How to multiply two numbers in base 3 using Karatsuba I want to multiply $a =  11201$ and $b =  1022$ which are in base $3$, so I have a problem I do not know how to start, normally I should split the numbers into other number of two digits, but I do not know how since a has $5$ digits.

Any help please ?
Sorry I did not read the question ... Use the Karatsuba algorithm. I use the notation in the wiki page.
So our base $B=3$ and the best split of the numbers is into three digit numbers thus $m=3$. (All numbers will be understood to be in base 3 from here (apart from the exponent of $B$))
\begin{eqnarray*}
x=11B^3+201 \\ y=1B^3+022
\end{eqnarray*}
Now the theory says $xy= z_2 B^6+z_1 B^3 +z_0$
\begin{eqnarray*}
z_2=x_1 y_1 \\ z_1=(x_1+x_0)(y_1+y_0)-z_2-z_0 \\ z_0=x_0y_0
\end{eqnarray*}
The finesse,of course is in the second equation, only one multiplication (computationally expensive) needs to be performed at the expense of a few more additions &amp; subtractions. We have
\begin{eqnarray*}
z_2=11 \\ z_1=(212)(100)-11-12122=1220 \\ z_0=12122
\end{eqnarray*}
Now put six zeros on the end of $z_2$  three zeros on the end of $z_1$ and add these onto $z_0$ and we have $20002122$.
You don't need the messy lattice. Just multiply exactly the same as in base 10, remembering the base 3 multiplication table:

$$0* 0 = 0 \ \ \ 0*1 = 0 \ \ \  0*2 = 0$$
$$1*0 = 0 \ \ \ 1*1 = 1 \ \ \  1*2 = 2$$
$$2*0 = 0 \ \ \  2 * 1 = 2 \ \ \ 2*2 =11_3$$

The only carrying you need to do is when you have $2*2$
Then remember base 3 addition, noting $2 + 1 = 10_3$