Lesson 6, Topic 3
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# Questions & Solutions – Modular Arithmetic

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### Example 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) For $$\scriptsize 2 \: \bigotimes \: x = 3$$

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

(iii) 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

From the table p = 4

### Example 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)

x = 4(mod 12)

Also, for  8x = 20(mod 12)

We can have

8x = 12 + 8(mod 12)

⇒   8x = 0 + 8(mod 12)

⇒ x = 1(mod 12)

In order to get the other solution, construct the table of 8x = 20(mod 12)

i.e. table of 8x = 0 + 8(mod 12)

Check for the number 8 on the residues column and the entries on its row, you will find out that entry number 8 tallies with 1, 4, 7 and 10 on the residues row.

Therefore, the solution for x: 1, 4, 7 and 10

Hint: Whenever the given modulus is not a prime number, it is possible to have more than one solution, therefore when constructing the table you should take adequate care in resolving the solutions.

(iii) 1 ⨸ 2(mod 4)

i.e. 2x = 1(mod 4)

x does not have a solution in modulo 4.

i.e. x does not exist.

(iv) 98 ⨸ 10 (mod 6)

⇒ 10x = 98(mod 6)

⇒    (6 + 4x) = (16 × 6 + 2)(mod 6)

⇒      (0 + 4)x = (0 + 2)(mod 6)

i.e. 4x = 2(mod 6)

⇒    4x + 6 = 2 + 6(mod 6)

⇒   4x + 0 = 8 mod 6

4x = 8 mod 6

i.e.     x = 2 mod 6

Also, 4x + 3(6) = 3(6) +2 (mod 6)

= 4x + 0 = 20(mod 6)

⇒  4x = 20(mod 6)

⇒   x = 5 mod 6

∴     the solution for x = 2 and 5

### Example 3

Find the multiplicative inverse of the following:

(i) 6(mod 8)
(ii) 3(mod5)
(iii) -2(mod 7)

Solution

(i) 6 mod 8 → the inverse of 6 in (mod 8) is the number such that when it is multiplied by 6 gives 1. However, the inverse of 6 in mod 8 does not exist.

(ii) 3 mod 5 → 3 x 2 = 6 = 1 mod 5

∴ the inverse of 3 in mod 5 is 2.

(iii) -2(mod 7)

i.e. -2 = -7 + 5(mod 7)

⇒ -2(mod 7) = 5 mod 7

Therefore, the inverse of 5 in mod 7 is given by

5 × 3 = 15 = 1mod 7

∴   The inverse of 5 in mod 7 is 3.

### Example 4

Work out the square root of the following:

(i) $$\scriptsize \sqrt{16} \: (mod 5)$$
(ii) $$\scriptsize \sqrt{64} \: (mod 6)$$

Solution

(i) $$\scriptsize \sqrt{16} = \pm \: 4 mod\: 5$$

4 (mod 5) or -4 (mod 5)

4 (mod 5) or (-5 + 1) (mod 5)

⇒ 4 (mod 5) or 1 (mod 5)

(ii) $$\scriptsize \sqrt{64} = \pm \: 8\: mod 6$$

⇒ 8(mod 6) or -8(mod 6)

⇒ (6 + 2)(mod 6)or (-12 + 4)(mod 6)

⇒ 2(mod 6) or 4(mod 6)

### Example 5

Solve the following equations for x

(i) 6x = 18 (mod 10)
(ii) 4x + 1= 5(mod 8)
(iii) 2x + 2 = 3 (mod 4)
(iv) 5x + 2 = 8(mod 11)

Solution

(i) 6x = 18 (mod 10) (÷both sides by 6)

i.e.           x = 3(mod 10)

Also, add 30 to both sides

⇒ (6x + 3 × 10) = 18 + 3 × 10(mod 10)

⇒  (6x + 0) = 48 (mod 10)

⇒  6x = 48 (mod 10) (÷both sides by 6)

i⇒   x = 8 (mod 10)

∴   the solution for x : 3 and 8

(ii) 4x + 1 = 5(mod 8) (add 7 to both sides)

i.e.   4x + 1 + 7 = (5 + 7)(mod 8)

⇒ 4x + 8 = 12 (mod 8)

⇒ 4x = 12 (mod 8)       (÷both sides by 4)

⇒   x = 3 (mod 8)

4x + 1 + 15 = (5+15) mod 8 (add 15 to both sides)

