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Is the integer 2b divisible by 6 ? 10 Apr 2017, 19:20
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Doesn't the question already say 2b is an integer? How can the answer be (C)?

From (1) 8b is divisible by 3. Therefore, b has to be divisible by 3. From the question stem 2b is an integer. Combining these two, b can be 3/2, 3, 6/2, 6, 9/2, 9, ... etc... for all these values of b, 2b is divisible by 6. Hence (A).
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Is the integer 2b divisible by 6 ? 10 Apr 2017, 22:05
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manhasnoname
Doesn't the question already say 2b is an integer? How can the answer be (C)?

From (1) 8b is divisible by 3. Therefore, b has to be divisible by 3. From the question stem 2b is an integer. Combining these two, b can be 3/2, 3, 6/2, 6, 9/2, 9, ... etc... for all these values of b, 2b is divisible by 6. Hence (A).
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Your doubt is addressed several times on previous page. For (1) consider b = 3/2 --> 8b = 12, so it's divisible by 3 but 2b = 3, so not divisible by 6.
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Is the integer 2b divisible by 6 ? 11 Apr 2017, 23:36
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Is the integer 2b divisible by 6 ?

(1) 8b is divisible by 3.

(2) 9b is divisible by 12

its clearly asked whether the Integer 2b is divisible by 6.....so we can only to take the values to b which satisfies 2b as integer even 3/2 is possible but not 3/4

from statement 1----> 8b =>4*2b when divisible by 3 the possible values from given and statement 1 combined are 3/2,3,6......
conclusion:- statement 1 is enough to solve the question

statement 2 is clearly not enough to solve the equation.

So my answer would be A
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Is the integer 2b divisible by 6 ? 11 Apr 2017, 23:42
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sreenu7464
Is the integer 2b divisible by 6 ?

(1) 8b is divisible by 3.

(2) 9b is divisible by 12

its clearly asked whether the Integer 2b is divisible by 6.....so we can only to take the values to b which satisfies 2b as integer even 3/2 is possible but not 3/4

from statement 1----> 8b =>4*2b when divisible by 3 the possible values from given and statement 1 combined are 3/2,3,6......
conclusion:- statement 1 is enough to solve the question

statement 2 is clearly not enough to solve the equation.

So my answer would be A
Show more

The answer is NOT A it's C. It's explained several times on previous pages:

For a NO answer in (1) consider b = 3/2 --> 8b = 12, so it's divisible by 3 but 2b = 3, so not divisible by 6.
For an YES answer in (1) consider b = 3 --> 8b = 24, so it's divisible by 3 and 2b = 6, so divisible by 6.
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Is the integer 2b divisible by 6 ? 13 May 2018, 01:57
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Option 1: 8b divisible for 3 ==> b divisible for 3 ==> From 1, do not have enough info for answer
Option 2: 9b divisible for 12 ==> b divisible for 2 ==> From 2, do not have enough info for answer

From 1 and 2==> b divisible for 6 => answer C

Please hit kudo if it helps! THanks.
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Is the integer 2b divisible by 6 ? 13 May 2018, 09:59
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Is the integer 2b divisible by 6 ?

(1) 8b is divisible by 3.

(2) 9b is divisible by 12

its clearly asked whether the Integer 2b is divisible by 6.....so we can only to take the values to b which satisfies 2b as integer even 3/2 is possible but not 3/4

from statement 1----> 8b =>4*2b when divisible by 3 the possible values from given and statement 1 combined are 3/2,3,6......
conclusion:- statement 1 is enough to solve the question

statement 2 is clearly not enough to solve the equation.

So my answer would be A

The answer is NOT A it's C. It's explained several times on previous pages:

For a NO answer in (1) consider b = 3/2 --> 8b = 12, so it's divisible by 3 but 2b = 3, so not divisible by 6.
For an YES answer in (1) consider b = 3 --> 8b = 24, so it's divisible by 3 and 2b = 6, so divisible by 6.
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Is it ok to make it from 2b/6 to b/3? Do we lose anything?
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Is the integer 2b divisible by 6 ? 13 May 2018, 23:40
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The answer is NOT A it's C. It's explained several times on previous pages:

For a NO answer in (1) consider b = 3/2 --> 8b = 12, so it's divisible by 3 but 2b = 3, so not divisible by 6.
For an YES answer in (1) consider b = 3 --> 8b = 24, so it's divisible by 3 and 2b = 6, so divisible by 6.[/quote]

Is it ok to make it from 2b/6 to b/3? Do we lose anything?[/quote]

Hello

You are correct. Asking 'whether 2b is divisible by 6' is same as asking 'whether b is divisible by 3'.
In any case wherever 2b is divisible by 6, b has to be divisible by 3.
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Is the integer 2b divisible by 6 ? 05 Jul 2020, 15:52
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Quote:
Is the integer 2b divisible by 6 ?
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Quote:
(1) 8b is divisible by 3.
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Thus, \(8b = 3k\), k is an integer
\(b=\frac{3}{8} k\)

But 2b is an integer,
Thus \(2b=\frac{3}{4} k\) is an integer

Thus, \(2b = 3k\)

Thus, \(b = \frac{3}{2} k\)

Not Sufficient

Quote:
(2) 9b is divisible by 12
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Thus, \(9b = 12m\), m is an integer
\(b=\frac{4}{3} m\)

But 2b is an integer
Thus \(2b= \frac{8}{3}m\) is an integer

Thus \(2b = 8m\)

Thus \(b= 4m\)

Not Sufficient

Combining:

b is a multiple of \(\frac{3}{2} and \frac{4}{1}\)

LCM of \(\frac{3}{2} and \frac{4}{1} = \frac{LCM(3,4)}{HCF(2,1)} = 12\)

Thus b is a multiple of 12.

Sufficient
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Is the integer 2b divisible by 6 ? 20 May 2022, 18:53
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Concept: Any number is divisible by 6 if
1. the unit digit is even
2. The summation of the digits is divisible by 3
We only need to check if b is even and 2+b is divisible by 3

Statement 1: 8b is divisible by 3
81 (b=1) is divisible by 3
84 (b=4) is divisible by 3
87 (b=7) is divisible by 3
B can hold the value 1,4 and 7 but remember that we need to check if all of these are even or not. Here it has both even and odd, statement 1 is not sufficient

Statement 2: 9b is divisible by 12
means it is divisible by 2*2*3
So the number have to nullify the entire denominator if 2*2*3
3 cuts 9, so 3 goes off from denominator
Now only if b has 4 or 8 then it can nullify 2*2 in the denominator
That means b can be 4 or 8
Now test for whether 2+b is divisible by 3
2+4=6 divisible by 3
2+8=10 not divisible by 3
So, statement 2 is not sufficient

Now if you check for statement A and B together, then we know from statement 1 that
b=1,4,7
Statement 2 that b=4,8
Means b=4 is only the common value
So the number 2b is 24
I can now determine whether it is divisible by 6
Answer C
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Is the integer 2b divisible by 6 ? 11 Jun 2023, 10:37
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Hello from the GMAT Club BumpBot!

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