MSC 17 – Base Multiplication: Multiplying “Teen” Numbers and Others

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 base multiplication, is shown below:
(x + a) * (x + b)
= x2 + (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

= 11,128Note 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

  1. 12 x 13 =
  2. 14 x 17 =
  3. 15 x 18 =
  4. 108 x 101=
  5. 116 x 102 =
  6. 108 x 112 =
  7. 112 x 114=
  8. 123 x 106 =
  9. 1021 x 1006 =
  10. 1432 x 1002 =

Answers to all exercises are found in the answer key.

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