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Bunuel
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Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
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Bunuel
Is x divisible by 12?


(1) x is divisible by 27.

(2) x is divisible by 6.

Question Stem -

To find if x is divisible by 12

This means x is to be divisible by its factors i.e. 1, 2, 4, 6, 12.

Option 1 -

x is divisible by 27.

This means x has factors 1, 3, 9, 27

But it does not say anything about factors that are multiple of 2 for x to be divisible by 12

Hence Option 1 itself not sufficient

Option 2 -

x is divisible by 6.

This means x has factors 1, 2, 3, 6.

But it does not count for the factor 4 for x to be divisible by 12.

Hence Option 2 itself not sufficient.

Combining Option 1 & 2 -

x is divisible by 6 and 27.

But again it does not count for the factor 4 for x to be divisible by 12.

Hence both options are also not sufficient.

Hence answer is (E)
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Hi,

This is a common type(Remainders and Divisibility) of Question in GMAT.

Let’s understand the question stem first,

Question: Is x divisible by 12?


For the number to be divisible by 12, it should be multiple of 12.

Like 12, 24,36,…

Or else if you see when we prime factorize 12 i.e., 12 = 2^2 * 3

It has two 2’s and one 3.

So, a number which has two or more 2’s and one or more three is divisible by 12.

Let’s check the statements now,

Statement I is insufficient:

x is divisible by 27.

So, x is a multiple of 27, like 27, 54, 81, 108 etc…

We can see that, 27,54,81 are not multiples of 12 but 108 is.

So, it is insufficient.

Also, Prime factorization of 27 = 3^3.

So, it has only 3’s but no 2’s.

So insufficient.

Statement II is insufficient:

x is divisible by 6.

So, x is a multiple of 6, like, 6, 12, 18, 24, 30 etc…

We can see that, 6, 18, 30 are not multiples of 12 but 12,24, etc are multiples of 12.

So, it is insufficient.

Also prime factorization of 6 = 2 *3

It has only one 2 and only one three. But for a multiple of 12, we need minimum two 2’s and one three.

So insufficient.

Together also it is not sufficient.

With both statements together, x should be a multiple of 6 and 27.

LCM of 6 and 27 is 54.

So, x is 54, 108, 162 etc…

54 is not a multiple of 12 but 108 is a multiple of 12.

So insufficient.

Even together, prime factorization of 54 = 3^3 * 2

But, we need minimum of two 2’s and a 3.

So answer is E.
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Posting a view from my side w.r.t this question. I hope the rest of the aspirants can learn from my mistake.

In my opinion, it's always not advisable to go with trying out values as suggested by some peers above.
This question appeared to me in multiple mock CAT's and I have been doing the same mistake time and again.

On analyzing my error log, which I recently started after suggestions by many achievers, I could see a pattern of mistakes in a time crunch situation.

Hence my approach would be

Q: x = 2*2*3*n ? (where n>= 1)

1) x= 3*3*3*n (where n>= 1) Not Suff as n can be 1(no) or 4(yes)

2) x= 2*3*n (where n>= 1) Not Suff as n can be 1(no) or 2(yes)

1) + 2)
x = 2*3*3*3*n (where n>= 1) Not Suff as n can be 1(no) or 2(yes)

With this post, I wanna put an end to this silly mistake during online CAT

+1 Kudos if you ever faced this and this helps.

Cheers,
H
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