When the integer $-32$ is divided by $3$, the quotient $q$ and remainder $r$ (with $0 \le r < 3$) satisfy:

By definition the remainder must satisfy $0 \le r < d$ where $d$ is the (positive) divisor. So $r$ cannot be negative — that rules out A and D. We need the **greatest** multiple of $3$ that is $\le -32$. Try $q = -10$: $-10 \times 3 = -30$, but $-30 > -32$, so $-30$ is **not** $\le -32$. Try $q = -11$: $-11 \times 3 = -33 \le -32$ ✓. Remainder $= -32 - (-33) = 1$. Check: $-32 = (-11)(3) + 1$ ✓.