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If the positive integer y is divisible by 3, 8, and 12, then which of

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If the positive integer y is divisible by 3, 8, and 12, then which of  [#permalink]

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New post 03 Sep 2014, 22:25
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If the positive integer y is divisible by 3, 8, and 12, then which of the following must y be divisible by?

I. 24
II. 36
III. 48

A. I only
B. II only
C. I and II only
D. II and III only
E. I, II, and III
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Re: If the positive integer y is divisible by 3, 8, and 12, then which of  [#permalink]

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New post 03 Sep 2014, 22:46
neeti1813 wrote:
If the positive integer y is divisible by 3, 8, and 12, then which of the following must y be divisible by?

I. 24
II. 36
III. 48

A. I only
B. II only
C. I and II only
D. II and III only
E. I, II, and III


Answer = A

LCM of 3, 8, 12 = 24

Answer = 24

It seems this problem is already present in the forums where Bunuel has explained difference between must & could used
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If the positive integer y is divisible by 3, 8, and 12, then which of  [#permalink]

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New post 04 Sep 2014, 09:26
PareshGmat wrote:
neeti1813 wrote:
If the positive integer y is divisible by 3, 8, and 12, then which of the following must y be divisible by?

I. 24
II. 36
III. 48

A. I only
B. II only
C. I and II only
D. II and III only
E. I, II, and III


Answer = A

LCM of 3, 8, 12 = 24

Answer = 24

It seems this problem is already present in the forums where Bunuel has explained difference between must & could used


I always search for the problems on GMAT club & then post my query. Sorry but I couldn't find anything similar to this & as Kaplan gave such a huge explanation I got a bit confused.

Oh ! Must & Could I will find it.

Thanks !!


Here is the detailed explanation:

Let’s look at the prime factors of 3, 8 and 12.

The prime factorization of 3 is 3.

Let's find the prime factorization of 8. We have 8 = 2 × 4 = 2 × 2 × 2.

Let's find the prime factorization of 12. We have 12 = 2 × 6 = 2 × 2 × 3.

Thus,


A multiple of 3, 8, and 12 must have at least three factors of 2 and one factor of 3, or 2 x 2 x 2 x 3.

Let's write the prime factorizations of 24, 36, and 48.


This is a Roman Numeral question, so let’s start with the numeral that appears most in the answer choices or Statement I. Since 24 has 3 factors of 2 and one factor of 3, y is a multiple of 24. So the correct answer will contain Statement I. We can eliminate choices (B) and (D) which do not contain Statement I.

Since 36 contains two factors of 3, but not three factors of 2, the correct answer will not contain II. We can eliminate choices (C) and (E) which contain Statement II. We have eliminated the 4 incorrect answer choices, so we know that choice (A) must be correct. Although we’ve eliminated all the wrong answer choices, we can that show that Statement III is incorrect. Since 48 contains four factors of 2, while y must contain at least three factors of 2, y does not have to be a multiple of 48, because y could contain exactly three factors of 2. The correct answer will not contain Statement III. Answer Choice (A) is correct.
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Re: If the positive integer y is divisible by 3, 8, and 12, then which of  [#permalink]

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New post 04 Sep 2014, 22:01
neeti1813 wrote:
If the positive integer y is divisible by 3, 8, and 12, then which of the following must y be divisible by?

I. 24
II. 36
III. 48

A. I only
B. II only
C. I and II only
D. II and III only
E. I, II, and III


The official solution is overly complicated for no reason.

y needs to be divisible by 3, 8 and 12. This means that it must be the LCM of 3, 8 and 12 or a multiple of the LCM. The LCM is the smallest number which is divisible by all 3 numbers.

LCM of 3, 8 and 12 will be 3*8 = 24

Now y must be either 24 or a multiple of 24. In any case it will always be divisible by 24.
Will it be divisible by 36 necessarily? No. Say if y is 24, it will not be divisible by 36.
Will it be divisible by 48 necessarily? No. Say if y is 24, it will not be divisible by 48.

Answer (A)
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Re: If the positive integer y is divisible by 3, 8, and 12, then which of  [#permalink]

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New post 09 Feb 2018, 11:09
neeti1813 wrote:
If the positive integer y is divisible by 3, 8, and 12, then which of the following must y be divisible by?

I. 24
II. 36
III. 48

A. I only
B. II only
C. I and II only
D. II and III only
E. I, II, and III


The LCM of 3, 8, and 12 is 24. So y must be 24 or a multiple of of 24. Let’s test each answer choice.

We see that Roman numeral I is correct; if y = 24, then y/24 = 24/24 = 1.

Roman numeral II is not a multiple of 24, so it is not correct.

You might think that Roman numeral III, the value 48, should be correct, but it isn’t. If y were equal to 24, then y is NOT divisible by 48, because y/48 = 24/48 does not yield an integer as the answer.

The only answer that MUST be correct is y = 24.

Answer: A
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Re: If the positive integer y is divisible by 3, 8, and 12, then which of &nbs [#permalink] 09 Feb 2018, 11:09
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