Most of us have learned by heart the multiplication table up to 10 x 10. A simple technique will enable us to extend our multiplication power up to 20 x 20.

Example 17.1: Compute **14 x 12**

- Cover one of the ten’s digit and add what remains to the other number:
**14 + 2**or**4 + 12**will both give**16**. This will be the first or left hand part of the product. - Cover both ten’s digit and multiply what remains:
**2 x 4 = 8.**This will be the second or right hand part of the answer. - Thus
**14 x 12 = 16 | 8**or**168**

What we actually did in the first step is to add the excess over the base (10) of a number to the other.In the second step we multiplied the excesses. The algebraic proof of this method, known as , is shown below:base multiplication(x + a) * (x + b) = x^{2} + (a + b) * x + a * b = (x + a + b) * x + a * b = [ (x + a) + b] * x + a * b |

In our example above, **x = 10, a = 4** and **b = 2. **So we have

**14 x 12
= [ (10 + 4) + 2 ] * 10 + 4 * 2**

**= 160 + 8**

**= 168**

In this case both multiplicands are **above** the base.

Example 17.2: Compute **16 x 13**

**(10 + 6) x (10 + 3)**

- Add
**3**(excess of**13**over**10)**to**16**to get**19**

=**( 10 + 6) + 3 |**

= 19 |

- Multiply the excesses
**6 x 3 = 18.**

**= 19 | 6 x 3**

**= 19 | 18** - Since there is only one zero in the base, only one place is allotted for the right hand part. Therefore the
**1**of**18**must be carried or added to**19**to get a final answer of**208**

**= 20 | 8 = 208**

Example 17.3: Find **107 x 104**

- The left part is
**107 + 4 = 111**, and the right part is**7 * 4 = 28**

**(100 + 7) * (100 + 4)**

= 107 + 4 | 7 x 4

**= 111 | 28**

** = 11,128**Note that in this example we did not write the two zeroes after **111** but two places are reserved for the **28.**

Example 17.4: Find **1,025 x 1,012**

- The left part is
**(1025 + 12 ) = 1037**and the right part is**25 * 12 = 300**

**1,025 x 1,012**

**= (1000 + 25) * (1000 + 12)**

= 1025 + 12 | 25 * 12

**= 1037 | 300**

**= 1,037,300**

Example 17.5: Find **115 x 111**

- The left part is
**(115 + 11)**and the right part**165**

**115 x 111**

= (100 + 15) (100 + 11)

= 115 + 11 | 15 * 11

= 126 | 165 - Since we have only two zeroes in our base, we can allot only two spaces for the right hand part of the answer, so we must “carry” the
**1**of**165**into the left side.

**= 12,765**

Example 17.6: Find **102 x 104**

- This is pretty straightforward with the left
**102 + 4 = 106**and the right is 2 * 4 = 8

**102 * 104**

= (100 + 2) (100 + 4)

= 102 + 4 | 2(4)

= 106 | 08 - In this case, the product of the excesses is
**8**but since the base is**100**, two spaces are allotted to it. We write it as**08.**

**= 10,608**

Example 17.7: **103 x 119**

- The left side is
**103 + 19 = 3 + 119 = 122**and the right side is**3 * 19 = 57**

**(100 + 3) * (100 + 19)**

= 119 + 3 | 3 * 19

= 12,257 - Note that in this example we chose to add the smaller excess to the other number. This always leads to a simpler calculation.

Exercise 17: Find the following products using base multiplication

- 12 x 13 =
- 14 x 17 =
- 15 x 18 =
- 108 x 101=
- 116 x 102 =
- 108 x 112 =
- 112 x 114=
- 123 x 106 =
- 1021 x 1006 =
- 1432 x 1002 =

Answers to all exercises are found in the answer key.

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