Till now, we have seen the cases wherein the numbers multiplied were near base of 10, 100, and so on. For the simplicity of this discussion, I will call them as base numbers.

However, there are some situations where numbers multiplied are a multiple of these  base numbers (near base of 10s) and the answer can be obtained by simply doubling, tripling, halving etc. of such base numbers.

Case 1: Numbers are a multiple of base numbers.

For example:

We have to solve: 212 x 104

212 can be written as 2 x 106

So, 212 x 104 = (2 x 106) x 104 = 2 x (106 x 104)

Now, 106 x 104 can be found easily using our method that we learnt in “Multiplying numbers near 100 – Part 1“.

And then we double the answer.

Below is representation of the above solution.

proportionately 1

Case 2: Numbers are half, one-fourth, and so on of base numbers.

For example:

We have to solve: 47 x 96

47 can be written as 94 ÷ 2

So, 47 x 96 = (94 ÷ 2) x 96 = (94 x 96) ÷ 2

Now, 94 x 96 can be found easily using our method that we learnt in “Multiplying numbers near 100 – Part 1“.

And then we half the answer.

Below is representation of the above solution.

proportionately 2

The same method can be used to solve below problems:

91 × 46                                66 × 104                              306 × 118

51 × 104                              206 × 54                              44 × 99

48 × 186                              224 × 216