Is the integer x divisible by 6?

Is the integer x divisible by 6?

(1) x + 3 is divisible by 3. This basically means that x is divisible by 3 (x = {a multiple of 3} - 3 = {a multiple of 3}), which is not sufficient to say whether it's divisible by 6.

(2) x + 3 is an odd number. This means that x is even (\(x=odd-3=odd-odd=even\)). Not sufficient.

(1)+(2) x is an even multiple of 3, hence it's a multiple of 6. Sufficient.

Answer: C.
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Bunuel : If x = 0 (which is an even no.) + 3 is divisible by 3.
But 0/6 is not divisible. Am i missing something?

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and equations ensures a solution.


Is the integer x divisible by 6?

(1) x + 3 is divisible by 3
(2) x + 3 is an odd number

In the original condition there is 1 variable and we need 1 equation to match the number of variable and equation. Since there is 1 each in 1) and 2), D is likely the answer.

In case of 1), x+3=3t(t:some integer) thus x=multiple of 3. If x=3 the answer is no, if x=6 the answer is yes. Therefore the condition is not sufficient.
In case of 2), if x+3=odd, x=odd-3=even, x=2 then the answer is no. If x=6 the answre is yes, therefore the condition is not sufficient.
Using both 1) & 2) together, if x=multiple of 3 and even, it must be a multiple of 6 and thus the answer is yes. Therefore the answer is C.
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0/6 = 0 = integer. 0 is divisible by every integer except 0 itself.
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Bunuel : Muchos Gracias Bunuel.

We need to determine whether x/6 = integer.

Statement One Alone:

x + 3 is divisible by 3.

Statement one is not sufficient to answer the question. If x = 3, then x is not divisible by 6; however, if x = 6, then x is divisible by 6.

Statement Two Alone:

x + 3 is an odd number.

Statement two is not sufficient to answer the question. If x = 2, then x is not divisible by 6; however, if x = 6, then x is divisible by 6.

Statements One and Two Together:

Using statements one and two, we see that x + 3 is an odd multiple of 3, such as 3, 9, 15, etc. Thus, x must be an even multiple of 3. Since all even multiples of 3 are multiples of 6, x/6 is an integer.

Answer: C
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Hi Bunuel, Do we need to consider negative integer for statement (1)?
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x could be a negative integer for (1) or/and (2) but it does not affect the answer. For example, when we consider the statements together, we get that x is an even multiple of 3, so x could be ..., -12, -6, 0, 6, 12, ... As you can see x is indeed divisible by 6 for all possible values of x.
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I have a doubt regarding this question because as per my approach I got answer D.

The question says x is divisible by 6 which means x can be 6,12,18,24 ......etc

Stat1: x+3 is divisible by 3
We plug in any value of x, say 12, so 12+3=15 which is divisible by 3. Likewise, put in the rest of the numbers. Sufficient.

Stat2: x+3 is an odd number.
We know any even+odd=odd so all the values are even for x so plugging in here we can prove that x+3 is indeed odd. Sufficient.

Where am I going wrong?

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Whenever you see a data sufficiency question, its a good idea to ask yourself, "What would I need to know in order to answer this question?" Stated differently, its a good idea to ask yourself, "What's the question behind the question?" Doing so will get your mental juices flowing before you take a look at the statements...

The question here is whether x is divisible by 6. The question behind the question is, "Is x divisible by both 2 and 3?" That's what you'd need to know in order to answer the original question. With that in mind, you know what to look for when you take a look at the two statements, so when you read in statement (1) that x + 3 is divisible by 3, it may pop out at you quickly that, "Oh, yes, so that means x itself is divisible by 3. Statement (1) isn't sufficient, but it does tell me half of what I need to know."

Same story for statement (2). When you see that x + 3 is odd, you may recall quickly that x must be EVEN - i.e. x is divisible by 2 - and you'll think, "Great. That's the second half of what I need to know. Answer C."
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hi, you failed to consider 2 cases which are very important for this question.
REMINDER question is asking if IF X IS DIVISIBLE BY 6?

you tried to make the statements sufficient whereas the job on DS is to make them insufficient.
what happens if you plugin x=3?
3+3 is divisible by 6 and 3 both.
but, 3 (x here) isn't divisible by 6.
Also, put in x=9
9 + 3 is divisible by 3, but x=9, which isn't divisible by 6.

Therefore, the statement is not sufficient.

Statement 2 says x+3 is odd...

X = any even value..
X= 2 +3 = 5, yes but (X =2) isn't divisible by 6 (this is what the question is asking)
so, it also gives us a yes and a no.

Combining them we can say that X is divisible by 6, as X is even (from 2) and X is either 6,12,18 (from 1) ... etc.

(1) x + 3 is divisible by 3

x+3 = 3p
x = 3p - 3

p=-1, x=-6
p=0, x=-3
p=1, x=0
p=2, x=3
p=3, x=6
p=4, x=9
p=5, x=12

We can see that when p is odd, x will be even and divisible by 6.
Not sufficient because we don't know the value of p or x.

(2) x + 3 is an odd number
If x = 2 then no, if x = 6 then yes, in other words we don't know if x is even or odd.
Not sufficient.

Together, (2) x + 3 is odd, so (1) x + 3 = 3p is odd, therefore 3p is odd and x will be divisible by 6.

Bunuel, nice jobe!

But what if x = 0 (even number) then x+3 = multiple of 3 but it's not multiple of 6??

0 is divisible by every integer (except 0 itself): 0/integer = 0.
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Answer: C

Question Stem --> Is x = 6n?

Stmt1:
x+3/3 = integer
If we take x as 0, 3, 6, 9; we get answers for our stem as Yes, No, Yes, No.
Hence not sufficient.

Stmt2:
x+3 = odd
Which means x=even
So x can be 6, 14, 16, and the answer to our stem would be Yes, No, No
Hence not sufficent.

Together:
x can take values 0, 6, 12, 18
and we get the answer to our stem as only Yes.
Hence sufficient.

Bunuel I just faced this question during an MBA mock GMAT and they give answer E as the correct option - any insights?

The correct answer is C. Either you are mistaken or the mock is wrong. Can you please post a screenshot of the question? Thank you.
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