{x}\) is supposed to be an integer as only integer can have factors . Based on this, statement 2 says product of 2 consecutive odd integers is odd (x-1)(x+1)= odd

1*3= 3 here x= 2 but \(\sqrt{2}\) is not an integer hence we cannot take x= 2

3*5 = 15 here x= 4 and \(\sqrt{4}\) is an integer hence we can take x= 4

5*7 = cannot take, as this is not in the form 3K

7*9=63 here x = 8 but \(\sqrt{8}\) is not an integer hence we cannot take x= 8

9*11=99 X=10 but \(\sqrt{10}\) is not an integer hence we cannot take x= 10

so from statement 2 after testing various numbers I felt that x = 4 was the only value that qualified hence B was the answer. Are there any values that I missed ? or \(\sqrt{x}\) need not be an integer?

It is very hard to prove something by just testing numbers. You may not have tested the right numbers or you may not have tested enough numbers. To prove something, use of logic is required. To disprove, it is much easier to use numbers since you only have to find one example where it doesn't hold.

From statement 2, you have [fraction]x^2 = 3x + 1[/fraction] You need an even perfect square which is 1 more than a multiple of 3. But since \(\sqrt{x}\) must be an integer too, we are looking for an even fourth power which is 1 more than a multiple of 3.

An even fourth power is \(16 ( = 2^4)\) which is 1 more than 15. \(\sqrt{x} = 2\) here. Another even fourth power is \(256 ( = 2^8)\) which is 1 more than 255 (multiple of 3). \(\sqrt{x} = 4\) here. Another even fourth power is \(4096 = 2^{12}\) which is 1 more than 4095 (multiple of 3). \(\sqrt{x} = 8\) here.

Hey, Last week I started a few new things in my life. That includes shifting from daily targets to weekly targets, 45 minutes of exercise including 15 minutes of yoga, making...

This week went in reviewing all the topics that I have covered in my previous study session. I reviewed all the notes that I have made and started reviewing the Quant...

I started running as a cross country team member since highshcool and what’s really awesome about running is that...you never get bored of it! I participated in...