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M28-34

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M28-34  [#permalink]

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New post 16 Sep 2014, 01:30
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The greatest common divisor of two positive integers is 25. If the sum of the integers is 350, then how many such pairs are possible?

A. \(1\)
B. \(2\)
C. \(3\)
D. \(4\)
E. \(5\)

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Re M28-34  [#permalink]

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New post 16 Sep 2014, 01:30
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Official Solution:

The greatest common divisor of two positive integers is 25. If the sum of the integers is 350, then how many such pairs are possible?

A. \(1\)
B. \(2\)
C. \(3\)
D. \(4\)
E. \(5\)


We are told that the greatest common factor of two integers is 25. So, these integers are \(25x\) and \(25y\), 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 \(25x\) and \(25y\) will be more that 25.

Next, we know that \(25x+25y=350\). Reducing by 25 gives \(x+y=14\). Now, since \(x\) and \(y\) don't share any common factor but 1 then (x, y) can be only (1, 13), (3, 11) or (5, 9) (all other pairs (2, 12), (4, 10), (6, 8) and (7, 7) do share common factor greater than 1).

So, there are only three pairs of such numbers possible:

\(25*1=25\) and \(25*13=325\);

\(25*3=75\) and \(25*11=275\);

\(25*5=125\) and \(25*9=225\).


Answer: C
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Re: M28-34  [#permalink]

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New post 28 Nov 2014, 22:38
wouldnt there be 6 pairs possible (13, 1) and (1,13) (11,3) (3,11) (5,9) and (9,5)?

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New post 29 Nov 2014, 05:24
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Re: M28-34  [#permalink]

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New post 05 Dec 2016, 22:53
Bunuel,

If you consider (1,13) and (13,1) as 2 different pairs, then yes the answer will be 6.
I believe these can be treated as 2 different pairs because you are going to get 2 different values of the 2 numbers whose sum is 350.
Please correct if i am thinking in the wrong direction.
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Re: M28-34  [#permalink]

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New post 05 Dec 2016, 23:46
shubham1985 wrote:
Bunuel,

If you consider (1,13) and (13,1) as 2 different pairs, then yes the answer will be 6.
I believe these can be treated as 2 different pairs because you are going to get 2 different values of the 2 numbers whose sum is 350.
Please correct if i am thinking in the wrong direction.


It's clear from the solution that we do not consider (25, 325) different from (325, 25).
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Re: M28-34  [#permalink]

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New post 21 Jun 2017, 19:25
Isnt 200 and 150 another pair ? which would make the answer D ?
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Re: M28-34  [#permalink]

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New post 21 Jun 2017, 20:01
NJ3110 wrote:
Isnt 200 and 150 another pair ? which would make the answer D ?



NO
We are looking for Greatest Common Divisor as 25..
But what happens in pair of 200 and 150..
Both are div by 50, so GCD is 50 BUT we are looking for 25..
So out of the choices.

Hope it helps
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Re: M28-34  [#permalink]

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New post 22 Apr 2018, 21:24
and what about (175,175)? Isn't that should be the solution as well?
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New post 22 Apr 2018, 21:32
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Re: M28-34 &nbs [#permalink] 22 Apr 2018, 21:32
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