Questions & Solutions – Modular Arithmetic

Example 6.3.1:

From the multiplication table for arithmetic (mod 6) find the following:  

(i) \(\scriptsize 2 \bigotimes x = 3 \)
(ii) \(\scriptsize 5 \bigotimes x = 1 \)
(iii) \(\scriptsize 0 ⨸ 3 \)
(iv) \(\scriptsize 2 ⨸ 5 \)

Solution

(i) \(\scriptsize 2 \bigotimes x = 3 \)

For \( \scriptsize 2 \: \bigotimes \: x = 3\)

From the table, x does not exist.

(ii) \( \scriptsize 5 \: \bigotimes \: x = 1 \)

For \( \scriptsize 5 \: \bigotimes \: x = 1 \)

From the table, x = 5

(iii) \(\scriptsize 0 ⨸ 3 \)

Let 0 ⨸ 3 = p

i.e. 3p = 0 (mod 6)

From the table, p = 0, 2 and 4

(iv) \(\scriptsize 2 ⨸ 5 \)

Let 2 ⨸ 5 = p

i.e.   5p = 2 (mod 6)

From the table p = 4

Example 6.3.2:

Find the value(s) of the following:

(i) \( \scriptsize 7 \: \bigotimes \: 9^{-1} \: (mod \:17) \)
(ii) \( \scriptsize 20 \: ⨸ \: 8 \: (mod \:12) \)
(iii)   \( \scriptsize 1 \: ⨸ \: 2 \: (mod \:4) \)
(iv) \( \scriptsize 98 \: ⨸ \: 10 \: (mod \:6) \)

Solution

(i) Recall \( \scriptsize 7 \: \bigotimes \: 9^{-1} = 7 \: ⨸ \: 9 \: (mod \:17) \)

Let  7 ⨸ 9 = x (mod 17)

i.e.  9x = 7 (mod 17)

 9x + 7(17) = 7 + 7(17) (mod 17)

⇒   9x + 0 = 7 + 119 (mod 17)

⇒    9x = 126 (mod 17)

  ∴      x = 14 (mod 17)

(ii) 20 ⨸ 8 (mod 12)

Let 8x = 20 (mod 12)

i.e.  8x + 12 = 20 + 12 (mod 12)

 8x + 0 = 32 (mod 12)   (÷ both sides by 8)