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Test Prep

by moneyman » Sun Mar 16, 2008 6:33 am
What is the remainder when 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 then the remainder is 0.

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

The OA is D

I just have one doubt. What if x is 3 according to statement 1 then how do you treat it ?
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Re: Test Prep

by gabriel » Mon Mar 17, 2008 1:00 am
moneyman wrote:What is the remainder when 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 then the remainder is 0.

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

The OA is D

I just have one doubt. What if x is 3 according to statement 1 then how do you treat it ?
The answer is wrong it cant be D. It should be B.
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by amit_gmatter2008 » Mon Mar 17, 2008 12:49 pm
My answer - D

Reason

Consider 1.
x=2k+1 and x=3p, where k and p are integers >=0
So what can be the possible values
from x=2k+1, x={1,3,5,7,9...}
from x=3p, x={3,6,9,12,15...}
from both x={3,9,15....}
hence, when x is divided by 6, u get remainder 3. Hence 1 is sufficient.

Consider 2
x=12q+3
Hence when x is divided by 6, it will always give the remainder 3.
Hence 2 is sufficient.

Hence the answer is D
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by ikant » Tue Mar 18, 2008 3:08 am
The explanation given by amit is correct. One needs to test the common subject in the above case.
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Re: Test Prep

by Stuart@KaplanGMAT » Tue Mar 18, 2008 2:51 pm
moneyman wrote:What is the remainder when 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 then the remainder is 0.

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

The OA is D

I just have one doubt. What if x is 3 according to statement 1 then how do you treat it ?
(1) the first piece of information tells us that X is odd. The second piece of information tells us that x is a multiple of 3. Together, we know that x is an odd multiple of 3, i.e. {3, 9, 15, 21, ...}. Every odd multiple of 3 divided by 6 has a remainder of 3: sufficient.

If we were to let x=3, we'd still get a remainder of 3, since 3 divided by 6 = 0rem3.
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by moneyman » Wed Mar 19, 2008 10:25 pm
Hi Stuart,
Can you please tell me how do you get a remainder of 3 when 3 is divided by 6 ?
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by Stuart@KaplanGMAT » Thu Mar 20, 2008 10:25 am
moneyman wrote:Hi Stuart,
Can you please tell me how do you get a remainder of 3 when 3 is divided by 6 ?
The remainder is how much is left over when you divide through as many times as you can.

So, how many times can we "shove" 6 into 3? None. So 6 "goes into" 3 zero times.

How much of the 3 is left over? All of it, so the remainder is 3.

Another way we can think of remainders is as the numerator of the fraction when we write the problem as a mixed fraction.

For example, 25/6 = 4 and 1/6. So, the remainder when we divide 25 by 6 is 1.

31/7 = 4 and 3/7. So, the remaineder when we divide 31 by 7 is 3.

Applying that to 3 divided by 6, as a mixed fraction it's simply 3/6. Therefore, 3 divided by 6 has a remainder of 3.

Note: when using the mixed fraction method, we never reduce the fraction.
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