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

Question

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

Bunuel · Accepted Answer

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).

geometric · Answer

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

Bunuel · Answer

I'd say it's about 650+ level question.

manulath · Answer

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.

farukqmul · Answer

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

Bunuel · Answer

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.

pavanpuneet · Answer

@ 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.

NYC5648 · Answer

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!

Bunuel · Answer

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.

pavanpuneet · Answer

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 · Answer

I got it!! Thanks you Bunuel!

Bunuel · Answer

Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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salsal · Answer

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

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.

Bunuel · Answer

Merging similar topics. Please refer to the solutions above.

avaneeshvyas · Answer

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?

Bunuel · Answer

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

samarpan.g28 · Answer

I took the two numbers as x and y.
x+y=588 and,
HCF(x,y)=49.
When I started solving, I thought that if x is an even multiple of 49, then y also becomes even since 588 is even. Then HCF(x,y) will not be just 49 but there may be one or more numbers of 2s or other prime numbers. However, we can still try if x is an odd multiple of 49.
At first, I took x=49 and y=539. The sum is 588 and HCF=49.
The next which followed was x=245 and y=343. HCF is 49.
After this, x is 343 which means it repeats. So, 2 pairs (49,539) and (245,343). Option (E) is correct.

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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

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