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.

The correct answer is D) Each statement alone is sufficient.

The above problem has already been solved, but there is a method by which you can solve the question that I do not clearly understand. I had already posted my queries on that post but since I did not get any reply I am starting a new thread. Sorry Mod, if at all I need to be.

Anyways, this is the part of the problem that I don't understand:

Let N be a number.
When divided by 2 it leaves a remainder 1.
When divided by 3 it leaves no remainder.

N = 2a + 1
N = 3b = (3b - 1) + 1

(3b - 1) is divisible by 2 if b = 1 and min of N is 3

So if N is divided by LCM of 2 & 3 the remainder would be 3

N = 6c + 3
Remainder = 3

My question:
1. What is the need of the part : "(3b - 1) is divisible by 2 if b = 1 and min of N is 3" in reaching N = 6c + 3?
2. The reason I am asking is how do I generalize for other cases. For instance:
N = 3a + 1
N = 5b
How do I generate the N taking into account both the statements?

PS. I knw the ans is: N = 15c + 10.

Thanks,
Jeevan

here's a quick solution:
post5248.html#p5248

if you have further questions, please post them on that thread, not this one (and search the forum next time please!).