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Determining factorizations of a number [#permalink]

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12 May 2010, 13:11

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83% (01:57) correct
17% (01:15) wrong based on 1 sessions

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Now, this might be very easy to a lot of people, but it's one of the things that really slows me down and I need to figure out a way to speed this up. For example.

A, B, and C are positive integers greater than 1. If A<B<C and abc = 286, what is c-b?

What you must do is break up 286 into it's prime factorizations, in this case, 2, 11, and 13. At that point it should be very easy to solve the problem. I can very easily determine that 286 is divisible by 2, yet figuring out that 143 is 11 x 13 is my problem; it's a large, unwieldy number that poses a significant time sync for me. BTW, the answer to the question is 2.

For figuring out larger multiples such as that, do you guys have any tips or tricks?

Re: Determining factorizations of a number [#permalink]

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12 May 2010, 14:24

3

This post received KUDOS

Dartastic wrote:

Now, this might be very easy to a lot of people, but it's one of the things that really slows me down and I need to figure out a way to speed this up. For example.

A, B, and C are positive integers greater than 1. If A<B<C and abc = 286, what is c-b?

What you must do is break up 286 into it's prime factorizations, in this case, 2, 11, and 13. At that point it should be very easy to solve the problem. I can very easily determine that 286 is divisible by 2, yet figuring out that 143 is 11 x 13 is my problem; it's a large, unwieldy number that poses a significant time sync for me. BTW, the answer to the question is 2.

For figuring out larger multiples such as that, do you guys have any tips or tricks?

Factorization is the quickest way..!! To find the factor of a given (no. in this case 143) it's important to know whether a given no.(143) is a prime no..??

If it's a prime no..then there can be only two factors 1 and the number itself.. otherwise look for the factors...

Check for prime.

Find the approximate square root of the number (143)... the square root should lie between 11 and 12..!! as \(11^2\) is 121 and \(12^2\) is 144... then look for all the prime numbers less than 11.. in this case 2,3,5,7,11 and check whether the given no. (143) is divisible by any of these (2,3,5,7,11)..If it's divisible..then the no. is not prime..!! and any one or more of these numbers are its factors..

BUT, if the number is not divisible then the number is prime...!! so 143 is divisible by 11...

In this way, we don't blindly look for all the factors of a given number..

Take example of 149... Its square root should lie between 12 and 13.. Try dividing 149 by all the prime numbers less than 12 >>>> 2,3,5,7,11... and u will find it's not divisible by any of these prime numbers...So 143 is prime..

Re: Determining factorizations of a number [#permalink]

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12 May 2010, 15:49

nverma wrote:

Dartastic wrote:

Now, this might be very easy to a lot of people, but it's one of the things that really slows me down and I need to figure out a way to speed this up. For example.

A, B, and C are positive integers greater than 1. If A<B<C and abc = 286, what is c-b?

What you must do is break up 286 into it's prime factorizations, in this case, 2, 11, and 13. At that point it should be very easy to solve the problem. I can very easily determine that 286 is divisible by 2, yet figuring out that 143 is 11 x 13 is my problem; it's a large, unwieldy number that poses a significant time sync for me. BTW, the answer to the question is 2.

For figuring out larger multiples such as that, do you guys have any tips or tricks?

Factorization is the quickest way..!! To find the factor of a given (no. in this case 143) it's important to know whether a given no.(143) is a prime no..??

If it's a prime no..then there can be only two factors 1 and the number itself.. otherwise look for the factors...

Check for prime.

Find the approximate square root of the number (143)... the square root should lie between 11 and 12..!! as \(11^2\) is 121 and \(12^2\) is 144... then look for all the prime numbers less than 11.. in this case 2,3,5,7,11 and check whether the given no. (143) is divisible by any of these (2,3,5,7,11)..If it's divisible..then the no. is not prime..!! and any one or more of these numbers are its factors..

BUT, if the number is not divisible then the number is prime...!! so 143 is divisible by 11...

In this way, we don't blindly look for all the factors of a given number..

