Modular Arithmetic

Welcome to Our Site

I greet you this day,

These are the solutions to the WASSCE Mathematics past questions on the topics on Modular Arithmetic.
When applicable, the TI-84 Plus CE calculator (also applicable to TI-84 Plus calculator) solutions are provided for some questions.
The link to the video solutions will be provided for you. Please subscribe to the YouTube channel to be notified of upcoming livestreams. You are welcome to ask questions during the video livestreams.
If you find these resources valuable and helpful in your passing the Additional Mathematics test of WASSCE, please consider making a donation:

Cash App: $ExamsSuccess or
cash.app/ExamsSuccess

PayPal: @ExamsSuccess or
PayPal.me/ExamsSuccess

Google charges me for the hosting of this website and my other educational websites. It does not host any of the websites for free.
Besides, I spend a lot of time to type the questions and the solutions well. As you probably know, I provide clear explanations on the solutions.
Your donation is appreciated.

Comments, ideas, areas of improvement, questions, and constructive criticisms are welcome.
Feel free to contact me. Please be positive in your message.
I wish you the best.
Thank you.

(2.) In what number base was the addition $1 + nn = 100$, where $n \gt 0$, done?

$ A.\;\; n - 1 \\[3ex] B.\;\; n \\[3ex] C.\;\; n + 1 \\[3ex] D.\;\; n + 2 \\[3ex] $

$ 1 + nn = 100 \\[3ex] 1 + 99 = 100 ...base\;\;ten \\[3ex] nn = 99 \implies n = 9 \\[3ex] 9 + 1 = 10 \\[3ex] \therefore base = n + 1 $

(4.) Find the product of 1101_two and 111_two

$ A.\;\; 1101011_{two} \\[3ex] B.\;\; 1011101_{two} \\[3ex] C.\;\; 1110011_{two} \\[3ex] D.\;\; 1011011_{two} \\[3ex] $

First Method: Operate in Base Two

$ \begin{align} 1\ \ 1\ \ 0\ \ 1\ \ \\ * \ \ \ \ \ \ 1\ \ 1\ \ 1\ \ \\ \hline 1\ \ 1\ \ 0\ \ 1\ \ \\ + \ \ \ \ \ \ \ \ \ \ 1\ \ 1\ \ 0\ \ 1~~~~\ \ \\ 1\ \ 1\ \ 0\ \ 1~~~~~~~~\ \ \\ \hline 1\ \ 0\ \ 1\ \ 1\ \ 0\ \ 1\ \ 1~~ \\ \hline \end{align} $

$1101_{two} * 111_{two} = 1011011_{two}$

Second Method: Convert to Base Ten, Operate in Base Ten, Convert to Base Two

$ \underline{Convert\;\;to\;\;Base\;\;Ten} \\[3ex] 1101_{two} \\[3ex] = 1 * 2^3 + 1 * 2^2 + 0 * 2^1 + 1 * 2^0 \\[3ex] = 1(8) + 1(4) + 0 + 1(1) \\[3ex] = 8 + 4 + 0 + 1 \\[3ex] = 13 \\[5ex] 111_{two} \\[3ex] = 1 * 2^2 + 1 * 2^1 + 1 * 2^0 \\[3ex] = 1(4) + 1(2) + 1(1) \\[3ex] = 4 + 2 + 1 \\[3ex] = 7 \\[5ex] \underline{Operate\;\;in\;\;Base\;\;Ten} \\[3ex] 13 * 7 = 91 \\[5ex] \underline{Convert\;\;to\;\;Base\;\;Two} \\[3ex] \begin{array}{c|c} 2 & 91 \\ \hline 2 & 45 \:R\: 1 \\ \hline 2 & 22 \:R\: 1 \\ \hline 2 & 11 \:R\: 0 \\ \hline 2 & 5 \:R\: 1 \\ \hline 2 & 2 \:R\: 1 \\ \hline 2 & 1 \:R\: 0 \\ \hline & 0 \:R\: 1 \end{array} Count \:the\: remainders\: upwards \\[3ex] 91 = 1011011_{two} \\[3ex] $ $1101_{two} * 111_{two} = 1011011_{two}$

