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SS1: MATHEMATICS - 1ST TERM

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  1. Number Base System I | Week 1
    6Topics
    |
    2 Quizzes
  2. Number Base System II | Week 2
    3Topics
    |
    1 Quiz
  3. Number Base System III | Week 3
    2Topics
    |
    1 Quiz
  4. Modular Arithmetic I | Week 4
    2Topics
  5. Modular Arithmetic II | Week 5
    2Topics
  6. Modular Arithmetic III | Week 6
    3Topics
    |
    1 Quiz
  7. Indices I | Week 7
    2Topics
  8. Indices II | Week 8
    1Topic
    |
    1 Quiz
  9. Logarithms I | Week 9
    3Topics
  10. Logarithms II
    1Topic
    |
    1 Quiz
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Screen Shot 2021 10 24 at 1.59.02 PM

Consider the circle above numbered 0 to 4. These digits are the same as those used in the modulo 5 system.

Hint

1. If the pointer is on 2 and is turned clockwise through 4 spaces, it will end up at 1. This is usually written as \( \scriptsize 2 + 4 \equiv 1 (mod5) \kern-1em\)  Where the sign \( \scriptsize \equiv \) means “it’s congruent to” i.e. \( \scriptsize 2 + 4 \equiv 1 (mod5) \kern-1em\)  means “2 plus 4 is congruent to 1 modulo 5”

2. If the pointer is on 2 and turns clockwise through 20 spaces, it will end up at 2. Thus, \( \scriptsize 2 + 20 \equiv 2 (mod5)\) 

Example 1

Simplify the following:

(i) 56 (mod 5)

(ii) 440 (mod 7)

Solution

(i) 56 (mod 5) =  \( \scriptsize 56 \: \div \: 5 = 11 \: remainder \: 1 \) 

\(\therefore \scriptsize 56 mod5 \equiv 1(mod5)\)

(ii) 440 (mod 7) = \( \scriptsize 440 \; \div \; 7 = 62 \; remainder \; 6 \) 

\(\therefore \scriptsize 440 (mod 7) \equiv 6 (mod 7)\)

Hint:

It is important to note that we focus our attention on the remainders rather than the multiplier of the moduli.

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