Multiplying very small numbers without the use of a calculator

Question

As part of my college course I have to sit some math exams and I cannot use a calculator. The problem is I suck at maths with a calculator not to mind how bad I am at it without using a calculator! Anyways I'm wondering if anyone has any tips for calculating answers in my head. Particularly answers to questions where very small numbers are multiplied together.

Eg: $$0.0024 \cdot 0.12, 0.000054 \cdot 0.002, 0.0725 \cdot 0.41 $$ Thanks in advance!

Answer 1

Multiply $24\times 12$ or $54\times 2$ or $725\times 41$ (though for the last one I would reach for my scratch paper), and then adjust the position of the decimal point afterwards.

In other words, for example, $$ 0.0024\times 0.12 = (24\times 10^{-4})\times (12\times 10^{-2}) = (24\times 12)\times 10^{-4}10^{-2} = 288 \times 10^{-6} = 0.000288 $$

Answer 2

The most obvious thing for me to do, and it is something I do regularly when making calculations in my head or in paper, is to just "ignore" the fact that the number is a decimal number. What do I mean by "ignoring" it?

Well, say you want to find $0.000054 \cdot 0.002$. From the decimal point $.$ up to the $4$, which is the last digit in $0.000054$, I can count $6$ digits. For $0.002$, from the $.$ to the $2$ (the last digit) I count $3$. Then I add those up, yielding $9$, and save it for later. Now I do the regular calculation

$$54\cdot2 = 108$$

And now I reverse the process of counting. From the decimal point $.$ to this $8$, I must be able to count $9$ digits. So, I need to have something like

$$0.0\cdots0108$$

It turns out I want $0.000000108$; This is the number that will make this counting right.

Another example:

$$0.0725 \cdot 0.41$$

For $0.41$ I count $2$ decimal places and for $0.0725$ I count $4$, adding up to $6$. I then calculate

$$725\cdot41 = 29725$$

in my head or in paper, and then correct it to have $6$ decimal places:

$$0.029725$$

The mathematical explanation for this "trick" is that decimal numbers are of the form $k\cdot10^n$ where $k$ is just some whole number. Then, when multiplying two together, you have

$$(k\cdot10^{-n}) \cdot (j\cdot10^{-p}) = (k\cdot j)\cdot(10^{-n}\cdot10^{-p}) = (k\cdot j)\cdot(10^{-n-p})$$

And thus the $10^{-n-p} = 10^{-(n+p)}$ is just the "correction" part where we add the decimal places we took.