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by sandranjeim » Mon Apr 26, 2010 9:09 am
What is the remainder when the positive integer x is divided by 6?

1) When x is divided by 2, the remainder is 1; and when x is divided by 3, the remainder is 0.

2) When x is divided by 12, the remainder is 3.


OA D
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Source: — Data Sufficiency |
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by iamseer » Mon Apr 26, 2010 11:30 am
From 1:

X=2n+1 and X=3m, where m and n are integers. So basically X takes values which are odd multiples of 3 e.g. 3,9,15 etc
when divided by 6 remainder is always 3
SUFFICIENT

FROM 2:
X=12n+3 So basically X take values like 15, 27 etc (12n is always divisible by 6 leaving 3 as remainder)
when divided by 6 remainder is always 3
SUFFICIENT

IMO answer D
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by tanviet » Tue Apr 27, 2010 4:16 am
How do you know (1) is sufficient?

only by plugging specific numbers?
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by liferocks » Tue Apr 27, 2010 4:32 am
duongthang wrote:How do you know (1) is sufficient?

only by plugging specific numbers?
from 1 we get x=3(2n+1) where n is any integer
or x=6n+3..so when x is divided by 6 reminder is 3
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by tpr-becky » Tue Apr 27, 2010 3:35 pm
if the remainder is 1 when divided by 2 then we know that x is odd - becuase when you divide by 2 you either get a remainder of 0 (for an even number) or a remainder of 1 (for an odd number) so the first part says that x is odd. If the remainder is 0 when divided by three that means that the number is divisible by three. So stmt 1 says that x is an odd multiple of three - odd multiples of three are 6 numbers apart which means they will all have the same remainder when divided by 6 (incidentally it is 3 in this case)

AD

Stmt 2 says that x is three more than a multiple of 12 - therefore it is also 3 more than a multiple of 6 (because 12 is a multiple of 6) Therefore sufficient.

D.
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by tanviet » Tue Apr 27, 2010 11:59 pm
liferocks wrote:
duongthang wrote:How do you know (1) is sufficient?

only by plugging specific numbers?
from 1 we get x=3(2n+1) where n is any integer
or x=6n+3..so when x is divided by 6 reminder is 3
from (1) you can not have x=3(2n+1)

from (1) you can have x=3m(2n+1)= 6mn+3m with m is odd

so when x is divided by 6, remainder is 3m-with m is odd

so 3m=3.(2a+1)=6a+3

only now you can say that remainder is 3.


I see this problem is hard. and plugging number is ok. we should not solve above way.

any idea of this problem. pls, remember , we have only 2 minutes for this problem
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by MBA » Mon Sep 27, 2010 9:27 pm
tpr-becky wrote:if the remainder is 1 when divided by 2 then we know that x is odd - becuase when you divide by 2 you either get a remainder of 0 (for an even number) or a remainder of 1 (for an odd number) so the first part says that x is odd. If the remainder is 0 when divided by three that means that the number is divisible by three. So stmt 1 says that x is an odd multiple of three - odd multiples of three are 6 numbers apart which means they will all have the same remainder when divided by 6 (incidentally it is 3 in this case)

AD

Stmt 2 says that x is three more than a multiple of 12 - therefore it is also 3 more than a multiple of 6 (because 12 is a multiple of 6) Therefore sufficient.

D.
HI

If we consider the 1st statement i.e
"X=2n+1 and X=3m, where m and n are int ... always 3 "

We need to consider first X=3,in that case 3 divided by 6 gives a reminder different from ,the reminder whenX=9 is divded by 6.So how the statement 1 is sufficient.
Thanks.
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