Take example of 149... Its square root should lie between 12 and 13.. Try dividing 149 by all the prime numbers less than 12 >>>> 2,3,5,7,11... and u will find it's not divisible by any of these prime numbers...So 143 is prime..

Re: Determining factorizations of a number [#permalink]

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14 May 2010, 13:57

what i would suggest you is to do lot of math questions so that will definately help you in remembering most common factors , roots LCM of few no. ,avg etc, and that will help you to quickly solve these sort of problems.

Re: Determining factorizations of a number [#permalink]

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21 May 2010, 04:27

nverma wrote:

Take example of 149... Its square root should lie between 12 and 13.. Try dividing 149 by all the prime numbers less than 12 >>>> 2,3,5,7,11... and u will find it's not divisible by any of these prime numbers...So 143 is prime..

Hope it helps..!!

Typos, instead of 143 , you probably have meant 149.

Re: Determining factorizations of a number [#permalink]

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10 Jun 2010, 08:53

Dartastic wrote:

Now, this might be very easy to a lot of people, but it's one of the things that really slows me down and I need to figure out a way to speed this up. For example.

A, B, and C are positive integers greater than 1. If A<B<C and abc = 286, what is c-b?

What you must do is break up 286 into it's prime factorizations, in this case, 2, 11, and 13. At that point it should be very easy to solve the problem. I can very easily determine that 286 is divisible by 2, yet figuring out that 143 is 11 x 13 is my problem; it's a large, unwieldy number that poses a significant time sync for me. BTW, the answer to the question is 2.

For figuring out larger multiples such as that, do you guys have any tips or tricks?

Knowing the tests for divisibility for the basic primes (2, 3, 5, 7, 11) would also help. Take 143 for example, numbers divisible by 11 have alternate digits adding to be the same (or sums differ by a multiple of 11) here: 1+3=4. So keep dividing it up, use a<b<c and voila, you have a, b, c.

The advantage with GMAT quant is that it wont give numbers that are too hard to factorize, nor multiples of primes 13 and above.

Re: Determining factorizations of a number [#permalink]

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10 Jun 2010, 11:22

nverma wrote:

Dartastic wrote:

Now, this might be very easy to a lot of people, but it's one of the things that really slows me down and I need to figure out a way to speed this up. For example.

A, B, and C are positive integers greater than 1. If A<B<C and abc = 286, what is c-b?

What you must do is break up 286 into it's prime factorizations, in this case, 2, 11, and 13. At that point it should be very easy to solve the problem. I can very easily determine that 286 is divisible by 2, yet figuring out that 143 is 11 x 13 is my problem; it's a large, unwieldy number that poses a significant time sync for me. BTW, the answer to the question is 2.

For figuring out larger multiples such as that, do you guys have any tips or tricks?

Factorization is the quickest way..!! To find the factor of a given (no. in this case 143) it's important to know whether a given no.(143) is a prime no..??

If it's a prime no..then there can be only two factors 1 and the number itself.. otherwise look for the factors...

Check for prime.

Find the approximate square root of the number (143)... the square root should lie between 11 and 12..!! as \(11^2\) is 121 and \(12^2\) is 144... then look for all the prime numbers less than 11.. in this case 2,3,5,7,11 and check whether the given no. (143) is divisible by any of these (2,3,5,7,11)..If it's divisible..then the no. is not prime..!! and any one or more of these numbers are its factors..

BUT, if the number is not divisible then the number is prime...!! so 143 is divisible by 11...

In this way, we don't blindly look for all the factors of a given number..

Take example of 149... Its square root should lie between 12 and 13.. Try dividing 149 by all the prime numbers less than 12 >>>> 2,3,5,7,11... and u will find it's not divisible by any of these prime numbers...So 143 is prime..

Hope it helps..!!

Nice. Another point in the problem is that it specifically states that three numbers are all greater than 1. So when you get to 143 and 2, you know that 143 cannot be prime.

gmatclubot

Re: Determining factorizations of a number
[#permalink]
10 Jun 2010, 11:22