Karatsuba Multiplication

Karatsuba's equation to reduce the amount of time it takes in brute force multiplication is as follows (I believe this is a divide-and-conquer algorithm):

$$x y = 10^n(ac) + 10^{n/2}(ad + bc) + bd$$

My question is this. Where did the $10^{n/2}$ and $10^n$ come from?

Thanks

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Maybe there is somme secret code in use in that area, for all those who don't know this is would be helpful if you could reveal the relations between $x, y, a, b, c, d$ and $n$ to us. –  user20266 Aug 4 '12 at 16:28

Let $x = a10^n + b$ and $y = c10^n + d$, and $a,b,c,d < 10^n$. Then to find the product $xy$, one notes that $xy = ac10^{2n} + (ad + bc)10^n + bd$. The advantage of the algorithm is that you can just calculate the products $ac, ad, bc$ and $bd$, all of which have much smaller sizes than the original (for large $n$).
You'll note that I use $2n$ and $n$ instead of $n$ and $n/2$, but the idea is the same.
@David Johnson: $n$ is just some number, which may be different for each multiplication. For example, if $x = 123456$ and $y = 654321$, maybe I would use $n=3$ to write this as $123\cdot 10^3 + 456$ and $654\cdot 10^3 + 321$. The idea of using $n$ to be half the number of digits as the larger of the two numbers is the general use (and it's usually done in binary). –  mixedmath Aug 4 '12 at 16:44
@mixedmath ohhh so you just foil it out into $xy = ac10^{2n}$ etc.. –  The Internet Aug 4 '12 at 16:52