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Re: Sum of the distinct factors of square root of a number [#permalink]

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26 Nov 2010, 21:32

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

(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]

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12 Sep 2013, 09:26

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If x is an integer, what is the sum of all distinct positive [#permalink]

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25 Jul 2014, 06:52

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?

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]

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25 Jul 2014, 16:12

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, 02:36, edited 1 time in total.

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

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]

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26 Jul 2014, 02:25

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 .

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.

Re: If x is an integer, what is the sum of all distinct positive [#permalink]

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01 Oct 2015, 16:49

Hello from the GMAT Club BumpBot!

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