Question

B3 a. Solve for x in this equation: 2x + 11 = 2 (mod 4). b. What are the sets of units and zero divisors in the ring of integers modulo 22? (Specify at least the smaller set using set-roster notation.) c. Find a formula for the quotient and the exact remainder when 534 is divided by 8. Hint: find the remainder first by modular arithmetic. Then subtract the remainder from the power and divide to find the quotient.

Answer 1

[I have helped with questions a. and b.].

a.

$2x+11\equiv2(\mod 4) \\=>2x+11=4m+2\ \textup{for some }m\in\mathbb Z \\=>2x=4m-9$

But, for all integers
2.772
, the number
20
is even and is odd. Since an odd number cannot equal an even number, there is no integer solution for this equation.

b.

The elements of the ring of integers modulo 22 ($\mathbb Z_{22}$) are [m], where $m\in\mathbb Z,0\le m<22$ .

Let us suppose that [m] is a unit element of $\mathbb Z_{22}$ . Then, m>0 and there is an [n] such that, $[m][n]=[1]$

$\\=>mn\equiv1(\mod 22) \\=>mn=22z+1\textup{ for some }z\in\mathbb N\cup\{0\} \\=>mn\in\{1,23,45,67,89,111,133,155,177,199,221,243, \\ {\color{Yellow} .}\qquad\qquad265,287,309,331,353,375,397,419,441\}\qquad[\because 21\times21=441]$, and 21 is the largest m such that [m] is in $\mathbb Z_{22}$ .

Since all of the above numbers are odd, neither of m,n can be even. Also, since they are all 1 more than a multiple of 22, and therefore of 11, none of m,n is divisible by 11.

• For m=1, n=1, mn=1.
• For m=3, n=15, mn=45.
• For m=5, n=9, mn=45.
• For m=7, n=19, mn=133.
• For m=13, n=17, mn=221.
• For m=21, m=21, mn=441.

So, for all odd m less than 22, and other than 11, [m] has an inverse, and therefore, is a unit.

The set of units of $\mathbb Z_{22}$ is .

Let us suppose that [m] is a zero divisor of $\mathbb Z_{22}$ . Then, m>0 and there is an [n] such that,

$\\=>mn\equiv0(\mod 22) \\=>mn=22z\textup{ for some }z\in\mathbb N\cup\{0\} \\=>mn=\{0,22,44,66,88,110,132,154,176,198,220,242,264, \\ {\color{Yellow} .}\qquad\qquad\qquad286,308,330,352,374,396,418,440\}\quad[\because 21\times21=441]$, and 21 is the largest m such that [m] is in $\mathbb Z_{22}$ .

Since all of the above numbers are divisible by 11, atleast one of m,n must be divisible by 11. So, atleast one of m,n must be 11, the only $m\in\mathbb Z,0< m<22$ that is divisible by 11.

• For m=11, n=10, mn=110

The set of zero divisors of $\mathbb Z_{22}$ is .