Macsen wrote:

What is the remainder when a 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.

I chose B, but the OA says wrong. Can anyone please explain how do we derive a remainder from statement 1? Because from stmt 1, x can be 3 or anything in the form 3n, where n is odd.

Stmnt 1: When x is divided by 2, the remainder is 1; - It means x is odd.

and when x is divided by 3, the remainder is 0. - It means x is divisible by 3

So basically, stmnt 1 tells us that x is an odd multiple of 3 e.g. 3, 9, 15 etc. When you divide these numbers by 6, you will always get 3 as remainder.

Deduce it logically - x i.e. an odd multiple of 3 will always be 3 more than an even multiple of 3 since odd and even multiples of 3 will alternate (e.g. 3 (O), 6(E), 9(O), 12(E), 15(O), 18(E) etc)

Every even multiple of 3 is divisible by 6 so we can say that odd multiples of 3 are always 3 more than multiples of 6. Hence the remainder when you divide x by 6 will always be 3.

Stmnt 2: x is 3 more than a multiple of 12. Since 12 is divisible by 6, x is 3 more than a multiple of 6 too. Hence, remainder is always 3.

Answer (D)

For further theory on divisibility, check:

http://www.veritasprep.com/blog/2011/04 ... unraveled/http://www.veritasprep.com/blog/2011/04 ... y-applied/http://www.veritasprep.com/blog/2011/05 ... emainders/
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[b]Karishma

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