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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

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]

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07 Jun 2012, 14:05

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, 08:49

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]

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15 Jun 2012, 00:15

@ 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]

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28 Jul 2012, 06:10

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]

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08 Sep 2013, 00:24

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]

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31 Oct 2015, 15:17

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