(5.) Given that $110_x = 40_{five}$, find the value of x

$ 110_x = 40_{five} \\[3ex] \text{Convert both bases to base ten} \\[3ex] 1 * x^2 + 1 * x^1 + 0 * x^0 = 4 * 5^1 + 0 * 5^0 \\[3ex] x^2 + x + 0 = 20 + 0 \\[3ex] x^2 + x = 20 \\[3ex] x^2 + x - 20 = 0 \\[3ex] (x + 5)(x - 4) = 0 \\[3ex] x = -5 \hspace{2em}OR\hspace{2em} x = 4 \\[3ex] \text{The base cannot be negative} \\[3ex] \therefore x = 4 $

(6.) The operation $\triangle$ is defined on the set $T = \{2, 3, 5, 7\}$ by $x \triangle y = (x + y + xy) \mod 8$

(i) Construct modulo $8$ table for the operation $\triangle$ on the set $T$
(ii) Use the table to find:

$ I.\;\;\; 2\triangle (5 \triangle 7) \\[3ex] II.\;\;\; 2 \triangle n = 5 \triangle 7 \\[3ex] $

$ T = \{2, 3, 5, 7\} \\[3ex] x \triangle y = (x + y + xy) \mod 8 \\[3ex] y \triangle x = (y + x + yx) \mod 8 \\[3ex] x \triangle y = y \triangle x ...Commutative\;\;Property \\[5ex] (i) \\[3ex] 2 \triangle 2 = [2 + 2 + 2(2)] \mod 8 \\[3ex] = (4 + 4) \mod 8 \\[3ex] = 8 \mod 8 = 0 \\[3ex] 2 \triangle 2 = 0 \\[3ex] 2 \triangle 3 = [2 + 3 + 2(3)] \mod 8 \\[3ex] = (5 + 6) \mod 8 \\[3ex] = 11 \mod 8 = 3 \\[3ex] 2 \triangle 3 = 3 \\[3ex] 2 \triangle 3 = 3 \triangle 2 = 3 \\[3ex] 2 \triangle 5 = [2 + 5 + 2(5)] \mod 8 \\[3ex] = (7 + 10) \mod 8 \\[3ex] = 17 \mod 8 = 1 \\[3ex] 2 \triangle 5 = 1 \\[3ex] 2 \triangle 5 = 5 \triangle 2 = 1 \\[3ex] 2 \triangle 7 = [2 + 7 + 2(7)] \mod 8 \\[3ex] = (9 + 14) \mod 8 \\[3ex] = 23 \mod 8 = 7 \\[3ex] 2 \triangle 7 = 7 \\[5ex] 2 \triangle 7 = 7 \triangle 2 = 7 \\[3ex] 3 \triangle 3 = [3 + 3 + 3(3)] \mod 8 \\[3ex] = (6 + 9) \mod 8 \\[3ex] = 15 \mod 8 = 7 \\[3ex] 3 \triangle 3 = 7 \\[3ex] 3 \triangle 5 = [3 + 5 + 3(5)] \mod 8 \\[3ex] = (8 + 15) \mod 8 \\[3ex] = 23 \mod 8 = 7 \\[3ex] 3 \triangle 5 = 7 \\[3ex] 3 \triangle 5 = 5 \triangle 3 = 7 \\[3ex] 3 \triangle 7 = [3 + 7 + 3(7)] \mod 8 \\[3ex] = (10 + 21) \mod 8 \\[3ex] = 31 \mod 8 = 7 \\[3ex] 3 \triangle 7 = 7 \\[3ex] 3 \triangle 7 = 7 \triangle 3 = 7 \\[5ex] 5 \triangle 5 = [5 + 5 + 5(5)] \mod 8 \\[3ex] = (10 + 25) \mod 8 \\[3ex] = 35 \mod 8 = 3 \\[3ex] 5 \triangle 5 = 3 \\[3ex] 5 \triangle 7 = [5 + 7 + 5(7)] \mod 8 \\[3ex] = (12 + 35) \mod 8 \\[3ex] = 47 \mod 8 = 7 \\[3ex] 5 \triangle 7 = 7 \triangle 5 = 7 \\[5ex] 7 \triangle 7 = [7 + 7 + 7(7)] \mod 8 \\[3ex] = (14 + 49) \mod 8 \\[3ex] = 63 \mod 8 = 7 \\[3ex] 7 \triangle 7 = 7 \\[3ex] $ The modulo table for the operation $\triangle$ on the set $T$ is:

