Topic Content:
- Modular Congruence
Consider the circle above numbered 0 to 4. These digits are the same as those used in the modulo 5 system.
Key-Point:
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 \( \scriptsize 2 + 4 \equiv 1 \:(mod\:5) \kern-1em \:\)
- Where the sign \( \scriptsize \equiv \) means “it’s congruent to”
⇒ \( \scriptsize 2 + 4 \equiv 1\: (mod\:5) \kern-1em\: \) 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, \( \scriptsize 2 + 20 \equiv 2 \:(mod\:5)\)
Example 4.2.1:
Simplify the following:
(i) 56 (mod 5)
(ii) 440 (mod 7)
Solution:
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easy ready for the quiz
It’s really helpful