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Problem on Multiples

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Problem on Multiples

by Uri » Thu Jan 15, 2009 8:50 am
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.

Official Answer: B

Please explain your methodology.
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Re: Problem on Multiples

by logitech » Thu Jan 15, 2009 11:18 am
Is m a multiple of 6?
(1) More than 2 of the first 5 positive integer multiples of m are multiples of 3.

m 2m 3m 4m 5m

since more than 2 are multiples of 3, m is divisible with 3

3m is the first one so there has to be another 3 somewhere :)

But is M an even ? we dont know so it is divisible with 3 but not sure of 6

INSUF

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

m 2m 3m 4m 5m

Fever than 2 = 1

so that would be m

m is a multiple of 12

so it is a multiple of 6

Choose B
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by 720dreaming » Thu Jan 15, 2009 11:33 am
Couldn't fewer than 2=0?
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by maihuna » Thu Jan 15, 2009 11:36 am
I have following reasoning:

Let us take 9 : 9 18 27 36 45 more than 3 are multiples of 3, not a multiple of 6 though.

for 12: 12, 24, 36, 48, 60 etc, more than 3 are multiple of 3 12 is a multiple of 3.

So A is out.

For B, though all example I can think of are negative, can one choose a number whose 0 or 1 multiple of the same are positive.

e.g. 15 is not 6 multiple and multiples are: 15 30 45 60 75, only 60 is a multiple of 12, and 15 is not a multiple of 6.

how to conclude there ae no such scenario where if at the most one is multiple of 12, it will have multiple of 6 too...

honestly, I would have choosen E in exam env's..
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Re: Problem on Multiples

by maihuna » Thu Jan 15, 2009 11:38 am
logitech wrote:Is m a multiple of 6?
(1) More than 2 of the first 5 positive integer multiples of m are multiples of 3.

m 2m 3m 4m 5m

since more than 2 are multiples of 3, m is divisible with 3

3m is the first one so there has to be another 3 somewhere :)

But is M an even ? we dont know so it is divisible with 3 but not sure of 6

INSUF

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

m 2m 3m 4m 5m

Fever than 2 = 1

so that would be m

m is a multiple of 12

so it is a multiple of 6

Choose B
See my above post, 15 doesn't hold on your explanation, you have given an reverse opposite it seems, can you put exact numerals supporting your claim.
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by maihuna » Thu Jan 15, 2009 11:47 am
I have decipher it correctly it seems, here follows the explanation:

A) I have already posted how two scenario are possible.

B) 6 has 2 and 3 as prime multiples,
If 1st five multiples contains at the most one multiple of 12 it will be: 2, 2 3

Pick a number and try to multiply it by 1 2 3 4 5, out of these a group of 6 will be possible with pairs of missing 3 only as 12 contains 2 2 3 and 6 requires 2 3 so a couple or more requires 2 3 2 3 that is pairs of 2's and 3's, but as it is saying out of the 1 2 3 4 5 only one multiple of 12 restricts our number or multiplier to 3 or 2

so any number will always lack a pair of 2 and 3 as otherwise we will get more than 2 multiples of 12, using either 3 or 2 and 4

So no number with 6 as a multiple can justify that. And so B says there are no such multiples of 6 ans is answer.
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Re: Problem on Multiples

by coffee5251 » Thu Jan 15, 2009 11:50 am
logitech wrote: (2) Fewer than 2 of the first 5 positive integer multiples of m are multiples of 12.

m 2m 3m 4m 5m

Fever than 2 = 1

so that would be m

m is a multiple of 12

so it is a multiple of 6

Choose B
Hi, maybe I'm not understanding the question, but why does m have to be a multiple of 12? What if m is 6, then 4m is 24 which is a multiple of 12 and still satisifies the statement that only 1 number is a multiple of 12.

Could you explain this please? Thanks!
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Re: Problem on Multiples

by Stuart@KaplanGMAT » Thu Jan 15, 2009 12:43 pm
Uri wrote: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.

Please explain your methodology.


Statement (2) is sufficient, but it actually gives us a definite "NO" answer.

What if m=7? Then the first 5 positive integer multiples of m are:

7, 14, 21, 28, 35.

Fewer than 2 of those are multiples of 12, so it fits the rule. Is 7 a multiple of 6? NO.

We've got a "no", let's try to get a "yes".

if m=6, then the first 5 positive integer multiples of m are:

6, 12, 18, 24, 30.

2 of those are multiples of 12, so it violates the rule. Therefore, we CANNOT pick m=6.

if m=12, then the first 5 positive integer multiples of m will ALL be multiples of 12, so we clearly cannot pick m=12. Further, no multiple of 12 will be a permissible value for m.

In fact, we should be able to see that no multiple of 6 is permissible, since the first 5 multiples of m are:

m, 2m, 3m, 4m and 5m

and if m is a multiple of 6, both 2m and 4m will always be multiples of 12.

Therefore, if FEWER than 2 of the first 5 multiples of m are multiples of 12, m CANNOT be a multiple of 6: sufficient.
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Thanks!

by Uri » Mon Jan 19, 2009 3:16 am
Thanks all for participating. I feel the first part of the first post by maihuna coupled with the post of Stuart provides the right direction. Thanks once again to all!
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