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Re: If a and b are positive integers divisible by 6, is 6 the greatest com
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03 Sep 2010, 06:20
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27
gmatbull wrote:
If a and b are positive integers divisible by 6, is 6 the greatest common divisor of a and b?
(1) a = 2b + 6
(2) a = 3b
What is the fastest (perhaps, algebra?) means of solving the question besides random plugging numbers under test condition?
If a and b are positive integers divisible by 6, is 6 the greatest common divisor of a and b?
Given: \(a=6x\) and \(b=6y\). Question: is \(GCD(a,b)=6\)? 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.
(1) \(a=2b+6\) --> \(6x=2*6y+6\) --> \(x=2y+1\) --> \(x\) and \(y\) do not share any factor >1, as if they were we would be able to factor out if from \(2y+1\). Sufficient.
(2) \(a=3b\) --> clearly insufficient.
Answer: A.
There is also a general rule: if \(a\) and \(b\) are multiples of \(k\) and are \(k\) units apart from each other then \(k\) is greatest common divisor of \(a\) and \(b\).
For example if \(a\) and \(b\) are multiples of 7 and \(a=b+7\) then 7 is GCD of \(a\) and \(b\).
So if we apply this rule to (1) \(a=2b+6\) --> both \(a\) and \(2b\) are multiples of 6 and are 6 apart, so GCD of \(a\) and \(2b\) is 6, hence GCD of \(a\) and \(b\) is also 6. Sufficient.
Re: If a and b are positive integers divisible by 6, is 6 the greatest com
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02 Sep 2010, 23:41
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note that a=3b does not mean that 6 will be the gcd. example 6, 18 is 6, but 12, 36 is 12 not suff
for a =2b+6= 2(b+3); take b=6,a=18, gcd =6 ; take b=12,a=30; gcd=6; take b=18,a=42; gcd=6 suff
for an algebraic proof: a=6r, b=6s 2. a=3b means 6r=18s or r=3s; a=6s b=18s; cannot conclude about gcd a=2b+6 means r=2s+1; a=6(2s+1); b=6s no common factors; you can conclude that 6 will be gcd.
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Re: If a and b are positive integers divisible by 6, is 6 the greatest com
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17 Feb 2011, 00:26
Bunuel wrote:
gmatbull wrote:
If a and b are positive integers divisible by 6, is 6 the greatest common divisor of a and b?
(1) a = 2b + 6
(2) a = 3b
What is the fastest (perhaps, algebra?) means of solving the question besides random plugging numbers under test condition?
If a and b are positive integers divisible by 6, is 6 the greatest common divisor of a and b?
Given: \(a=6x\) and \(b=6y\). Question: is \(GCD(a,b)=6\)? 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.
(1) \(a=2b+6\) --> \(6x=2*6y+6\) --> \(x=2y+1\) --> \(x\) and \(y\) do not share any factor >1, as if they were we would be able to factor out if from \(2y+1\). Sufficient.
(2) \(a=3b\) --> clearly insufficient.
Answer: A.
There is also a general rule: if \(a\) and \(b\) are multiples of \(k\) and are \(k\) units apart from each other then \(k\) is greatest common divisor of \(a\) and \(b\).
For example if \(a\) and \(b\) are multiples of 7 and \(a=b+7\) then 7 is GCD of \(a\) and \(b\).
So if we apply this rule to (1) \(a=2b+6\) --> both \(a\) and \(2b\) are multiples of 6 and are 6 apart, so GCD of \(a\) and \(2b\) is 6, hence GCD of \(a\) and \(b\) is also 6. Sufficient.
Hope it helps.
Thanks for the rule. I picked A as I knew there is "some" rule for common multiples a an integer with that integer as the difference between the common multiples but was not recollecting it..
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My GMAT debrief: http://gmatclub.com/forum/from-620-to-710-my-gmat-journey-114437.html
Re: If a and b are positive integers divisible by 6, is 6 the greatest com
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05 Feb 2012, 02:43
Bunuel wrote:
gmatbull wrote:
If a and b are positive integers divisible by 6, is 6 the greatest common divisor of a and b?
(1) a = 2b + 6
(2) a = 3b
What is the fastest (perhaps, algebra?) means of solving the question besides random plugging numbers under test condition?
If a and b are positive integers divisible by 6, is 6 the greatest common divisor of a and b?
Given: \(a=6x\) and \(b=6y\). Question: is \(GCD(a,b)=6\)? 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.
(1) \(a=2b+6\) --> \(6x=2*6y+6\) --> \(x=2y+1\) --> \(x\) and \(y\) do not share any factor >1, as if they were we would be able to factor out if from \(2y+1\). Sufficient.
(2) \(a=3b\) --> clearly insufficient.
Answer: A.
There is also a general rule: if \(a\) and \(b\) are multiples of \(k\) and are \(k\) units apart from each other then \(k\) is greatest common divisor of \(a\) and \(b\).
For example if \(a\) and \(b\) are multiples of 7 and \(a=b+7\) then 7 is GCD of \(a\) and \(b\).
So if we apply this rule to (1) \(a=2b+6\) --> both \(a\) and \(2b\) are multiples of 6 and are 6 apart, so GCD of \(a\) and \(2b\) is 6, hence GCD of \(a\) and \(b\) is also 6. Sufficient.
Hope it helps.
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!!
Re: If a and b are positive integers divisible by 6, is 6 the greatest com
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05 Feb 2012, 03:02
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Saurajm wrote:
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!!
