### 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)** \( \scriptsize (A3B_{12})^2 \)

**(i)** 1101_{2} x 101_{2}

**Solution**

The following rules apply when multiplying binary numbers:

0 x 0 = 0

1 x 0 = 0

0 x 1 = 0

1 x 1 = 1

Answer = 1000001_{2}

**(ii)** 111.01_{2} x 1.11_{2}

**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:** 111.01_{2} x 1.11_{2} = 1100.1011_{2}

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

**Solution**

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

**Explanation of 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 3 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 \( \scriptsize 1660_8 \)

âˆ´ 1304_{8} + 2610_{8} + 1660_{8} = 215404_{8}

**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**

Use the same steps as above to multiply 45_{7} and 63_{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 Ã— B = 11 Ã— 11 = 121. 121 divided by 12 = 10 remainder 1. We write 1 and carry over 10

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

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

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

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

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

A Ã— A = 10 Ã— 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!

âˆ´ 9571_{12} + 26B9_{12} + 8732_{12} = 8A7741_{12}

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