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The sum of two number is 588 and their HCF is 49. How many

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The sum of two number is 588 and their HCF is 49. How many [#permalink] New post 07 Jun 2012, 11:44
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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
[Reveal] Spoiler: OA

Last edited by Bunuel on 08 Jun 2012, 05:34, edited 2 times in total.
Edited the question and added the OA
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Re: The sum of two number is 588 and their HCF is 49. How many [#permalink] New post 07 Jun 2012, 12:14
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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.

Answer: E.

P.S. Please read and follow: 11-rules-for-posting-133935.html (points 3 and 8).
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Re: The sum of two number is 588 and their HCF is 49. How many [#permalink] New post 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.

Answer: E.

P.S. Please read and follow: 11-rules-for-posting-133935.html (points 3 and 8).


Awesome explanation! I was wondering what difficulty level you found this question? This seems pretty complicated to me.
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Re: The sum of two number is 588 and their HCF is 49. How many [#permalink] New post 07 Jun 2012, 13:11
Expert's post
vandygrad11 wrote:
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.

Answer: E.

P.S. Please read and follow: 11-rules-for-posting-133935.html (points 3 and 8).


Awesome explanation! I was wondering what difficulty level you found this question? This seems pretty complicated to me.


I'd say it's about 650+ level question.
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Re: The sum of two number is 588 and their HCF is 49. How many [#permalink] New post 08 Jun 2012, 05:33
I choose the wrong option A. :oops:

The moment I clicked and saw the correct as E, I realised my mistake.

I hope to not do it in my test.

A simple and nice explanation Bunuel.
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Re: The sum of two number is 588 and their HCF is 49. How many [#permalink] New post 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.

Answer: E.

P.S. Please read and follow: 11-rules-for-posting-133935.html (points 3 and 8).


Hi,Can you please explain how you got the value 49x and 49y?if you give me an example that will be great...Thanks
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Re: The sum of two number is 588 and their HCF is 49. How many [#permalink] New post 08 Jun 2012, 07:53
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farukqmul wrote:
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.

Answer: E.

P.S. Please read and follow: 11-rules-for-posting-133935.html (points 3 and 8).


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.

Hope it's clear.
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Re: The sum of two number is 588 and their HCF is 49. How many [#permalink] New post 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.
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Re: The sum of two number is 588 and their HCF is 49. How many [#permalink] New post 28 Jul 2012, 04:53
Hi Bunuel,

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!
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Re: The sum of two number is 588 and their HCF is 49. How many [#permalink] New post 28 Jul 2012, 05:05
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NYC5648 wrote:
Hi Bunuel,

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.

Hope it's clear.
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Re: The sum of two number is 588 and their HCF is 49. How many [#permalink] New post 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.

I still don't get it.

Many thanks!

Great weekend!
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Re: The sum of two number is 588 and their HCF is 49. How many [#permalink] New post 28 Jul 2012, 06:25
I got it!! Thanks you Bunuel!
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Re: The sum of two number is 588 and their HCF is 49. How many [#permalink] New post 26 Jun 2013, 01:25
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The sum of two numbers is 588 and their HCF is 49. How many [#permalink] New post 06 Sep 2013, 13:11
The sum of two numbers is 588 and their HCF is 49. How many such pairs of numbers can be formed?

(a) 1
(b) 2
(c) 3
(d) 4
(e) 5


[Reveal] Spoiler:
My approach:

49 = 7^2 is the greatest common factor, thus both numbers must contain 7^2.

So, one pair would be: 49 and 539 (from 588-49). Both contain 7^2 if you prime factor them. The prime factorization of 539 is 7^2 x 11.

Another pair would be: 7^3 = 343 and 245 (from 588-343). The prime factorization of 245 is 7^2 x 5 and thus contains 7^2.

Since 7^4 exceeds 588, we can stop looking for further pairs.
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Re: The sum of two numbers is 588 and their HCF is 49. How many [#permalink] New post 06 Sep 2013, 13:28
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salsal wrote:
The sum of two numbers is 588 and their HCF is 49. How many such pairs of numbers can be formed?

(a) 1
(b) 2
(c) 3
(d) 4
(e) 5


[Reveal] Spoiler:
My approach:

49 = 7^2 is the greatest common factor, thus both numbers must contain 7^2.

So, one pair would be: 49 and 539 (from 588-49). Both contain 7^2 if you prime factor them. The prime factorization of 539 is 7^2 x 11.

Another pair would be: 7^3 = 343 and 245 (from 588-343). The prime factorization of 245 is 7^2 x 5 and thus contains 7^2.

Since 7^4 exceeds 588, we can stop looking for further pairs.


Merging similar topics. Please refer to the solutions above.
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Re: The sum of two number is 588 and their HCF is 49. How many [#permalink] New post 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.

Answer: E.

P.S. Please read and follow: 11-rules-for-posting-133935.html (points 3 and 8).



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?
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Re: The sum of two number is 588 and their HCF is 49. How many [#permalink] New post 08 Sep 2013, 05:20
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avaneeshvyas wrote:
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.

Answer: E.

P.S. Please read and follow: 11-rules-for-posting-133935.html (points 3 and 8).



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)?
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Re: The sum of two number is 588 and their HCF is 49. How many [#permalink] New post 28 Oct 2014, 13:09
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Re: The sum of two number is 588 and their HCF is 49. How many   [#permalink] 28 Oct 2014, 13:09
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