If a and b are positive integers, what is the greatest commo

Question

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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

Kudos for a correct  solution .

Bunuel · Answer

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  -->  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.

Answer: D.

Try NEW divisibility  PS question .

JusTLucK04 · Answer

D
I am not sure if there is any standard approach for these...Just hit out some numbers

For I: I see that when A is odd say 3 then B is even 52..As B is an integer..the value of A will go down by 3 slots for each increase in B..So there is no case such that A and B will have any common factor other than 1
For II: You can again try a few..you can see that it is again 1...

manpreetsingh86 · Answer

Let k be the greatest common divisor of a and b such that a=kx and b=ky, where x and y are co-prime positive integers

St. 1; a+3b=61;
kx+3ky=61;
x+3y=(61/k)
now since 61 is a prime number therefore maximum value of k is 1
hence sufficient

st.2
5a-b=1
5kx-ky=1
5x-y=(1/k) since x and y are integers, therefore 5x-y will only be integer if k=1

hence sufficient

therefore answer should be D

Bunuel · Answer

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  -->  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.

Answer: D.

Kudos points given to correct solutions above.

Try NEW divisibility  PS question .

PathFinder007 · Answer

in st2 we can also say that bis multiple of 3. so this can be odd or even. so at a time b can be odd then a is even. or if b is even then a is odd. so GCD of a odd and even no. is 1.
 st2 is straight . numbers are co-prime

manhasnoname · Answer

Is there a compilation of such 700 level questions?

I really like these. Sometimes some of them are tricky.

Could someone please point me to any collection that you are aware of.

Many thanks!

Bunuel · Answer

Please browse the site: viewforumtags.php 150-hardest-and-easiest-questions-for-ds-204132.html ds-question-directory-by-topic-and-difficulty-128728.html 150-hardest-and-easiest-questions-for-ps-204134.html gmat-ps-question-directory-by-topic-difficulty-127957.html

GMATGuruNY · Answer

Let k = the GCD of  a  and  b , implying that  \frac{a}{k} =  integer and  \frac{b}{k} =  integer.
Question stem, rephrased:
What is the value of  k ?

Statement 1:  a+3b=61 
Since  a  and  b  are positive,  a<61  and  b<61 , with the result that their greatest common factor --  k  -- must also be less than 61.
Reversing the equation and dividing both sides by k, we get:
 \frac{61}{k} = \frac{a}{k} + 3\frac{b}{k} 
 \frac{61}{k} =  integer + 3(integer)
 \frac{61}{k} =  integer
For the left side to yield an integer,  k  must be a factor of 61.
Factors of 61: 1 and 61
Since  k<61 , only one option is possible:
k=1
SUFFICIENT.

Statement 2:  5a-b=1 
Reversing the equation and dividing both sides by k, we get:
 \frac{1}{k} = 5\frac{a}{k} - \frac{b}{k} 
 \frac{1}{k} =   5(integer) - (integer)
 \frac{1}{k} =  integer
For the left side to yield an integer,  k  must be a factor of 1.
Only one option is possible:
k=1
SUFFICIENT.

D

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If a and b are positive integers, what is the greatest commo

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