If a and b are positive integers divisible by 6, is 6 the greatest com

Yes, both are divisible by 6, which means that they are divisible by 2 and 3.

Next, we have that \(a=6x\) and \(b=6y\).

Consider two cases:
1. \(x\) and \(y\) share some common factor >1: for example \(x=2\) and \(y=4\) then \(a=12\) and \(b=24\) --> \(GCD(a,b)=12>6\);
2. \(x\) and \(y\) DO NOT share any common factor >1: for example \(x=5\) and \(y=2\) then \(a=30\) and \(b=12\) --> \(GCD(a,b)=6\).

From (1) we have that --> \(x=2y+1\) --> \(x\) is one more than multiple of \(y\). For example: \(x=3\) and \(y=1\) OR \(x=5\) and \(y=2\) OR \(x=7\) and \(y=3\) ... as you can see in all these cases x and y do not share any common factor more than 1. Now, as we concluded above if \(x\) and \(y\) DO NOT share any common factor >1, then \(GCD(a,b)=6\).

Or another way: \(b=6y\) and \(a=6(2y+1)\). \(2y\) and \(2y+1\) are consecutive integers and consecutive integers do not share any common factor 1. As \(2y\) has all factors of \(y\) then \(y\) and \(2y+1\) also do not share any common factor but 1, which means that 6 must GCD of \(a\) and \(b\)

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Hope it helps.

Given a and b are positive integers divisible by 6

From st 1, we have a = 2b+ 6
b is of the form b= 6c where c is a positive integer

therefore we have a = 2*6c+6 ----> a= 6 (2c+1)
b= 6c

Now HCF of a and b will be 6 as
a= 2*3*(2c+1)
b=2*3
Only common factor is 6

So, Ans is A alone sufficient. We can remove options B, C and E as possible answers

Not st 2 says a= 3b
taking b = 6c we have a=18c

a= 2*3^2*c
b=2*3 c
HCF will be 2*3*c that is 6c

Now if c=1 then 6 is a the HCF of both a and b
but if c=2 then 12 is HCF
Since 2 ans choices are possible.

Ans will be A to this Question

We have that \(a=6x\) and \(b=6y\).

Consider two cases:
1. \(x\) and \(y\) share some common factor >1: for example \(x=2\) and \(y=4\) then \(a=12\) and \(b=24\) --> \(GCD(a,b)=12>6\);
2. \(x\) and \(y\) DO NOT share any common factor >1: for example \(x=5\) and \(y=2\) then \(a=30\) and \(b=12\) --> \(GCD(a,b)=6\).

From (1) we have that --> \(x=2y+1\) --> \(x\) is one more than multiple of \(y\). For example: \(x=3\) and \(y=1\) OR \(x=5\) and \(y=2\) OR \(x=7\) and \(y=3\) ... as you can see in all these cases x and y do not share any common factor more than 1. Now, as we concluded above if \(x\) and \(y\) DO NOT share any common factor >1, then \(GCD(a,b)=6\).

Hope it's clear.

Take b = 12, you will get a = 36.

Now try to calculate the HCF of a and b, you will get your answer, why B is insufficient.

Question:
6 * x =a and 6 * y =b
we need to find if 6 is the GCD? YES or NO question.
so, basically if we can find a single common factor in x and y, thats it its not a GCD, or if we cant find one then that should also work for us.

(1) a = 2b + 6

6x=12y + 6

x = 2y +1 =>No matter what you do, this will always result in an no common factor.

Thus 6 is the only GCD =>Sufficient

(2) a = 3b

6x = 18y
x=3y

Take y=8, and Y=2 =>This is clearly Not sufficient.

Ans: A

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Hope it helps.

Recall that the GCD of two numbers must be a factor of their difference.
6 is a factor of a and b.

(1) a = 2b + 6

a - 2b = 6
This means that GCD of a and 2b is almost 6. Then GCD of a and b cannot be more than 6 either (because b has a subset of factors of 2b).
Since both a and b have 6 as a factor, 6 must be their GCD.
Sufficient.

(2) a = 3b
If a = 3b, we need the GCD of 3b and b. That will simply be 'b'. If b is 6, the GCD will be 6. If b is 12, the GCD will be 12 etc.
Not sufficient.

Answer (A)

Hi Bunuel, could you kindly elaborate on the following statement:

"x=2y+1 --> x and y do not share any factor >1, as if they were we would be able to factor out if from 2y+1. Sufficient."

I'm obviously missing some fundamental insight, but I don't understand why not being able to factor something out of the 2y+1 means that 6 is the GCD.

Thanks!

Crystal clear, thanks so much!

Perfect explanation

But if a and b are both divisible of 6, means that both are even, therefore at least both of them should be divisible by 2.... I am right????? I do not understand why (1) is valid... thanks!!

Hi Bunuel, Can you explain the following rule with few examples.

Given: and . Question: is ? Now, If x and y share any common factor >1then GCD (a,b) will be more than 6 if not then GCD (a,b) will be 6.

Sure. Both \(a\) and \(b\) are multiples of 6 --> \(a=6x\) and \(b=6y\). Consider two cases:

A. \(x\) and \(y\) do not share any common factor >1, for example \(a=6*2=12\) and \(b=6*3=18\) --> GCD(a,b)=6. As you can see 2 and 3 did not contribute any common factor to the GCD;

B. \(x\) and \(y\) share some common factor >1, for example \(a=6*2=12\) and \(b=6*4=24\) --> GCD(a,b)=12, here 2 and 4 contributes common factor 2 to the GCD.

Hope it's clear.

Hi Bunuel,

Can you please explain

x and y do not share any factor >1, as if they were we would be able to factor out if from 2y+1. Sufficient.


so if we have x=2(2y+1) then x does have a factor of 2,but how do we know about the factors of y.

x=2y+1 means that x and 2y are co-prime (x and 2y do not share any common factor but 1), which, on the other hand, means that x and y are co-prime. How else? If x and 2y do not share any common factor greater than 1, how can x and y share any common factor greater than 1?

It would be great if one could find more questions involving GCD/GCF as they seem to be little tricky more often than not.

If x and 2y do not share any common factor greater than 1, how can x and y share any common factor greater than 1?
Please Bunuel clarify this comment more. what does it mean.
another question is that how frequently DS questions of divisibility come in GMAT test ?