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Bunuel
Is m a multiple of 6?

(1) More than 2 of the first 5 positive integer multiples of m are multiples of 3.
(2) Fewer than 2 of the first 5 positive integer multiples of m are multiples of 12.

Hi, can you explain why A is not suff?

First 5 positive integer multiples of m are: m, 2m, 3m, 4m, 5m
If m = 3: 3, 6, 9, 12, 15 --> only 2 are multiples of 6 (6, 12)
If m = 9: 9, 18, 27, 36, 45 --> only 2 are multiples of 6 (18, 36)
For more than 2 to be multiples of 6, m itself should be even and divisible by 3. Thus A suff.

What am I missing here?
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ailyninja

For more than 2 to be multiples of 6, m itself should be even and divisible by 3. Thus A suff.

What am I missing here?

You don't know that more than two are multiples of 6 -- if you did, then yes, you're correct, that would be sufficient information. But Statement 1 only tells you that more than two are multiples of 3, not of 6.
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ailyninja

For more than 2 to be multiples of 6, m itself should be even and divisible by 3. Thus A suff.

What am I missing here?

You don't know that more than two are multiples of 6 -- if you did, then yes, you're correct, that would be sufficient information. But Statement 1 only tells you that more than two are multiples of 3, not of 6.

Ahhh!! Right!! Thanks for pointing out that I totally mis-read the question
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can someone explain this question by steps, how is it that we are using two particular numbers, those numbers can be anything right, it is mentioned more than 2 ?
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Bunuel
Is m a multiple of 6?

(1) More than 2 of the first 5 positive integer multiples of m are multiples of 3.
(2) Fewer than 2 of the first 5 positive integer multiples of m are multiples of 12.

Statement 1:
Case 1: m=3 --> first 5 positive multiples of m are 3, 6, 9, 12, 15 --> five multiples of 3
In this case, m is not a multiple of 6, so the answer to the question stem is NO.
Case 2: m=6 --> first 5 positive multiples of m are 6, 12, 18, 24, 30 --> five multiples of 3
In this case, m is a multiple of 6, so the answer to the question stem is YES.
Since the answer is NO Case 1 but YES in Case 2, INSUFFICIENT.

Statement 2:
Notice that Case 2 does not satisfy Statement 2, since the yielded list of values -- 6, 12, 18, 24, 30 -- includes two multiples of 12.
Test where greater multiples of 6 will satisfy Statement 2.
Case 2: m=12 --> first 5 positive multiples of m are 12, 24, 36, 48, 60 --> five multiples of 12
Case 3: m=18 --> first 5 positive multiples of m are 18, 36, 54, 72, 90 --> two multiples of 12
None of these cases is viable, since each yields a list that includes at least two multiples of 12.
Implication:
For Statement 2 to be satisfied, m CANNOT be a multiple of 6.
Thus, the answer to the question stem is NO.
SUFFICIENT.

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Bunuel
Is m a multiple of 6?

(1) More than 2 of the first 5 positive integer multiples of m are multiples of 3.
(2) Fewer than 2 of the first 5 positive integer multiples of m are multiples of 12.

Does this mean that:

(1) m is any number that is a multiple of 3?
If 3, then 3, 6, 9, 12, 15 are multiples of m.
If 6, then 6, 12, 18, 24, 30 are multiples of m.
If so, then m could be any multipe of 3, and all of them are not multiples of 6.

(2) 12, 24, 36, 48, 60 are multiples of 12.

What am i missing here? If m is 1 then fewer than 2 of the first 5 are multiples of 12, obviously. The first five is 1, 2, 3, 4, 5.
If m is 2, then fewer than 2 of the first 5 are multiples of 12. The first five is 2, 4, 6, 8, 10.

How could we tell anything from this information?

I feel that I dont understand the word multiple correctly even after reading definitions about it.


Edit: I think I get it.

1) if 3, then 3, 6, 9, 12, 15 are multiples of m. if 6, then (6), 9, (12), 15, (18) and thus more than 2.

2) says that if m is 6, then the five multiples would have been 6, 12, 18, 24, 30, i.e. at least 2 multiples of 12 (12, 24), so m can not be 6. For every other number than 6, or multiples of 6 or 12, statement 2 holds true. So the answer is No, m is not a multiple of 6.

Good lord, this took me 30 minutes to grab. However, many a mickle makes a muckle!
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Bunuel
Is m a multiple of 6?

(1) More than 2 of the first 5 positive integer multiples of m are multiples of 3.
(2) Fewer than 2 of the first 5 positive integer multiples of m are multiples of 12.

For (1), look at 6, 12, 18, 24, 30. All of these multiples of 6 are also multiples of 3. However, m = 3 is also a multiple of 3, but not a multiple of 6.

For (2), look at 6, 12, 18, 24, 30. Both 12 and 24 are multiples of 12. This means that m can not be 6. m can not be 12 either, or not even 18, as all of these have at least two of five first multiples that are also multiples of 12. Lets return to low numbers:

Both 3 and 4 satisfies statement (2). 3, 6, 9, 12, 15 = fewer than two. 4, 8, 12, 16, 20 = fewer than two. No matter if we choose m = 3 or m = 4, m can not be a multiple of 6.

Edit: Oh, I already answered this once, but Im going for repetition round now. :blushing:
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Bunuel
Is m a multiple of 6?

(1) More than 2 of the first 5 positive integer multiples of m are multiples of 3.
(2) Fewer than 2 of the first 5 positive integer multiples of m are multiples of 12.

Hi, can you explain why A is not suff?

First 5 positive integer multiples of m are: m, 2m, 3m, 4m, 5m
If m = 3: 3, 6, 9, 12, 15 --> only 2 are multiples of 6 (6, 12)
If m = 9: 9, 18, 27, 36, 45 --> only 2 are multiples of 6 (18, 36)
For more than 2 to be multiples of 6, m itself should be even and divisible by 3. Thus A suff.

What am I missing here?

Hi,

The number can be 3 or 9. Hence not a multiple of 6

hope I am correct>
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