I got something method like the simplest multiplying methods when I googling.
If you had a number, like 123.456 and you wanted to multiply by 100 you'd just ... Move the decimal point to the right! Exactly. Move it 2 places to the right, since 100 = 10^2 And to multiply by 1/10000 = 10^-4, you'd move the decimal point 4 places to the left. These have to be the world's easiest multiplications,
But actually I am unfamiliar with this explain. So would you please rephrasing this?
This example shows why it works:
$$ 10^2\times 1.749=10^2\times(\overbrace{1\times 10^0+7\times 10^{-1}+4\times 10^{-2}+9\times10^{-3}}^{\textrm{definition of }1.749}) $$
$$ =1\times (10^2\times 10^0)+7\times (10^2\times 10^{-1})+4\times (10^2\times 10^{-2})+9\times(10^2\times 10^{-3})$$
$$=\underbrace{1\times 10^2+7\times 10^{1}+4\times 10^{0}+9\times10^{-1}}_{\textrm{definition of }174.9} $$
$$=174.9$$
The reason this works is because of the base of our number system. In elementary school, we refer to the digits of a number by the "place" they are in. For example, in the number 5,463, there is a 3 in the "one's place" and a 6 in the "ten's place" and a 4 in the "hundred's place" etcetera.
The mathematical way of saying this is that each new digit counts the coefficient in front of a higher power of 10. So, going back to our example:
$\begin{align} 5,463 &= 5 \times 10^3 + 4 \times 10^2 + 6 \times 10^1 + 3 \times 10^0\\ &= 5 \times 1000 + 4 \times 100 + 6 \times 10 + 3 \times 1 \\ \end{align}$
The second equality is why we say the digits are in the "one's place", "ten's place", etcetera.
Now, when we go to multiply by a power of 10, we can use laws of exponents to make our work much easier.
$\begin{align} 18 \times 100 &= 18 \times 10^2\\ &= (1 \times 10^1 + 8 \times 10^0) \times 10^2\\ &= 1 \times 10^3 + 8 \times 10^2 \\ &= 1 \times 1000 + 8 \times 100 \\ &= 1000 + 800 = 1800 \end{align}$
The same holds for negative powers of 10, one example of which is the one you give in your question.
A number written in some basis $b$ can be written as:
$$\sum_{k=-\infty}^{\infty}a_kb^k$$
Where the $a_k \in \{0,\cdots,b-1\}$.
So any multiplication with a monomial in $b$ actually becomes a convolution with a time shift / or lazy filter. Because it moves every $a_k$ a fixed amount of "steps".
