Karatsuba multiplication for $x = 123 , y = 100$ Given the numbers $x = 123$ and $y = 100$ how to apply the Karatsuba algorithm to multiply these numbers ?
The formula is 
xy=10^n(ac)+10^n/2(ad+bc)+bd

As I understand $n = 3$ (number of digits) and I tried writing the numbers as 
x = 10*12+3 , y = 10*10 +0 thus a = 12 , b = 3 , c = 10 , d = 0

or
x = 100*1+23 , y = 100*1 +0 thus a = 1 , b = 23 , c = 1 , d = 0

I looked at some explanations of the algorithm and tried it successfully for other numbers , but I don't know how to solve it in this particular case.
Is this example part of a more general case of the algorithm (like 3-digit numbers)? I found a question that may be related (Karatsuba multiplication with integers of size 3) and from the answer I gather that it's impossible , so is it that Karatsuba can't multiply $2$ numbers of $3$-digits or is there a way to do this ?

$xy=123\cdot 100$
$(12\cdot 10+3)(10\cdot 10+0)$
$xy=10^n(ac)+10^n/2(ad+bc)+bd$, where $n=2
a=12, b=3, c=10, d=0$
subcategory 1:
$ac=12\cdot 10 = 120$
$bd=3\cdot 0 = 0$
$ad+bc= (12+3)(10+0)=150-120-0 = 30..$
$120\cdot 10^2 + 30\cdot 10^1 + 0$
subcategory 2:
$12\cdot 10$
$a=1,
b=2,
c=1,
d=0$
$ac=1$
$bd=0$
$ad+bc=(a+b)(b+d)-ac-bd$
$ad+bc= (1+2)(1+0)-1-0 = 2$
$ac\cdot 10^2 + (ad+bc)\cdot 10 + bd =120..$
