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DS: Remainder

Problem Solving — algebra and arithmetic (GMAT Focus Edition)
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DS: Remainder

by haidgmat » Sun Nov 21, 2010 11:23 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 but how come??? I answered B
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by shovan85 » Sun Nov 21, 2010 10:19 pm
haidgmat wrote: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 but how come??? I answered B
Remainder when the positive integer x is divided by 6

Option 1:

When x is divided by 2, the remainder is 1
So we can write in general x = 2k + 1 (k is an integer) [1,3,5,7,.... (Odds)]

when x is divided by 3, the remainder is 0
So we can write in general x = 3k (k is integer) [0,3,6,9,12,15....]

Combine both,
from second set we need to find the odds as those will be div by 3 not by 2.

So those are 3,9,15,...

When these are divided by 6 we get a Remainder 3 (always)

Thus Sufficient.

Option 2: When x is divided by 12, the remainder is 3

In general x = 12k + 3
When this is divided by 6 remainder is always 3.

Thus Sufficient.

IMO D
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by haidgmat » Mon Nov 22, 2010 1:04 am
But 3 isnt divisible by 6! Isn't remainder the number which is left behind after a number is divided by another number?
shovan85 wrote:
haidgmat wrote: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 but how come??? I answered B
Remainder when the positive integer x is divided by 6

Option 1:

When x is divided by 2, the remainder is 1
So we can write in general x = 2k + 1 (k is an integer) [1,3,5,7,.... (Odds)]

when x is divided by 3, the remainder is 0
So we can write in general x = 3k (k is integer) [0,3,6,9,12,15....]

Combine both,
from second set we need to find the odds as those will be div by 3 not by 2.

So those are 3,9,15,...

When these are divided by 6 we get a Remainder 3 (always)

Thus Sufficient.

Option 2: When x is divided by 12, the remainder is 3

In general x = 12k + 3
When this is divided by 6 remainder is always 3.

Thus Sufficient.

IMO D
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by shovan85 » Mon Nov 22, 2010 2:02 am
haidgmat wrote:But 3 isnt divisible by 6! Isn't remainder the number which is left behind after a number is divided by another number?
Yes 3 is not divisible by 6. But, when 3 is divided by 6 the remainder is 3 and quotient is 0 (zero). And thats what has been proved in Option number 1.

When you consider x as any of the set (3, 9, 15, 21....) and divide the x by 6 you will ALWAYS get the remainder 3.

AS you are always getting remainder as 3 this option is sufficient.

If still not clear ask yourself is 15 divisible by 6? (No. So as 3 but remainder is 3 for both the case )
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by haidgmat » Mon Nov 22, 2010 6:17 am
gotcha. I got part B but wasn't sure why A is also suff. Thanks bud!
shovan85 wrote:
haidgmat wrote:But 3 isnt divisible by 6! Isn't remainder the number which is left behind after a number is divided by another number?
Yes 3 is not divisible by 6. But, when 3 is divided by 6 the remainder is 3 and quotient is 0 (zero). And thats what has been proved in Option number 1.

When you consider x as any of the set (3, 9, 15, 21....) and divide the x by 6 you will ALWAYS get the remainder 3.

AS you are always getting remainder as 3 this option is sufficient.

If still not clear ask yourself is 15 divisible by 6? (No. So as 3 but remainder is 3 for both the case )
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