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 500,000 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:

The sum of two numbers is 588 and their HCF is 49. How many such pairs of numbers can be formed?

A 6 B. 5 C. 4 D. 3 E. 2

The question should read:

The sum of two positive integers is 588 and their greatest common factor is 49. How many such pairs of numbers can be formed?

We are told that the greatest common factor of two integers is 49. So, these integers are \(49x\) and \(49y\), for some positive integers \(x\) and \(y\). Notice that \(x\) and \(y\) must not share any common factor but 1, because if they do then GCF of \(49x\) and \(49y\) will be more that 49.

Next, we know that \(49x+49y=588\) --> \(x+y=12\) --> since \(x\) and \(y\) don't share any common factor but 1 then (x, y) can be only (1, 11) and (5, 7) (all other pairs (2, 10), (3, 9), (4, 8), (6, 6) do share common factor greater than 1). So, there are only two pairs of such numbers possible: 49*1=49 and 49*11=539 AND 49*5=245 and 49*7=343.

Re: The sum of two number is 588 and their HCF is 49. How many [#permalink]

Show Tags

07 Jun 2012, 13:05

Bunuel wrote:

NYC5648 wrote:

The sum of two numbers is 588 and their HCF is 49. How many such pairs of numbers can be formed?

A 6 B. 5 C. 4 D. 3 E. 2

The question should read:

The sum of two positive integers is 588 and their greatest common factor is 49. How many such pairs of numbers can be formed?

We are told that the greatest common factor of two integers is 49. So, these integers are \(49x\) and \(49y\), for some positive integers \(x\) and \(y\). Notice that \(x\) and \(y\) must not share any common factor but 1, because if they do then GCF of \(49x\) and \(49y\) will be more that 49.

Next, we know that \(49x+49y=588\) --> \(x+y=12\) --> since \(x\) and \(y\) don't share any common factor but 1 then (x, y) can be only (1, 11) and (5, 7) (all other pairs (2, 10), (3, 9), (4, 8), (6, 6) do share common factor greater than 1). So, there are only two pairs of such numbers possible: 49*1=49 and 49*11=539 AND 49*5=245 and 49*7=343.

The sum of two numbers is 588 and their HCF is 49. How many such pairs of numbers can be formed?

A 6 B. 5 C. 4 D. 3 E. 2

The question should read:

The sum of two positive integers is 588 and their greatest common factor is 49. How many such pairs of numbers can be formed?

We are told that the greatest common factor of two integers is 49. So, these integers are \(49x\) and \(49y\), for some positive integers \(x\) and \(y\). Notice that \(x\) and \(y\) must not share any common factor but 1, because if they do then GCF of \(49x\) and \(49y\) will be more that 49.

Next, we know that \(49x+49y=588\) --> \(x+y=12\) --> since \(x\) and \(y\) don't share any common factor but 1 then (x, y) can be only (1, 11) and (5, 7) (all other pairs (2, 10), (3, 9), (4, 8), (6, 6) do share common factor greater than 1). So, there are only two pairs of such numbers possible: 49*1=49 and 49*11=539 AND 49*5=245 and 49*7=343.

Re: The sum of two number is 588 and their HCF is 49. How many [#permalink]

Show Tags

08 Jun 2012, 07:49

Bunuel wrote:

NYC5648 wrote:

The sum of two numbers is 588 and their HCF is 49. How many such pairs of numbers can be formed?

A 6 B. 5 C. 4 D. 3 E. 2

The question should read:

The sum of two positive integers is 588 and their greatest common factor is 49. How many such pairs of numbers can be formed?

We are told that the greatest common factor of two integers is 49. So, these integers are \(49x\) and \(49y\), for some positive integers \(x\) and \(y\). Notice that \(x\) and \(y\) must not share any common factor but 1, because if they do then GCF of \(49x\) and \(49y\) will be more that 49.

Next, we know that \(49x+49y=588\) --> \(x+y=12\) --> since \(x\) and \(y\) don't share any common factor but 1 then (x, y) can be only (1, 11) and (5, 7) (all other pairs (2, 10), (3, 9), (4, 8), (6, 6) do share common factor greater than 1). So, there are only two pairs of such numbers possible: 49*1=49 and 49*11=539 AND 49*5=245 and 49*7=343.

The sum of two numbers is 588 and their HCF is 49. How many such pairs of numbers can be formed?

A 6 B. 5 C. 4 D. 3 E. 2

The question should read:

The sum of two positive integers is 588 and their greatest common factor is 49. How many such pairs of numbers can be formed?

We are told that the greatest common factor of two integers is 49. So, these integers are \(49x\) and \(49y\), for some positive integers \(x\) and \(y\). Notice that \(x\) and \(y\) must not share any common factor but 1, because if they do then GCF of \(49x\) and \(49y\) will be more that 49.

Next, we know that \(49x+49y=588\) --> \(x+y=12\) --> since \(x\) and \(y\) don't share any common factor but 1 then (x, y) can be only (1, 11) and (5, 7) (all other pairs (2, 10), (3, 9), (4, 8), (6, 6) do share common factor greater than 1). So, there are only two pairs of such numbers possible: 49*1=49 and 49*11=539 AND 49*5=245 and 49*7=343.

