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is m a multiple of 6?

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Source: — Data Sufficiency |
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by raghavsarathy » Sun Jul 19, 2009 8:15 am
Statement 1

Example m =3

Here m is not a multiple of 6 but satisties 1.

Counter example m = 6 Satisfies both the main statement and statement 1

Hence 1 is insufficient

Statement 2

Surely not sufficient

Eg m= 11(not a multiple of 6) and m= 12(multiple of 6)

Hence 2 is insufficient

Combining 1 and 2

Eg m=3 (satisfies both the statements)

First 5 multiples of 3
3
6
9
12 -- multiple 0f 12
15

No other value of m satisfies both statements. Take m=6 (does not satisfy statement 2)

Hence the only values of m are 3 and -3... Both these values are not multiples of 3..

Hence both statements are together sufficient
C
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by Sher1 » Sun Jul 19, 2009 9:57 am
I would think its B

if a number is a multiple of 12...12,24,36,,48,60, it will be a multiple of 6.
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by EbrahimHashem » Sun Jul 19, 2009 11:43 am
Think Harder, Guys :D
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by yogami » Sun Jul 19, 2009 12:26 pm
I would go with E
Because 1 is obviously insuff 3 x 5 x 7 x 9 x 27 = whatever number. More than 2 are multiples of 3 but we dont have a 2 in there so no multiples of 6

As per (2) It says fewer than 2 are multiples of 12. But fewer than 2 can imply none too right unless otherwise specified that fewer than 2 but at least one.
So this is insuff too. If it was at least one then B alone could have been suff

combining yields nothing so E
200 or 800. It don't matter no more.
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by kaulnikhil » Sun Jul 19, 2009 11:33 pm
should be E
3*9*27*81*13
satisfy both answer NO
3*9*12*81*13
answer YES
both together not sufficient ..
IMO E
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by real2008 » Mon Jul 20, 2009 2:16 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.

stmt1:

one case: 6,12,18,24,30 (m=6)[1m,2m,3m,4m,5m]; more than 2 of the nos. are divisible by 3

another case: 3,6,9,12,15 (m=3) ; more than 2 of the nos. are divisible by 3

not sufficient.

stmt2:

if m=6

then nos. are 6, 12, 18,24,30
but then stmt2 can't be true.

hence m is not equal to 6

sufficient

Hence B.
Last edited by real2008 on Mon Jul 20, 2009 12:37 pm, edited 1 time in total.
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by axat » Mon Jul 20, 2009 3:14 am
The question talks for the first 5 positive integer multiples.
These are 1xm, 2xm, 3xm, 4xm, 5xm

Now for statement one, think of the first five multiples of 3, namely, 3,6,9,12,15.
Here more than 2 are multiples of 3(well, they all are multiples of 3,) but that doesn't make 3 a multiples of 6).

For statement 2: think of the table of 6, namely, 6,12,18,24,30. or of any other multiple of 6(say w), it can't have less than 2 multiples of 12 because wx2 and wx4 will always be multiples of 12.
Thus statement 2 assures us that m cannot possibly be a multiple of 6.

I hope this helps.
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by Domnu » Mon Jul 20, 2009 5:12 pm
I agree with B. Here's a logic statement:

m is a multiple of 6 implies that at least 2 of m, 2m, 3m, 4m, 5m

are divisible by 12; particularly, the numbers in question are 2m and 4m. Taking the contrapositive of the above,

Fewer than 2 of m, 2m, 3m, 4m, 5m being multiples of 12 implies that m is not a multiple of 6,

which is definite.
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by tohellandback » Mon Jul 20, 2009 6:55 pm
IMO B

1) numbers are m,2m,3m,4m,5m
more than two numbers are multiples of 3. it can be possible when m =3 or m=6
so INSUFF

2) m,2m,3m,4m,5m
case1: no multiple of 12
possible when m is not a multiple of 3 or 6
case2: 1 multiple of 12
possible when m is multiple of 3 but not a multiple of 6

so m not a multiple of 6
SUFFICIENT
The powers of two are bloody impolite!!
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