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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
    3 Topics
  6. Modular Arithmetic III | Week 6
    4 Topics
    |
    1 Quiz
  7. Indices I | Week 7
    3 Topics
    |
    1 Quiz
  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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Topic Content:

  • Addition & Subtraction in Modular Arithmetic

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 5.2.1:

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

Solution

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

i.e.  

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SUCCESS GRABBERS SCHOOL
SUCCESS GRABBERS SCHOOL
11/08/2023 6:43 AM

This is superb

Reply
Agnes Ediale
Agnes Ediale
13/05/2024 8:04 PM

Is 67 and 94 (mod 9) equivalent?

Reply
Christopher Idu
Christopher Idu
17/07/2024 1:52 AM

I love this

Reply
Nicole Ita
Nicole Ita
06/08/2024 6:22 PM

This is great

Reply

Solution (i)

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

Step 1

Add 54 and 65

i.e. 54 + 65 = 119 (mod 6)

Step 2

Take the answer of the addition to mod 6

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

= 19 r 5

Step 3

119 = 19 × 6 + 5 (mod 6)

= 0 + 5 (mod 6)

∴ 119 (mod 6) = 5 (mod 6)

Solution (ii)

⇒ – 25 (mod 8)

=  (– 4 × 8) + 7 (mod 8)

= – 32 + 7 (mod 8)

= 0 + 7 (mod 8)

∴  – 25 (mod 8) = 7 mod 8

Solution (iii)

⇒ – 56 (mod 12)

= (– 5 × 12) + 4 (mod 12)

= – 60 + 4 (mod 12)

= 0 + 4 (mod 12)

∴   – 56 (mod 12) = 4 (mod 12)

Solution (iv)

⇒ 2 ⊖ 11 (mod 5)

2 – 11 (mod 5) = – 9 (mod 5)

=  – 5  × 2 + 1 (mod 5)

=  – 10 + 1 (mod 5)

= 0 + 1 (mod 5)

∴  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)

∴   12 – 5 (mod 6) = 1 (mod 6)

Solution (vi)

⇒ 5 ⊖ 20 (mod 11)

5 ⊖ 20 (mod 11) = – 15 (mod 11)

          =  –2 × 11 + 7 (mod 11)

          =  – 22 + 7 (mod 11)

           = 0 + 7 mod 11

∴     5 ⊖ 20 (mod 11) = 7 (mod 11)

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