Hi,Can you please explain how you got the value 49x and 49y?if you give me an example that will be great...Thanks

Since 49 is the greatest common factor of two integers, then 49 is a factor of both, so we can express one of them as 49x and another as 49y. For example one of them can be 49x=49*1=49 or 49*x=49*5=245.

Re: The sum of two number is 588 and their HCF is 49. How many [#permalink]

Show Tags

14 Jun 2012, 23:15

@ farukqmul: To answer your question, 49x+49y: We know that the GCF of the two number x and y is 49, which means that these number definitely have 49 common in both, thus, it translates to that they are definitely multiples of 49. Thus, x+y=588 can be rewritten as 49x + 49y = 588, I hope I have answered your question.

could you please explain why (1,11) and (5,7) are the only possible pairs.

I still don't get it.

Many thanks!

Great weekend!

Because if \(x\) and \(y\) are for example 2 and 10, respectively, the greatest common factor of \(49x=98\) and \(49y=490\) will be 98, so more than 49. That's because \(x\) and \(y\), in this case, share a common factor greater than 1. Only if \(x\) and \(y\) are 1 and 11 OR 5 and 7 (so only when \(x\) and \(y\) do not share any common factor more than 1) the greatest common factor of \(49x\) and \(49y\) will be 49.

Re: The sum of two number is 588 and their HCF is 49. How many [#permalink]

Show Tags

28 Jul 2012, 05:10

To answer your question why only these two pairs are possible:

we know that 49 is the GCF of two numbers and their sum is 588, so let it be x+y = 588. Further, given that their GCF is 49, they can be written as 49x + 49y = 588.

In other words, x+y = 12. Now the important thing is that x and y must be selected in such a way that their should be no additional factor, otherwise, the GCF will not remain 49. For example, let x and y be 4 and 8..sum being 12 but the GCF of the numbers (49*4 and 49*8) will be more than 49, hence you cannot take this pair. and if you write down the pairs then only 1,11 and 5,7 satisfy as they do not have any factor in common.

I hope this helps.

NYC5648 wrote:

Hi Bunuel,

could you please explain why (1,11) and (5,7) are the only possible pairs.

Re: The sum of two number is 588 and their HCF is 49. How many [#permalink]

Show Tags

07 Sep 2013, 23:24

Bunuel wrote:

NYC5648 wrote:

The sum of two numbers is 588 and their HCF is 49. How many such pairs of numbers can be formed?

A 6 B. 5 C. 4 D. 3 E. 2

The question should read:

The sum of two positive integers is 588 and their greatest common factor is 49. How many such pairs of numbers can be formed?

We are told that the greatest common factor of two integers is 49. So, these integers are \(49x\) and \(49y\), for some positive integers \(x\) and \(y\). Notice that \(x\) and \(y\) must not share any common factor but 1, because if they do then GCF of \(49x\) and \(49y\) will be more that 49.

Next, we know that \(49x+49y=588\) --> \(x+y=12\) --> since \(x\) and \(y\) don't share any common factor but 1 then (x, y) can be only (1, 11) and (5, 7) (all other pairs (2, 10), (3, 9), (4, 8), (6, 6) do share common factor greater than 1). So, there are only two pairs of such numbers possible: 49*1=49 and 49*11=539 AND 49*5=245 and 49*7=343.

Hi Bunuel.... aren't we supposed to count the two pairs (1,11) and (5,7) twice? "unique pair" is not mentioned in the question. Aren't we supposed to mark C(4 pairs) as the right choice?

The sum of two numbers is 588 and their HCF is 49. How many such pairs of numbers can be formed?

A 6 B. 5 C. 4 D. 3 E. 2

The question should read:

The sum of two positive integers is 588 and their greatest common factor is 49. How many such pairs of numbers can be formed?

We are told that the greatest common factor of two integers is 49. So, these integers are \(49x\) and \(49y\), for some positive integers \(x\) and \(y\). Notice that \(x\) and \(y\) must not share any common factor but 1, because if they do then GCF of \(49x\) and \(49y\) will be more that 49.

Next, we know that \(49x+49y=588\) --> \(x+y=12\) --> since \(x\) and \(y\) don't share any common factor but 1 then (x, y) can be only (1, 11) and (5, 7) (all other pairs (2, 10), (3, 9), (4, 8), (6, 6) do share common factor greater than 1). So, there are only two pairs of such numbers possible: 49*1=49 and 49*11=539 AND 49*5=245 and 49*7=343.

Hi Bunuel.... aren't we supposed to count the two pairs (1,11) and (5,7) twice? "unique pair" is not mentioned in the question. Aren't we supposed to mark C(4 pairs) as the right choice?

How is (49, 539) different from (539, 49)?
_________________

Re: The sum of two number is 588 and their HCF is 49. How many [#permalink]

Show Tags

28 Oct 2014, 13:09

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.
_________________

Re: The sum of two number is 588 and their HCF is 49. How many [#permalink]

Show Tags

31 Oct 2015, 14:17

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.
_________________

Happy New Year everyone! Before I get started on this post, and well, restarted on this blog in general, I wanted to mention something. For the past several months...

It’s quickly approaching two years since I last wrote anything on this blog. A lot has happened since then. When I last posted, I had just gotten back from...

Happy 2017! Here is another update, 7 months later. With this pace I might add only one more post before the end of the GSB! However, I promised that...