doubt here

Just adding one thing here, something that may be useful in terms of generalizing this concept to other problems. If you are ever told that "when x is divided by 2, the remainder is 1," that's the exact same thing as being told that "x is odd." (And of course, being told that "when x is divided by n, the remainder is 0" is the same as being told that "x is a multiple of n".) So in this particular problem, we are effectively being told that "x is an odd multiple of 3." Multiples of *any* odd number n will flip-flop odd to even to odd to even to odd and so on. The even ones will always be multiples of 2n; the odd ones will always yield a remainder of n when divided by 2n.

For instance, considering the remainders when multiples of 7
7 14 21 28 35 42 49 56 63 70 77 84 ...
are divided by 14, I'll get
7 0 7 0 7 0 7 0 7 0 7 0 ...

Works every time

here we have that x is divisible by 3 and when divided by 2 we will get the remainder 1 so for example choose the numbers for this condition = 3, 9,15,21,27,33,39.....


when we divide these numbers by 6 in every case we will get the remainder as 3.

so the answer should be 3