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 \(b\)
Re: If a and b are positive integers divisible by 6, is 6 the greatest com
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14 Mar 2012, 09:08
pavanpuneet wrote:
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.
Re: If a and b are positive integers divisible by 6, is 6 the greatest com
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26 Jun 2013, 08:14
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gmatbull wrote:
If a and b are positive integers divisible by 6, is 6 the greatest common divisor of a and b?
(1) a = 2b + 6
(2) a = 3b
What is the fastest (perhaps, algebra?) means of solving the question besides random plugging numbers under test condition?
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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Re: If a and b are positive integers divisible by 6, is 6 the greatest com
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29 Sep 2014, 14:09
Bunuel wrote:
gmatbull wrote:
If a and b are positive integers divisible by 6, is 6 the greatest common divisor of a and b?
(1) a = 2b + 6
(2) a = 3b
What is the fastest (perhaps, algebra?) means of solving the question besides random plugging numbers under test condition?
If a and b are positive integers divisible by 6, is 6 the greatest common divisor of a and b?
Given: \(a=6x\) and \(b=6y\). Question: is \(GCD(a,b)=6\)? 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.
(1) \(a=2b+6\) --> \(6x=2*6y+6\) --> \(x=2y+1\) --> \(x\) and \(y\) do not share any factor >1, as if they were we would be able to factor out if from \(2y+1\). Sufficient.
(2) \(a=3b\) --> clearly insufficient.
Answer: A.
There is also a general rule: if \(a\) and \(b\) are multiples of \(k\) and are \(k\) units apart from each other then \(k\) is greatest common divisor of \(a\) and \(b\).
For example if \(a\) and \(b\) are multiples of 7 and \(a=b+7\) then 7 is GCD of \(a\) and \(b\).
So if we apply this rule to (1) \(a=2b+6\) --> both \(a\) and \(2b\) are multiples of 6 and are 6 apart, so GCD of \(a\) and \(2b\) is 6, hence GCD of \(a\) and \(b\) is also 6. Sufficient.
Hope it helps.
It would be great if one could find more questions involving GCD/GCF as they seem to be little tricky more often than not.
Re: If a and b are positive integers divisible by 6, is 6 the greatest com
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30 Sep 2014, 00:48
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earnit wrote:
Bunuel wrote:
gmatbull wrote:
If a and b are positive integers divisible by 6, is 6 the greatest common divisor of a and b?
(1) a = 2b + 6
(2) a = 3b
What is the fastest (perhaps, algebra?) means of solving the question besides random plugging numbers under test condition?
If a and b are positive integers divisible by 6, is 6 the greatest common divisor of a and b?
Given: \(a=6x\) and \(b=6y\). Question: is \(GCD(a,b)=6\)? 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.
(1) \(a=2b+6\) --> \(6x=2*6y+6\) --> \(x=2y+1\) --> \(x\) and \(y\) do not share any factor >1, as if they were we would be able to factor out if from \(2y+1\). Sufficient.
(2) \(a=3b\) --> clearly insufficient.
Answer: A.
There is also a general rule: if \(a\) and \(b\) are multiples of \(k\) and are \(k\) units apart from each other then \(k\) is greatest common divisor of \(a\) and \(b\).
For example if \(a\) and \(b\) are multiples of 7 and \(a=b+7\) then 7 is GCD of \(a\) and \(b\).
So if we apply this rule to (1) \(a=2b+6\) --> both \(a\) and \(2b\) are multiples of 6 and are 6 apart, so GCD of \(a\) and \(2b\) is 6, hence GCD of \(a\) and \(b\) is also 6. Sufficient.
Hope it helps.
It would be great if one could find more questions involving GCD/GCF as they seem to be little tricky more often than not.
Re: If a and b are positive integers divisible by 6, is 6 the greatest com
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31 Mar 2016, 12:01
Bunuel, a question if you don't mind..
Why do you say that 12 and 18 don't share any common factors >1 ? they do. 12=2*2*3 18=3*3*2, so 2 and 3 are shared. and they form the GCF in this case.
More, if a and b are multiples of 6, and a=2b+6, if b=6, doesn't this mean that a=3b? (or a=18 simply).. if b=12 then a=5b etc. in any case, a and b are NOT 6 apart (or are more than 6 apart) from each other, so we get a NO answer to the question, so Sufficient.
Re: If a and b are positive integers divisible by 6, is 6 the greatest com
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31 Mar 2016, 12:15
iliavko wrote:
Bunuel, a question if you don't mind..
Why do you say that 12 and 18 don't share any common factors >1 ? they do. 12=2*2*3 18=3*3*2, so 2 and 3 are shared. and they form the GCF in this case.
More, if a and b are multiples of 6, and a=2b+6, if b=6, doesn't this mean that a=3b? (or a=18 simply).. if b=12 then a=5b etc. in any case, a and b are NOT 6 apart (or are more than 6 apart) from each other, so we get a NO answer to the question, so Sufficient.
Thank you!
Where did I say that 12 and 18 do not share any common factor?
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Re: If a and b are positive integers divisible by 6, is 6 the greatest com
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31 Mar 2016, 13:16
I am referring to this reply you gave: A) says it.
Bunuel wrote:
pavanpuneet wrote:
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.
Re: If a and b are positive integers divisible by 6, is 6 the greatest com
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31 Mar 2016, 13:21
iliavko wrote:
I am referring to this reply you gave: A) says it.
Bunuel wrote:
pavanpuneet wrote:
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.
Thank you!
Please read carefully:\(x\) and \(y\) do not share any common factor >1, NOT a and b.
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