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SS1: MATHEMATICS - 1ST TERM

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  1. Number Base System I | Week 1
    6 Topics
    |
    2 Quizzes
  2. Number Base System II | Week 2
    3 Topics
  3. Number Base System III | Week 3
    2 Topics
    |
    1 Quiz
  4. Modular Arithmetic I | Week 4
    2 Topics
  5. Modular Arithmetic II | Week 5
    2 Topics
  6. Modular Arithmetic III | Week 6
    3 Topics
    |
    1 Quiz
  7. Indices I | Week 7
    3 Topics
  8. Indices II | Week 8
    1 Topic
    |
    1 Quiz
  9. Logarithms I | Week 9
    3 Topics
  10. Logarithms II | Week 10
    4 Topics
    |
    1 Quiz



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Screenshot 2022 08 25 at 12.54.40

The diagram above shows addition in modulo 6 in which an element on the extreme left is added to an element on the top row, in that order. You will observe that the first column on the extreme left and the top row contain the same set of integers.

Example

\( \scriptsize 3 \: \bigoplus \: 2 \)

Solution

look for the number 3 on row 4 and the number 2 on column 3, they intersect at number 5

i.e.  \( \scriptsize 3 \: \bigoplus \: 2 = 5(mod6) \)

In the same vein, subtraction can be considered as the inverse operation of addition. So, in mod6, we can use the addition table to work out subtraction

Example

\( \scriptsize 2 \: \circleddash \: 5 \) is a number y such that 5 + y = 2, i.e. 5 plus a number gives 2.

From the residue columns and rows, 5 and 3 intersect at 2, therefore y = 3.

It is important to know that in subtraction the first number is picked from the residues column and the 2nd is picked from the residues row.

More Examples: Addition in mod 6

(i) \( \scriptsize 5 \: \bigoplus \: 4 = 3 \: (mod 6) \)

(ii) \( \scriptsize 4 \: \bigoplus \: 1 = 5 \: (mod 6) \)

(iii) \( \scriptsize 3 \: \bigoplus \: 5 = 2 \: (mod 6) \)

(iv) \( \scriptsize 1 \: \bigoplus \: 5 = 0 \: (mod 6) \)

Example

Evaluate \( \scriptsize 4 \: \bigoplus \: 7 \: (mod4) \)

Solution

Step 1: First step add 4 and 7 = 4 + 7 = 11

Step 2: Convert the answer to modulus 4 (Remember numbers in modulus 4 can only be 0, 1, 2, or 3)

\( \frac{11}{4} \)

= 2 remainder 3

= 3 (mod 4)

Worked Examples

Evaluate the following in the given moduli

(i) 54 ⊕ 65 (mod6)

(ii) – 25 (mod8)

(iii) – 56 (mod 12)

(iv) 2 ⊖ 11 (mod 5)

(v) 12 ⊖ 5 (mod 6)

(vi) 5 ⊖ 20 (mod 11)

Responses

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SOLUTION (i)

:> \( \scriptsize 54 \: \bigoplus \:  65 \: (mod 6) \)

Step 1

Add 54 and 65, i.e 54 + 65 = 119 (mod6)

Step 2

Take the answer of the addition to mod 6

= \( \frac{119}{6} \)

= 19 r 5

Step 3

119 = 19 x 6 + 5 (mod6)

= 0 + 5 (mod6)

i.e. 119 (mod 6) = 5(mod 6)

SOLUTION(ii)

– 25 (mod8) =  -4 x 8 + 7 (mod8)

= – 32 + 7 (mod8)

= 0 + 7(mod8)

i.e. -25 (mod 8) = 7 mod 8

SOLUTION(iii)

– 56 (mod 12) = -5 x 12 + 4 (mod 12)

= -60 + 4 (mod 12)

= 0 + 4 (mod 12)

i.e.        -56 (mod 12) = 4 (mod 12)

SOLUTION(iv)

2 ⊖ 11 (mod 5)

2 – 11 (mod 5) = -9 (mod 5)

= -5 x 2 + 1 (mod 5)

= -10 + 1 (mod 5)

= 0 + 1 (mod 5)

i.e.  2 – 11 (mod 5) = 1 (mod 5)

SOLUTION(v)

12 ⊖ 5 (mod 6)

12 – 5 (mod 6) = 7 mod 6

= 6 + 1 (mod 6)

= 0 + 1 (mod 6)

i.e.       12 – 5 (mod 6) = 1 (mod 6)

SOLUTION(vi)

5 ⊖ 20 (mod 11)

5 20 (mod 11) = -15 (mod 11)

          = -2 x 11 + 7(mod 11)

          = -22 + 7 (mod 11)

           = 0 + 7 mod 11

i.e.      5 – 20 (mod 11) = 7 (mod 11)

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