Lesson 4, Topic 2
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Modular Congruence

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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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