Multiplication of Number Bases

Example 2.1.1:

Evaluate the following:

(i) \( \scriptsize 1101_2 \: \times \: 101_2 \)

(ii) \( \scriptsize 111.01_2 \: \times \: 1.11_2 \)

(iii) \( \scriptsize 354_8 \: \times \: 463_8 \)

(iv) \( \scriptsize 45_7 \: \times \: 63_7 \)

(v) \( \scriptsize (A3B_{12})^2 \)

(i) 1101₂ × 101₂

Solution

Answer = 1000001₂

(ii) 111.01₂  ×  1.11₂

Solution

First, ignore the decimal point and multiply.

Explanation:

11101 * 1 = 11101

11101 * 1 = 11101

11101 * 1 11101

Sum up. 11101 + 11101 + 11101

Similar to base 10 multiplication, count 4 decimal places from right to left.

Answer: 111.01₂  ×  1.11₂ = 1100.1011₂

(iii) \( \scriptsize 354_8 \times 463_8 \)

Solution

Note: 3 will multiply 354, 6 will multiply 354, 4 will multiply 354

Explanation of Multiplication:

3 × 4 = 12. 12 divided by 8 = 1 remainder 4. We write 4 and carry over 1
3 × 5 = 15 + 1 = 16. 16 divided by 8 = 2 remainder 0. We write 0 and carry over 2
3 × 3 = 9 + 2 = 11. 11 divided by 8 = 1 remainder 3. We write 3 and carry 1.

Therefore we have \( \scriptsize 1304_8 \)

6 × 4 = 24. 24 divided by 8 = 3 remainder 0. We write 0 and carry over 3
6 × 5 = 30 + 3 = 33. 33 divided by 8 = 4 remainder 1. We write 1 and carry over 4
6 × 3 = 18 + 4 = 22. 22 divided by 8 = 2 remainder 6. We write 6 and carry 2.

Therefore we have \( \scriptsize 2610_8 \)

4 × 4 = 16. 16 divided by 8 = 2 remainder 0. We write 0 and carry over 2
4 × 5 = 20 + 2 = 22. 22 divided by 8 = 2 remainder 6. We write 6 and carry over 2
4 × 3 = 12 + 2 = 14. 14 divided by 8 = 1 remainder 6. We write 6 and carry 1.

Therefore we have \( \scriptsize 1660_8 \)

∴ 1304₈ + 2610₈ + 1660₈ = 215404₈

Explanation of Addition:

4 + 0 = 4. 4 is less than 8 so we write 4
0 + 0 = 0
3 + 1 = 4
1 + 6 + 6 = 13. 13 divided by 8 = 1 remainder 5. We write 5 and carry over 1
2 + 6 = 8 + 1 = 9. 9 divided by 8 = 1 remainder 1. We write 1 and carry over 1
1 + 1 = 2

Therefore our answer is:

Answer: \( \scriptsize 215404_8 \)

(iv) \( \scriptsize 45_7 \: \times \: 63_7 \)

Solution

Explanation of Multiplication:

3 × 5 = 15. 15 divided by 7 = 2 remainder 1. We write 1 and carry over 2
3 × 4 = 12 + 2 = 14. 14 divided by 7 = 2 remainder 0. We write 0 and carry over 2

Therefore we have \( \scriptsize 201_7 \)

6 × 5 = 30. 30 divided by 7 = 4 remainder 2. We write 1 and carry over 4
6 × 4 = 24 + 4 = 28. 28 divided by 7 = 4 remainder 0. We write 0 and carry over 4

Therefore we have \( \scriptsize 402_7 \)

Explanation on Addition:

1 + 0 = 1. 1 is less than 7 so we write 7
0 + 2 = 2
2 + 0 = 2
4 + 0 = 4

Therefore we have \( \scriptsize 4221_7 \)

Answer: \( \scriptsize 4221_7 \)

(v) \( \scriptsize (A3B_{12})^2 \)

Solution

Explanation on Multiplication:

B × B = 11 × 11 = 121. 121 divided by 12 = 10 remainder 1. We write 1 and carry over 10
B × 3 = 11 × 3 = 33. 33 + 10 = 43. 43 divided by 12 = 3 remainder 7. We write 7 and carry over 3
B × A = 11 × 10 = 110. 110 + 3 = 113. 113 divided by 12 = 9 remainder 5. We write 5 and carry 9 to the last column.

Therefore we have \( \scriptsize 9571_{12} \)

3 × B = 3 × 11 = 33. 33 divided by 12 = 2 remainder 9. We write 9 and carry over 2
3 × 3 = 9. 9 + 2 = 11. 11 is less than 12. We write 11 which is = B.
3 × A = 3 x 10 = 30. 30 divided by 12 = 2 remainder 6. We write 6 and carry 2 to the last column.

Therefore we have \( \scriptsize 26B9_{12} \)

A × B = 10 × 11 = 110. 33 divided by 12 = 9 remainder 2. We write 2 and carry over 9

A × 3 = 10 × 3 = 30. 30 + 9 = 39. 39 divided by 12 = 3 remainder 3. We write 3 and carry over 3

A × A = 10 × 10 = 100. 100 + 3 = 103. 103 divided by 12 = 8 remainder 7. We write 7 and carry 8 to the last column.

Therefore we have \( \scriptsize 8732_{12} \)

Try the addition yourself!

∴ 9571₁₂ + 26B9₁₂ + 8732₁₂ = 8A7741₁₂