4x + 16 = 20 mod 8

⇒ 4x = 20 mod 8

⇒   x = 5 mod 8

4x + 1 + 23 = 5 + 23 (mod 8) (add 23 to both sides)

⇒ 4x + 24 = 28 (mod 8)

⇒ 4x = 28 (mod 8) (÷both sides by 4)

⇒  x = 7 mod 8

4x + 1 + 31 = (5 + 31) (mod 8) (add 31 to both sides)

⇒   4x + 32 = 36 (mod 8)

⇒   x = 9 (mod 8)

⇒   x = (8 + 1) (mod 8)

⇒  x = 1 (mod 8)

∴   the solution for x: 1, 3, 5 and 7

(iii) 2x + 2 = 3 (mod 4)

For any attempt, you will find out that x does not have a solution for mod 4 arithmetic, therefore x does not exist.

(iv)    5x + 2 = 8 (mod 11) (add 42 to both sides)

i.e. 5x + 2 + 42 = (8 + 42)(mod 11)

⇒ 5x + 44 = 50 (mod 11)

⇒ 5x + 44 = 50 (mod 11)

⇒ 5x + 4(11) = 50(mod 11)

⇒      5x + 0 = 50(mod 11)            ( ÷both sides by 5)

⇒      x = 10(mod 11)

### Example 6

Simplify the following:

(i) (0 – 4) × (4 + 2)(mod 5)
(ii) 4x + 5 + 8x + 3(mod 7)

Solution

(i) (0 – 4) × (4 + 2) (mod 5)

i.e. -4 × 6(mod 5)

⇒      (-5 + 1) × (5 + 1)(mod 5)

⇒      (0 + 1) × (0 + 1)(mod 5)

⇒       (1 × 1 )(mod 5)

∴   Answer = 1 (mod 5)

(ii)  4x + 5 + 8x + 3(mod 7)

12x + 8(mod 7)

(7 + 5)x + (7 + 1)(mod 7)

(0+ 5)x + (0 + 1)(mod 7)

5x + 1(mod 7)

∴    Answer = 5x + 1 (mod 7)

Evaluation:

1. Use the 12-hour clock time to find:

(i)  8 ⊕ 47
(ii) 12 ⊕ 68
(iii) 10 ⊕ 9

2. Write out the set of integers or residues for the following modular operations

(i) mod 8
(ii)  mod 9
(iii)  mod 11

3. Write out the set of integers in the number cycle below and use the diagram to find:

(i) 0 ⊕ 25        (ii) 2 ⊕ 11        (iii) 3 ⊕ 14

4. State whether the following pairs of numbers are equivalent in the moduli given

(i) 109 and 97 (mod 11)
(ii) 605 and 199 (mod 12)
(iii) 258 and 82 (mod 8)

5. Find the values of the following in the given modulo

(i)   60 ⊗ 9 mod 11
(ii) 21 ⊗ 8 (mod 5)
(iii) 55 ⊗ 14 mod 3
(iv) 12 ⊗ 7 (mod 11)
(v) 90 ⊗ 11 mod 8
(vi) 28 ⊗ 7 (mod 3)

6.  (i) Copy and complete the multiplication table given below in modulo 6 operation

7. Find the multiplicative inverse of the following residues:

(i) 4(mod 7)        (ii) 5 (mod 9)    (iii) -2 (mod 7)    (iv) 2(mod 9)

8. Simplify the following:

(i) $$\scriptsize 2 \: \times \: 3^{-1} (mod\: 7)$$
(ii) $$\scriptsize 9 \: \times \: 12^{-1} (mod\: 10)$$

9. Find the values of the following given modulo:

(i) 8 ⨸ 2 (mod 4)        (ii) 0 ⨸ 4 (mod 6)        (iii) 68 ⨸ 10 (mod 6)    (iv) 19 ⨸ 5 (mod 3)

10. Evaluate the square root of 25 and 64 in mod 5.

11. Solve the following equations for x in the given modulo

(i) 2x = 8(mod 5)
(ii) 3x = 4(mod5)
(iii) 7x = 3(mod 11)
(iv) 5x = 7(mod 8)
(v)   4x +1=2(mod 9)
(vi) 2x + 3 = 5 (mod 4)
(vii) 3x + 4 = 5(mod 8)
(viii)  2x + 2 = 4 (mod 7)

12. Simplify the following:

(i) 16 + 9 + 16 + 13(mod 5)

(ii) 8(4x – 5) + 6x (mod 4)

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