GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 22 Feb 2019, 05:17

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

## Events & Promotions

###### Events & Promotions in February
PrevNext
SuMoTuWeThFrSa
272829303112
3456789
10111213141516
17181920212223
242526272812
Open Detailed Calendar
• ### Free GMAT RC Webinar

February 23, 2019

February 23, 2019

07:00 AM PST

09:00 AM PST

Learn reading strategies that can help even non-voracious reader to master GMAT RC. Saturday, February 23rd at 7 AM PT
• ### FREE Quant Workshop by e-GMAT!

February 24, 2019

February 24, 2019

07:00 AM PST

09:00 AM PST

Get personalized insights on how to achieve your Target Quant Score.

# M30-16

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 53066

### Show Tags

16 Sep 2014, 00:45
2
14
00:00

Difficulty:

95% (hard)

Question Stats:

40% (01:52) correct 60% (01:16) wrong based on 181 sessions

### HideShow timer Statistics

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

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

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

_________________
Math Expert
Joined: 02 Sep 2009
Posts: 53066

### Show Tags

16 Sep 2014, 00:45
4
3
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.

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

### Show Tags

09 Nov 2015, 22:12
1
I think this is a high-quality question and I agree with explanation.
Manager
Joined: 02 Nov 2014
Posts: 187
GMAT Date: 08-04-2015

### Show Tags

10 Nov 2015, 04:54
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
Joined: 22 Nov 2014
Posts: 29

### Show Tags

17 Dec 2016, 05:28
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
Joined: 17 May 2016
Posts: 29
GMAT 1: 740 Q46 V46

### Show Tags

05 Jan 2017, 03:00
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

Math Expert
Joined: 02 Sep 2009
Posts: 53066

### Show Tags

05 Jan 2017, 03:06
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

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.
_________________
Intern
Joined: 25 Jan 2015
Posts: 20

### Show Tags

10 Feb 2017, 01:20
From this problem I derive that the GCD of an odd number and an even number is always 1. Is this correct?
Math Expert
Joined: 02 Sep 2009
Posts: 53066

### Show Tags

10 Feb 2017, 01:23
Shiridip wrote:
From this problem I derive that the GCD of an odd number and an even number is always 1. Is this correct?

Please check the post just above yours.
_________________
SVP
Joined: 26 Mar 2013
Posts: 2068

### Show Tags

10 Feb 2017, 23:58
1
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

### Show Tags

16 Jan 2019, 02:32
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
Posts: 1

### Show Tags

18 Jan 2019, 07:51
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.
Re: M30-16   [#permalink] 18 Jan 2019, 07:51
Display posts from previous: Sort by

# M30-16

Moderators: chetan2u, Bunuel

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.