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Solved Examples on Binary Operations, Modulo, and Modulo Tables

Samuel Dominic Chukwuemeka (SamDom For Peace) Verify your answers as applicable with the Modulo Arithmetic and Algorithms Calculators

Prerequisites:
(1.) Integers and Integer Operations
(2.) Exponents
(3.) Factoring

For ACT Students
The ACT is a timed exam...$60$ questions for $60$ minutes
This implies that you have to solve each question in one minute.
Some questions will typically take less than a minute a solve.
Some questions will typically take more than a minute to solve.
The goal is to maximize your time. You use the time saved on those questions you solved in less than a minute, to solve the questions that will take more than a minute.
So, you should try to solve each question correctly and timely.
So, it is not just solving a question correctly, but solving it correctly on time.
Please ensure you attempt all ACT questions.
There is no negative penalty for a wrong answer.

For JAMB Students
Calculators are not allowed. So, the questions are solved in a way that does not require a calculator.

For WASSCE Students
Any question labeled WASCCE is a question for the WASCCE General Mathematics
Any question labeled WASSCE:FM is a question for the WASSCE Further Mathematics/Elective Mathematics

For GCSE and Malta Students
All work is shown to satisfy (and actually exceed) the minimum for awarding method marks.
Calculators are allowed for some questions. Calculators are not allowed for some questions.

For NSC Students
For the Questions:
Any space included in a number indicates a comma used to separate digits...separating multiples of three digits from behind.
Any comma included in a number indicates a decimal point.
For the Solutions:
Decimals are used appropriately rather than commas
Commas are used to separate digits appropriately.

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(1.) ACT Among the following arithmetic operations, which could the symbol $\diamond$ represent given that the equation $(2 \diamond 1)^4 + (6 \diamond 3)^2 = 10$ is true?
    I. Addition
    II. Subtraction
    III. Division
F. I only
G. II only
H. III only
J. I and II only
K. I, II, and III


Let us test each operation to determine if the LHS equals the RHS

$ \underline{Test\;\;for\;\;Addition} \\[3ex] (2 + 1)^4 + (6 + 3)^2 \stackrel{?}{=} 10 \\[3ex] 3^4 + 9^2 \gt 10 \\[3ex] NO \\[3ex] \underline{Test\;\;for\;\;Subtraction} \\[3ex] (2 - 1)^4 + (6 - 3)^2 \stackrel{?}{=} 10 \\[3ex] 1^4 + 3^2 \stackrel{?}{=} 10 \\[3ex] 1 + 9 = 10 \\[3ex] YES \\[3ex] \underline{Test\;\;for\;\;Division} \\[3ex] (2 \div 1)^4 + (6 \div 3)^2 \stackrel{?}{=} 10 \\[3ex] 1^4 + 2^2 \stackrel{?}{=} 10 \\[3ex] 1 + 4 \lt 10 \\[3ex] NO \\[3ex] $ This is Subtraction only
II only
(2.) WASSCE Evaluate 6 - 36(mod 9)

$ A.\;\; 3 \\[3ex] B.\;\; 4 \\[3ex] C.\;\; 5 \\[3ex] D.\;\; 6 \\[3ex] $

$ 6 - 36(mod\;\;9) \\[3ex] 6 - 0 \\[3ex] 6 $
(3.) WASSCE 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.) ACT The operation $\otimes$ is defined on the set of positive integers by the rule $a \otimes b = a + b^2$.
What is the value of $(4 \otimes 3) \otimes 5$?

$ A.\;\; 34 \\[3ex] B.\;\; 38 \\[3ex] C.\;\; 44 \\[3ex] D.\;\; 50 \\[3ex] E.\;\; 74 \\[3ex] $

We have to do the parenthesis first

$ (4 \otimes 3) \otimes 5 \\[3ex] a \otimes b = a + b^2 \\[3ex] 4 \otimes 3 \\[3ex] = 4 + 3^2 \\[3ex] = 4 + 9 \\[3ex] = 13 \\[3ex] 13 \otimes 5 \\[3ex] = 13 + 5^2 \\[3ex] = 13 + 25 \\[3ex] = 38 $
(5.) WASSCE 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 $
(6.) ACT Suppose the operation $\oplus$ is defined by $a \oplus b = \dfrac{a + b}{7}$
What is the value of (98 $\oplus$ 77) $\oplus$ (21 $\oplus$ 49)?

$ A.\;\; 5 \\[3ex] B.\;\; 35 \\[3ex] C.\;\; 113 \\[3ex] D.\;\; 245 \\[3ex] E.\;\; 250 \\[3ex] $

$ a \oplus b = \dfrac{a + b}{7} \\[5ex] Parenthesis\;\;first \\[3ex] 98 \oplus 77 = \dfrac{98 + 77}{7} = \dfrac{175}{7} = 25 \\[5ex] 21 \oplus 49 = \dfrac{21 + 49}{7} = \dfrac{70}{7} = 10 \\[5ex] Then\;\;do\;\;the\;\;operation\;\;on\;\;the\;\;2\;\;results \\[3ex] 25 \oplus 10 = \dfrac{25 + 10}{7} = \dfrac{35}{7} = 5 $
(7.) WASSCE
$\oplus$ $0$ $1$ $2$ $3$ $4$
$0$ $0$ $1$ $2$ $3$ $4$
$1$ $1$ $2$ $3$ $4$ $0$
$2$ $2$ $3$ $4$ $0$ $1$
$3$ $3$ $4$ $0$ $1$ $2$
$4$ $4$ $0$ $1$ $2$ $3$
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 $
(8.)


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(10.) ACT A new operation, $\blacklozenge$, is defined on pairs of ordered pairs of integers as follows: $(a, b) \blacklozenge (c, d) = \dfrac{ac + bd}{ab - cd}$
What is the value of $(2, 1) \blacklozenge (4, 6)$ ?

$ F.\;\; -\dfrac{7}{11} \\[5ex] G.\;\; -\dfrac{7}{4} \\[5ex] H.\;\; \dfrac{7}{4} \\[5ex] J.\;\; 7 \\[3ex] K.\;\; 14 \\[3ex] $

$ (a, b) \blacklozenge (c, d) = \dfrac{ac + bd}{ab - cd} \\[5ex] (2, 1) \blacklozenge (4, 6) = \dfrac{2(4) + 1(6)}{2(1) - 4(6)} \\[5ex] = \dfrac{8 + 6}{2 - 24} \\[5ex] = \dfrac{14}{-22} \\[5ex] = -\dfrac{7}{11} $
(11.)


(12.) WASCCE If 20 mod 9 is equivalent to y mod 6, find y

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

$ 20\mod 9 \equiv y\mod 6 \\[3ex] 2 \equiv y\mod 6 \\[3ex] 2 \equiv 2\mod 6 \\[3ex] y = 2 $
(13.) WASCCE A binary operation $\otimes$ is defined on the set of real numbers, R, by $m \otimes n = mn - n - 2m$, where $m,\; n\; \in R$.
If $5 \otimes x = 22$, find the value of x


$ m \otimes n = mn - n - 2m \\[3ex] 5 \otimes x = 5(x) - x - 2(5) \\[3ex] = 5x - x - 10 \\[3ex] = 4x - 10 \\[3ex] 5 \otimes x = 22 \\[3ex] \implies \\[3ex] 4x - 10 = 22 \\[3ex] 4x = 22 + 10 \\[3ex] 4x = 32 \\[3ex] x = \dfrac{32}{4} \\[5ex] x = 8 $
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