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# Is the integer x divisible by 6?

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Manager
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Is the integer x divisible by 6? [#permalink]

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14 Feb 2008, 12:37
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Is the integer x divisible by 6?

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

[Reveal] Spoiler:
I get C.
As, say x = 6.
x+3 = 9 which is odd and divisible by 3.

But what is we put in x as "0".
x + 3 is 3 which is also odd and divisible by 3 but not 6.

Any idea what am I missing?
[Reveal] Spoiler: OA

Last edited by Bunuel on 12 Jun 2014, 11:19, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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14 Feb 2008, 13:16
jackychamp wrote:
Is the integer X divisible by 6?
1. x+3 is divisible by 3
2. x+3 is an odd number.

I get C.
As, say x = 6.
x+3 = 9 which is odd and divisible by 3.

But what is we put in x as "0".
x + 3 is 3 which is also odd and divisible by 3 but not 6.

Any idea what am I missing?

you are right, the answer is C. 0 is divisibly by 6.
1 -> x is divisibly by 3
2 -> x is an even number (divisible by 2)
1&2 -> x is divisible by 6
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14 Feb 2008, 13:27
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I. If x+3 divisible by 3 => x divisible by 3 but it doesn't say that is divisible by 2 also. Not suff
II. x+3 odd => x must be even. Alone is not sufficient

I and II say X is an even number (which we know must be divisible by 2) and x divisible by 3. From here we can conclude that x is divisible by 6.

C
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Re: Is the integer X divisible by 6? 1. x+3 is divisible by 3 2. [#permalink]

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12 Jun 2014, 11:10
jackychamp wrote:
Is the integer X divisible by 6?
1. x+3 is divisible by 3
2. x+3 is an odd number.

I get C.
As, say x = 6.
x+3 = 9 which is odd and divisible by 3.

But what is we put in x as "0".
x + 3 is 3 which is also odd and divisible by 3 but not 6.

Any idea what am I missing?

In II) You do not know that X+3 is divisible by 3, so, if you put 0, the answer is no, because 0+3 is odd but is not divisible by 6. However, you can put 6+3=9 that is odd and divisible by 6 (X=6) so, that's why II) is not sufficient. We have found two numbers that fit with the statement but we can not answer the main question.

Hope this helps.

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Re: Is the integer x divisible by 6? [#permalink]

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12 Jun 2014, 11:25
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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.

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Is the integer x divisible by 6? [#permalink]

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08 Sep 2015, 07:30
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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Re: Is the integer x divisible by 6? [#permalink]

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31 Oct 2016, 09:36
Bunuel : If x = 0 (which is an even no.) + 3 is divisible by 3.
But 0/6 is not divisible. Am i missing something?
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Re: Is the integer x divisible by 6? [#permalink]

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31 Oct 2016, 09:38
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Vishvesh88 wrote:
Bunuel : If x = 0 (which is an even no.) + 3 is divisible by 3.
But 0/6 is not divisible. Am i missing something?

0/6 = 0 = integer. 0 is divisible by every integer except 0 itself.
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Re: Is the integer x divisible by 6? [#permalink]

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31 Oct 2016, 09:42
Bunuel : Muchos Gracias Bunuel.
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Re: Is the integer x divisible by 6? [#permalink]

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15 Jan 2017, 23:57
Great Official Question.
From statement we get => x is a multiple of 3.=> Not sufficient.
From statement 2 we get => x is even. Not sufficient.

Combining the two statements => x is even multiple of 3 => multiple of 6.
Hence sufficient.
Hence C.

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Is the integer x divisible by 6? [#permalink]

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24 Jun 2017, 22:24
Given
Is the integer x divisible by 6?

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

Question asks whether x is divisible by 6 i.e. x is divisible by 3 AND x is divisible by 2.

Statement 1, x + 3 is divisible by 3 => i.e . x is divisible by 3, but no information about x divisible by 2. So insufficient.

