1. The problem is to simplify the given modular expressions: (a) $-3 \bmod 4$ (b) $-2 \bmod 5$ (c) $-10 \bmod 6$ (d) $-20 \bmod 12$ 2. Recall the definition of modulo operation: for any integer $a$ and positive integer $m$, $a \bmod m$ is the remainder when $a$ is divided by $m$. If $a$ is negative, add $m$ repeatedly until the result is non-negative and less than $m$. 3. Solve each part: (a) $-3 \bmod 4$: Since $-3$ is negative, add $4$ once: $$-3 + 4 = 1$$ So, $-3 \bmod 4 \equiv 1$. (b) $-2 \bmod 5$: Add $5$ once: $$-2 + 5 = 3$$ So, $-2 \bmod 5 \equiv 3$. (c) $-10 \bmod 6$: Add $6$ twice: $$-10 + 6 = -4$$ Still negative, add $6$ again: $$-4 + 6 = 2$$ So, $-10 \bmod 6 \equiv 2$. (d) $-20 \bmod 12$: Add $12$ twice: $$-20 + 12 = -8$$ Still negative, add $12$ again: $$-8 + 12 = 4$$ So, $-20 \bmod 12 \equiv 4$. Final answers: (a) 1 (b) 3 (c) 2 (d) 4