when a number A is divided by 6, the remainder is 3 and when another

We can write A as \(A = 6i + 3\) where \(i\) is any nonegative integer. Similarly \(B = 12j + 9\) (Note: do not use \(i\) here as A and B are unrelated). Next we can write \(A^2 + B^2\) as:

\((6i + 3)^2 + (12j+ 9)^2 \)

Now this next step is important, we are trying to find the remainder after dividing by 12. We can immediately eliminate any term that is a multiple of 12 since those terms will be divisible by 12, and do not affect the remainder. Whatever left that is not divisible by 12 will be part of our remainder.

To demonstrate the first step: \((12j+ 9)^2\) will give us three terms but we already know two of them are multiples of 12. The only useful term we will keep is \(9^2 = 81\).

Thus \((6i + 3)^2 + (12j+ 9)^2 => 36i^2 + 2 * 6i * 3 + 9 + 81 => 9 + 81 => 90 => 6\).

90/12 gives us a remainder of 6.

Ans: C