Modular Congruence

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.