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Re: If x and y are integers and x<y, is the greatest common factor (GCF) o [#permalink]

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28 Apr 2016, 03:08

1

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chetan2u wrote:

If x and y are integers and x<y, is the greatest common factor (GCF) of x and y greater than 1?

(1) x = 40! (2) y = 40! + 1

Modified version of Bunuel's Q OA after 2-3 replies

x < y. GCF(x,y) > 1?

St1: x = 40! --> Insufficient as we do not know the value of y

St2: y = 40! + 1 --> Sufficient 40! contains all the factors from 1 to 40. If all the integers from 1 to 40 are factors of 40!, the integers from 1 to 40 cannot be the factors of 40! + 1 Since x < y, x < 40! + 1 i.e x <= 40! x can contain all the factors upto 40! But will always be coprime with 40! + 1 i.e say x = 37 or x = 37*2. x is factor of 40! so it cannot be a factor of 40! + 1 Hence GCF(x, y) = 1

Re: If x and y are integers and x<y, is the greatest common factor (GCF) o [#permalink]

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29 Sep 2016, 05:29

chetan2u - I am unable to understand the solution for this problem. It seems the ans is wrongly mentioned as B while it should be C. Can you pls check & confirm.

The reason for my concern is, that would the same answer still holds if in option B, we change y= 5! + 1 just for the sake of easy calculation. Since x can be any integer less than y, let say x = 5 then gcf would be 5, while if x = 24 then gcf would be 1. Hence B is insufficient.

Re: If x and y are integers and x<y, is the greatest common factor (GCF) o [#permalink]

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29 Sep 2016, 05:42

nitswat wrote:

chetan2u - I am unable to understand the solution for this problem. It seems the ans is wrongly mentioned as B while it should be C. Can you pls check & confirm.

The reason for my concern is, that would the same answer still holds if in option B, we change y= 5! + 1 just for the sake of easy calculation. Since x can be any integer less than y, let say x = 5 then gcf would be 5, while if x = 24 then gcf would be 1. Hence B is insufficient.

Let me know your thoughts.

Thanks, Nitin

Pls share Kudos if you like my post

Hi Nitin,

The answer is B and the solution is explained above. If you can point out the portion that you were not able to understand I can try to answer your queries.

In your first case, y = 5! + 1 = 121 You have taken x = 5 GCF(5, 121) = 1

In the second case, y = 121 and x = 24 GCF(24, 121) = 1

Re: If x and y are integers and x<y, is the greatest common factor (GCF) o [#permalink]

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29 Sep 2016, 06:29

Vyshak wrote:

nitswat wrote:

chetan2u - I am unable to understand the solution for this problem. It seems the ans is wrongly mentioned as B while it should be C. Can you pls check & confirm.

The reason for my concern is, that would the same answer still holds if in option B, we change y= 5! + 1 just for the sake of easy calculation. Since x can be any integer less than y, let say x = 5 then gcf would be 5, while if x = 24 then gcf would be 1. Hence B is insufficient.

Let me know your thoughts.

Thanks, Nitin

Pls share Kudos if you like my post

Hi Nitin,

The answer is B and the solution is explained above. If you can point out the portion that you were not able to understand I can try to answer your queries.

In your first case, y = 5! + 1 = 121 You have taken x = 5 GCF(5, 121) = 1

In the second case, y = 121 and x = 24 GCF(24, 121) = 1

So B is sufficient

for simplicity,

y= 4! + 1 = 25 and if x<y, lets take x as 5. In this case GCF is 5. if x = 4, GCF is 1

On the other hand, in the question if it results in a prime number, the answer is always GCF=1

Re: If x and y are integers and x<y, is the greatest common factor (GCF) o [#permalink]

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29 Sep 2016, 07:12

nitswat wrote:

Vyshak, How about this one - y = 5! +1 = 121 and x = 11, then gcf (x,y) would be 11.

Yes you are right. But 5! + 1 is composite. I am not sure whether 40! + 1 is prime or composite. All I know is 40! + 1 can be written in the form of 6k + 1 and may be prime. The answer can be concluded as C if 40! + 1 is composite.

Re: If x and y are integers and x<y, is the greatest common factor (GCF) o [#permalink]

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29 Sep 2016, 07:25

Vyshak wrote:

nitswat wrote:

Vyshak, How about this one - y = 5! +1 = 121 and x = 11, then gcf (x,y) would be 11.

Yes you are right. But 5! + 1 is composite. I am not sure whether 40! + 1 is prime or composite. All I know is 40! + 1 can be written in the form of 6k + 1 and may be prime. The answer can be concluded as C if 40! + 1 is composite.

How were you able to conclude that it's of the form 6n+1? Besides,i don't think there is a way to conclude whether it's prime or composite.

Re: If x and y are integers and x<y, is the greatest common factor (GCF) o [#permalink]

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29 Sep 2016, 07:34

rahulkashyap wrote:

Vyshak wrote:

nitswat wrote:

Vyshak, How about this one - y = 5! +1 = 121 and x = 11, then gcf (x,y) would be 11.

Yes you are right. But 5! + 1 is composite. I am not sure whether 40! + 1 is prime or composite. All I know is 40! + 1 can be written in the form of 6k + 1 and may be prime. The answer can be concluded as C if 40! + 1 is composite.

How were you able to conclude that it's of the form 6n+1? Besides,i don't think there is a way to conclude whether it's prime or composite.

Posted from my mobile device

40! is divisible by 6 --> 40! = 6k --> 40! + 1 = 6k + 1

Re: If x and y are integers and x<y, is the greatest common factor (GCF) o [#permalink]

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16 Oct 2017, 08:44

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If x and y are integers and x<y, is the greatest common factor (GCF) o [#permalink]

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21 Oct 2017, 04:04

Vyshak correct me if I'm wrong. I think I know why the OA is C.

There is no indication in this problem that x or y are positive integers. Just that x<y

In the case of the Statement 2: If x is positive will have to be a factor of y! and therefore there won't be any common factor besides 1 BUT if x is negative it could be equal to: x= -1 * (40! + 1) yielding a gcf of (40! + 1) therefore insufficient

When analyzing both statements together we get only one possible answer that is gcf=1 since both positive integers are co-primes