Lesson 2, Topic 1
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# Multiplication of Number Bases

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### Examples: Multiplication of Number Bases

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

Solution

(i) 11012 x 1012

The following rules apply when multiplying binary numbers:

0 x 0 = 0

1 x 0 = 0

0 x 1 = 0

1 x 1 = 1

Solution

(ii) 111.012  x  1.112

First ignore the decimal point and multiply

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  x  1.112 = 1100.10112

Solution

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

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

Explanation on Multiplication:

4 x 3 = 12. 12 divided by 8 = 1 remainder 4. We write 4 and carry over 1

3 x 5 = 15 + 1 = 16. 16 divided by 8 = 2 remainder 0. We write 0 and carry over 2

3 x 3 = 9 + 2 = 11. 11 divided by 8 = 1 remainder 3. We write three and carry 1.

Therefore we have $$\scriptsize 1304_8$$

6 x 4 = 24. 24 divided by 8 = 3 remainder 0. We write 0 and carry over 3

6 x 5 = 30 + 3 = 33. 33 divided by 8 = 4 remainder 1. We write 1 and carry over 4

6 x 3 = 18 + 4 = 22. 22 divided by 8 = 2 remainder 6. We write 6 and carry 2.

Therefore we have $$\scriptsize 2610_8$$

4 x 4 = 16. 16 divided by 8 = 2 remainder 0. We write 0 and carry over 2

4 x 5 = 20 + 2 = 22. 22 divided by 8 = 2 remainder 6. We write 6 and carry over 2

4 x 3 = 12 + 2 = 14. 14 divided by 8 = 1 remainder 6. We write 6 and carry 1.

Therefore we have $$\scriptsize 1660_8$$

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 we have $$\scriptsize 215404_8$$

Answer: $$\scriptsize 215404_8$$

Solution

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

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: $$\scriptsize 4221_7$$

Solution

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

Explanation on Multiplication:

B x B = 11 x 11 = 121. 121 divided by 12 = 10 remainder 1. We write 1 and carry over 10

B x 3 = 11 x 3 = 33. 33 + 10 = 43. 43 divided by 12 = 3 remainder 7. We write 7 and carry over 3

B x A = 11 X 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 x B = 3 x 11 = 33. 33 divided by 12 = 2 remainder 9. We write 9 and carry over 2

3 x 3 = 9. 9 + 2 = 11. 11 is less than 12. We write 11 which is = B.

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

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

A x A = 10 x 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 $$\scriptsize 8732_{12}$$