$\triangle$	$2$	$3$	$5$	$7$
$2$	$0$	$3$	$1$	$7$
$3$	$3$	$7$	$7$	$7$
$5$	$1$	$7$	$3$	$7$
$7$	$7$	$7$	$7$	$7$

(ii)
$ (I.) \\[3ex] 2 \triangle (5 \triangle 7) \\[3ex] = 2 \triangle 7 \\[3ex] = 7 \\[3ex] (II.) \\[3ex] 2 \triangle n = 5 \triangle 7 \\[3ex] 2 \triangle n = 7 \\[3ex] 2 \triangle 7 = 7 \\[3ex] \implies n = 7 $

(7.)

$\oplus$	$0$	$1$	$2$	$3$	$4$
$0$	$0$	$1$	$2$	$3$	$4$
$1$	$1$	$2$	$3$	$4$	$0$
$2$	$2$	$3$	$4$	$0$	$1$
$3$	$3$	$4$	$0$	$1$	$2$
$4$	$4$	$0$	$1$	$2$	$3$
Fig. 1

$\otimes$	$0$	$1$	$2$	$3$	$4$
$0$	$0$	$0$	$0$	$0$	$0$
$1$	$0$	$1$	$2$	$3$	$4$
$2$	$0$	$2$	$4$	$1$	$3$
$3$	$0$	$3$	$1$	$4$	$2$
$4$	$0$	$4$	$3$	$2$	$1$
Fig. 2

Fig. 1 and Fig. 2 are the addition and multiplication tables respectively in modulo 5.
Use these tables to solve the equation $(n \otimes 4) \oplus 3 = 0(mod\;\; 5)$

$ A.\;\; 1 \\[3ex] B.\;\; 2 \\[3ex] C.\;\; 3 \\[3ex] D.\;\; 4 \\[3ex] $

$ \underline{RHS} \\[3ex] 0 \mod 5 = 0 \\[3ex] \underline{LHS} \\[3ex] (n \otimes 4) \oplus 3 \\[3ex] start\;\;from\;\;the\;\;left \\[3ex] what?\;\; \oplus 3 = 0 \\[3ex] 2 \oplus 3 = 0...from\;\;Fig.\;1 \\[3ex] this\;\;means\;\;that: \\[3ex] n \otimes 4 = 2 \\[3ex] 3 \otimes 4 = 2...from\;\;Fig.\;2 \\[3ex] n = 3 \\[3ex] \therefore (3 \otimes 4) \oplus 3 = 0 $

Top

$\triangle$	$2$	$3$	$5$	$7$
$2$	$0$	$3$	$1$	$7$
$3$	$3$	$7$	$7$	$7$
$5$	$1$	$7$	$3$	$7$
$7$	$7$	$7$	$7$	$7$

$\triangle$	$2$	$3$	$5$	$7$
$2$	$0$	$3$	$1$	$7$
$3$	$3$	$7$	$7$	$7$
$5$	$1$	$7$	$3$	$7$
$7$	$7$	$7$	$7$	$7$

$\triangle$	$2$	$3$	$5$	$7$
$2$	$0$	$3$	$1$	$7$
$3$	$3$	$7$	$7$	$7$
$5$	$1$	$7$	$3$	$7$
$7$	$7$	$7$	$7$	$7$