Statement 2, x+3 is odd number => i.e. x must be even number. So any even number must be divisible by 2. But no information about x divisible by 3. Clearly insufficient.

combining statement 1 & 2, X is divisible by 3 AND x is divisible by 2. So Answer is D.
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Re: Is the integer x divisible by 6? [#permalink]

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25 Jun 2017, 07:34
[quote="jackychamp"]Is the integer x divisible by 6?

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

Solution :

Statement 1:
Case 1: x=6. We get x to be a divisible of 6.

Case 2: x=15. We get x not to be a divisible of 6.
Therefore this statement is not sufficient on its own.

Statement 2:
Case 1: If x=10. We get x not to be a divisible of 6.
Case 2: If x=12. We get x not to be a divisible of 6.
Therefore this statement is not sufficient on its own.

Combining St1 and St2,
We get x to be divisible by 6.

Therefore the answer is Option C.
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Re: Is the integer x divisible by 6? [#permalink]

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25 Jun 2017, 07:36
JTR wrote:
Given
Is the integer x divisible by 6?

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

Question asks whether x is divisible by 6 i.e. x is divisible by 3 AND x is divisible by 2.

Statement 1, x + 3 is divisible by 3 => i.e . x is divisible by 3, but no information about x divisible by 2. So insufficient.

Statement 2, x+3 is odd number => i.e. x must be even number. So any even number must be divisible by 2. But no information about x divisible by 3. Clearly insufficient.

combining statement 1 & 2, X is divisible by 3 AND x is divisible by 2. So Answer is D.

No its wrong. You are assuming things. Try to get a scenario that makes the statements invalid. Please go through my solution if required. Thanks
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Re: Is the integer x divisible by 6? [#permalink]

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17 Jul 2017, 22:57
jackychamp wrote:
Is the integer x divisible by 6?

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

[Reveal] Spoiler:
I get C.
As, say x = 6.
x+3 = 9 which is odd and divisible by 3.

But what is we put in x as "0".
x + 3 is 3 which is also odd and divisible by 3 but not 6.

Any idea what am I missing?

St 1
(X +3) /3= some integer k this just implies that X has to be a multiple of 3 but it could be exactly 3- in which case X is too small to be divisible by 6
insufficient
St 2
X+3 = odd - this just means that X has to be an even number
insufficient

St 1 and St 2

Using inferences from St 1 and St 2 - X must be an even multiple of 3- the smallest of which is 6 which is clearly divisible by 6 - any even multiple of 3 is divisible by 6
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Re: Is the integer x divisible by 6? [#permalink]

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30 Jul 2017, 17:29
jackychamp wrote:
Is the integer x divisible by 6?

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

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.

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Re: Is the integer x divisible by 6? [#permalink]

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30 Jul 2017, 17:42
Bunuel wrote:
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.

Hi Bunuel, Do we need to consider negative integer for statement (1)?
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Re: Is the integer x divisible by 6? [#permalink]

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30 Jul 2017, 20:33
hazelnut wrote:
Bunuel wrote:
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.

Hi Bunuel, Do we need to consider negative integer for statement (1)?

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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Re: Is the integer x divisible by 6? [#permalink]

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25 Oct 2017, 20:08
jackychamp wrote:
Is the integer x divisible by 6?

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

[Reveal] Spoiler:
I get C.
As, say x = 6.
x+3 = 9 which is odd and divisible by 3.

But what is we put in x as "0".
x + 3 is 3 which is also odd and divisible by 3 but not 6.

Any idea what am I missing?

Statement 1
This just means that X must be a multiple of 3: 3,6,9 , etc

insuff

Statement 2
This just means X must be even : 2,4,6,8, etc

insuff

Statement 1 & 2
X must be an even multiple of 3 and because X must be an integer, X must at least be 6

suff
Re: Is the integer x divisible by 6?   [#permalink] 25 Oct 2017, 20:08
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