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Math Expert V
Joined: 02 Sep 2009
Posts: 58320

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16 00:00

Difficulty:   95% (hard)

Question Stats: 41% (02:19) correct 59% (01:47) wrong based on 144 sessions

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If $$a$$ and $$b$$ are positive integers, what is the greatest common divisor of $$a$$ and $$b$$?

(1) $$a + 3b = 61$$

(2) $$5a - b = 1$$

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Math Expert V
Joined: 02 Sep 2009
Posts: 58320

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Official Solution:

If $$a$$ and $$b$$ are positive integers, what is the greatest common divisor of $$a$$ and $$b$$?

Notice that two statements together are obviously sufficient to answer the question. When you see such question you should be extremely cautious when choosing C for an answer. Chances are that the question is a "C trap" question ("C trap" is a problem which is VERY OBVIOUSLY sufficient if both statements are taken together).

(1) $$a + 3b = 61$$.

Let the greatest common divisor of $$a$$ and $$b$$ be $$d$$, then $$a=md$$ and $$b=nd$$, for some positive integers $$m$$ and $$n$$. So, we'll have $$(md)+3(nd)=d(m+3n)=61$$. Now, since 61 is a prime number (61=1*61) then $$d=1$$ and $$m+3n=61$$ (vice versa is not possible because $$m$$ and $$n$$ are positive integers and therefore $$m+3n$$ cannot equal to 1). Hence we have that the $$GCD(x, y)=d=1$$. Sufficient.

(2) $$5a - b = 1$$. Rearrange: $$5a=b+1$$.

$$5a$$ and $$b$$ 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, $$5a$$ and $$b$$ don't share any common factor but 1, thus $$a$$ and $$b$$ also don't share any common factor but 1. Hence, the $$GCD(x, y)$$ is 1. Sufficient.

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Intern  Joined: 01 Jul 2015
Posts: 5
Concentration: Entrepreneurship, Technology
Schools: Babson '19

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I think this is a high-quality question and I agree with explanation.
Manager  Joined: 02 Nov 2014
Posts: 182
GMAT Date: 08-04-2015

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I got this problem right, but it took me 3 mins.
I plugged in different values of a and b: a=1, 4, 7, 10, .. then b= 20, 19, 18, 17, .. but it was lengthy and tiresome.

The explanation by Bunuel is just awesome. Such a time saver, but it needs an eye of a mathematician I think.

Thanks.
Intern  B
Joined: 22 Nov 2014
Posts: 29

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The easiest way is the concept of ODD EVEN..
st 1 -61 is ODD and also prime so numbers should be ODD and EVEN and hence GCD is 1
similarly for st 2 both should be ODD and EVEN
Intern  B
Joined: 17 May 2016
Posts: 27
GMAT 1: 740 Q46 V46 Show Tags

yaash

Sorry if I misunderstood, what do you exactly mean by should be odd and even ?

Your solution seems really fast and efficient but I did not fully got it

Many thanks for your help,
Math Expert V
Joined: 02 Sep 2009
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nickimonckom wrote:
yaash

Sorry if I misunderstood, what do you exactly mean by should be odd and even ?

Your solution seems really fast and efficient but I did not fully got it

Many thanks for your help,

That solution is not correct. yaash implies that the GCD of an odd number and of an even number is always 1, which is obviously wrong. For example, the GCD of 3 and 6 is 3. This is a hard question and has no silver bullet solution. Please refer to the official solution above.
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SVP  V
Joined: 26 Mar 2013
Posts: 2341

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My way to solve this question:

If $$a$$ and $$b$$ are positive integers, what is the greatest common divisor of $$a$$ and $$b$$?

(1) $$a + 3b = 61$$

a =61 -3b ..........Plug in numbers

b=1 ......a=58.......GCD= 1
b=2.......a=55.......GCD= 1
b=3.......a=52.......GCD= 1
b=4.......a=49.......GCD= 1
b=5.......a=46.......GCD= 1

It is safe to conclude that GCD=1

Sufficient

(2) $$5a - b = 1$$

b =5a -1

a=1 ........b=4.......GCD= 1
a=2.........b=9.......GCD= 1
a=3.........b=14.....GCD= 1
a=4.........b=19.....GCD= 1

It is safe to conclude that GCD=1

Sufficient

Intern  Joined: 16 Jan 2019
Posts: 1

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I answered this question incorrectly. I understood my error after reading Bunuel solution.
Please allow me to provide an alternative explanation of the second part. Which is easier for me to understand.

...Alternative explanation to Bunuel's solution of part 2...

(2) 5a−b=1. Rearrange: 5a=b+1.

Still using a=md and b=nd. We have 5(md)-(nd)=d(5d-n)=1. Since d and (5d-n) need to be positive Integers, the only possible solution to the equation is that both are equal to 1. Therefore d=1 and GCD(x,y)=d=1

Bunuel. Many Thanks for your solution
Intern  Joined: 30 Nov 2018
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If d is a divisor of both a and b, then it is also a divisor of any linear combination ma+nb of these numbers a and b. Therefore it is a divisor of 5a + (-1)b, and hence of 1 (as 5a−b=1). The only positive divisor of 1 is 1.
Senior Manager  G
Joined: 13 Feb 2018
Posts: 453
GMAT 1: 640 Q48 V28 Show Tags

For the sake of quant gods

How a mortal, like me, can solve this on real GMAT, even ignoring the time issue.

I'll try to remember the method used in 1 option. Re: M30-16   [#permalink] 12 Jun 2019, 01:59
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