Q.$-41 \bmod 9$ is (A) $5$ (B) $4$ (C) $3$ (D) $0$

$-41 = 9(-5)+4$, so the least non-negative remainder is $4$, giving $-41 \bmod 9 = 4$.

Divide 282 by 11 and the remainder, 7, is the smallest positive integer it is congruent to mod 11.

The unit digit of $2^n$ repeats in the cycle $2,4,8,6$ (period 4); $12$ is a multiple of 4, so $(22)^{12}$ ends in $6$.

The order of $\omega^2$ in the multiplicative group of cube roots of unity is $3$.

1. The cube roots of unity are $1,\omega,\omega^2$ where $\omega^3=1$ and this set forms a group under multiplication.
2. The order of an element $g$ is the smallest positive integer $m$ such that $g^m=1$.
3. Compute powers of $\omega^2$: $(\omega^2)^1=\omega^2\neq1$.
4. $(\omega^2)^2=\omega^4=\omega^{3+1}=\omega\neq1$.
5. $(\omega^2)^3=\omega^6=(\omega^3)^2=1^2=1$.
6. The smallest such $m$ is $3$, so the order of $\omega^2$ is $3$.

In the group of $n^{th}$ roots of unity, $(\omega^k)^{-1}=\omega^{n-k}$.

1. The $n^{th}$ roots of unity form a cyclic group of order $n$ under multiplication, with identity element $1=\omega^0=\omega^n$.
2. For an element $\omega^k$ ($k<n$), we need $m$ such that $\omega^k\cdot\omega^m=\omega^n=1$, i.e. $k+m\equiv0\pmod n$, so $m=n-k$.
3. Hence the inverse of $\omega^k$ is $\omega^{n-k}$; equivalently, since all $n^{th}$ roots of unity lie on the unit circle, $\omega^{n-k}=\overline{\omega^k}=(\omega^k)^{-1}$.
4. Options (a) and (d) generally are not even elements of the group (fractional exponents), and (b) $\omega^{-1}$ equals $\omega^{n-1}$, correct only when $k=1$, not in general.

In the cyclic group $(\mathbb Z_n,+_n)$, the order of element $[k]$ equals $n/\gcd(k,n)$; here $n=9,k=7$ are coprime, so $[7]$ generates the whole group and has order $9$.

1. $(\mathbb Z_9,+_9)=\{[0],[1],\dots,[8]\}$ is the cyclic group of integers modulo 9 under addition, of order 9.
2. For a cyclic group of order $n$ generated by $[1]$, the order of the element $[k]$ is $\dfrac{n}{\gcd(k,n)}$.
3. Here $k=7$, $n=9$. Compute $\gcd(7,9)$: since 7 is prime and does not divide 9, $\gcd(7,9)=1$.
4. So order of $[7] = \dfrac{9}{1}=9$.
5. Verification: repeatedly adding $7$ mod $9$: $7,14\equiv5,21\equiv3,28\equiv1,35\equiv8,42\equiv6,49\equiv4,56\equiv2,63\equiv0$ — this takes exactly $9$ steps to return to $[0]$, confirming order $9$.
6. This matches option (a).

The set $\{x=5k+2 : k\in\mathbb Z\}$ is, by definition, precisely all integers congruent to $2$ modulo $5$, i.e. the equivalence class $[2]$.

1. Congruence modulo 5 partitions $\mathbb Z$ into equivalence classes $[0],[1],[2],[3],[4]$, where $[r]=\{x\in\mathbb Z : x\equiv r\pmod5\} = \{x = 5k+r : k\in\mathbb Z\}$.
2. The given set is $\{x\in\mathbb Z : x=5k+2,\ k\in\mathbb Z\}$, i.e. all integers of the form $5k+2$: $\dots,-8,-3,2,7,12,\dots$
3. By the definition in step 1 with $r=2$, this set is exactly $[2]$.
4. This matches option (d).

Direct computation of successive powers of $-i$ shows it first returns to $1$ at the 4th power, so its order is $4$.

1. The 4th roots of unity are the solutions of $z^4=1$: $\{1,\ i,\ -1,\ -i\}$, forming a group of order $4$ under multiplication.
2. The order of an element $g$ in a finite group is the smallest positive integer $k$ such that $g^k=e$ (here $e=1$).
3. Compute powers of $-i$:
   • $(-i)^1=-i\neq1$.
   • $(-i)^2=(-i)(-i)=i^2=-1\neq1$.
   • $(-i)^3=(-i)^2\cdot(-i)=(-1)(-i)=i\neq1$.
   • $(-i)^4=(-i)^3\cdot(-i)=(i)(-i)=-i^2=1$.
4. The smallest $k$ for which $(-i)^k=1$ is $k=4$, so the order of $-i$ is $4$.
5. (This is consistent with the general fact that $-i$ is a primitive 4th root of unity, since $\gcd(3,4)=1$ where $-i=i^3$ — a primitive root of a cyclic group of order $n$ always has order exactly $n$.)
6. Distractors (b) $3$, (c) $2$, (d) $1$ correspond to $-i$ prematurely equalling $1$ at those powers, which the direct computation in step 3 disproves.