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Here I'm not sure that the answer is C because is true that we need of both statement to find possible values for X and Y. Infact statement 1 and 2 we do not have values for the variables (can be everything).

But it seems to be a trap answer......

A very simple way to solve this problem will be:

1. 2x is even -> even +y=odd -> this mean y is odd -> even and odd GCD is 1 -> sufficient. 2. 5x-3y=1 -> 2 options: - 5x is even -> [odd -3y=odd] and [x is even] -> 3y must be odd -> y must be odd -> X even & Y odd -> even and odd GCD is 1 - 5x is odd -> [odd-3y=odd] and [x is odd] -> 3y must be even -> Y must be even -> X is odd & Y is even -> even and odd GCD is 1

Re: If x and y are positive integers, what is the greatest [#permalink]

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20 Jun 2016, 05:12

manishkhare wrote:

If x and y are positive integers, what is the greatest common divisor of x and y?

1. 2x + y = 73 2. 5x – 3y = 1

Another way of solving it .Both x and y are integers . From 1: x =(73-y)/2 .Since x is a integer it implies 73-y =even number .73 is Odd so y is also Odd .X is even so GCM will be 1.Sufficient

From 2 :5x-3y =1 .They are consecutive numbers i.e .odd-even or even -odd .so the GCM in this case =1 .Sufficient

Option D is correct .

Press Kudos if you like the solution.

just a general thing .y is odd and x is even take y =15 and x=30 .GCD N.E. 1

so there's gotta be other approach just by saying y is odd and x even won't get GCD=1 always

i guess if this type of question encounters u better skip taking a hard guess .Dont waste time(BTW gmat won't give this type of problem involving so much calculations.the paper always play with tricks which you have to find out)

Re: If x and y are positive integers, what is the greatest [#permalink]

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02 Apr 2017, 02:30

Bunuel wrote:

carcass wrote:

Thanks bunuel.

From your explanation can we suppose to apply the reasoning from statement two to the first one or is not possible ?? would be a big mistake because 1 is NOT prime??'

I suppose you mean whether we can apply the reasoning from (1) to statement (2). Yes, we can:

(2) \(5x-3y=1\) --> Suppose GCD(x, y) is some integer \(d\), then \(x=md\) and \(y=nd\), for some positive integers \(m\) and \(n\). So, we'll have \(5(md)-3(nd)=d(5m-3n)=1\) --> \(d\) is a factor of 1, so \(d\) must equal 1. Sufficient.

Bunuel one silly question but please help with this :

Through the above logic for point 2 ,how can we be sure that d =1 .What about d= 1/(5M-3N) and that being another value apart from 1

From your explanation can we suppose to apply the reasoning from statement two to the first one or is not possible ?? would be a big mistake because 1 is NOT prime??'

I suppose you mean whether we can apply the reasoning from (1) to statement (2). Yes, we can:

(2) \(5x-3y=1\) --> Suppose GCD(x, y) is some integer \(d\), then \(x=md\) and \(y=nd\), for some positive integers \(m\) and \(n\). So, we'll have \(5(md)-3(nd)=d(5m-3n)=1\) --> \(d\) is a factor of 1, so \(d\) must equal 1. Sufficient.

Bunuel one silly question but please help with this :

Through the above logic for point 2 ,how can we be sure that d =1 .What about d= 1/(5M-3N) and that being another value apart from 1

Thanks a ton for all the help!!!

We have \(d(5m-3n)=1\). Both d and (5m-3n) are positive integers, there is only one value of d possible, d = 1.
_________________

Because factors are always integer. so D has to be taken as an integer only.

d= 1/(5M-3N) -- here (5M-3N) is nothing but 1. if (5M-3N) is anything other than 1 then D will not be an integer. and that would violate the rule "factors are always integer".
_________________

Re: If x and y are positive integers, what is the greatest [#permalink]

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18 Jun 2017, 22:28

Bunuel wrote:

zest4mba wrote:

If x and y are positive integers, what is the greatest common divisor of x and y?

1. 2x + y = 73 2. 5x – 3y = 1

This is a classic "C trap" question: "C trap" is a problem which is VERY OBVIOUSLY sufficient if both statements are taken together. When you see such question you should be extremely cautious when choosing C for an answer.

(1) \(2x+y=73\). Suppose GCD(x, y) is some integer \(d\), then \(x=md\) and \(y=nd\), for some positive integers \(m\) and \(n\). So, we'll have \(2(md)+(nd)=d(2m+n)=73\). Now, since 73 is a prime number (73=1*73) then \(d=1\) and \(2m+n=73\) (vice versa is not possible since \(m\) and \(n\) are positve integers and therefore \(2m+n\) cannot equal to 1). Hence we have that GCD(x, y)=d=1. Sufficient.

(2) \(5x-3y=1\) --> \(5x=3y+1\). So \(5x\) and \(3y\) are consecutive integers. Two consecutive integers are co-prime, which means that they don't share ANY common factor but 1. For example 20 and 21 are consecutive integers, thus only common factor they share is 1. So \(5x\) and \(3y\) don't share any common factor but 1, thus \(x\) and \(y\) also don't share any common factor but 1. Hence, GCD(x, y) is 1. Sufficient.

Answer: D.

Hope it's clear.

Hello Buñuel,

I was wondering why can't we apply the same logic given by you for statement 2 in statement 1, i.e. even 2 and 3 don't have any common factor other than 1?

Thanks and Regards.
_________________

------------------------------ "Trust the timing of your life" Hit Kudus if this has helped you get closer to your goal, and also to assist others save time. Tq

Here I'm not sure that the answer is C because is true that we need of both statement to find possible values for X and Y. Infact statement 1 and 2 we do not have values for the variables (can be everything).

But it seems to be a trap answer......

A very simple way to solve this problem will be:

1. 2x is even -> even +y=odd -> this mean y is odd -> even and odd GCD is 1 -> sufficient. 2. 5x-3y=1 -> 2 options: - 5x is even -> [odd -3y=odd] and [x is even] -> 3y must be odd -> y must be odd -> X even & Y odd -> even and odd GCD is 1 - 5x is odd -> [odd-3y=odd] and [x is odd] -> 3y must be even -> Y must be even -> X is odd & Y is even -> even and odd GCD is 1