Will the Russian Peasant work with anything other than base 2? The Russian peasant method involves doubling and halving by 2. Therefore you will get exact remainders of either 1 or 0, which perfectly represents one of the multiplicands in binary form.
I just wanted to make sure if this is strictly true just for base 2.
 A: No, it only applies to base 2.
Look at this link: http://mathforum.org/dr.math/faq/faq.peasant.html#binary
It explains in depth for the Russian Peasant method of multiplication, and indeed it only applies to base 2. Normally you can use it to multiply two numbers together, yet converting to base 2 is just something you can do with it. 
Here is a quick quote: "Russian peasant multiplication is actually a quick way to convert two numbers to binary form, multiply them together, and convert back to our number system. The connection is not surprising, because binary numbers use base two, and Russian peasant multiplication depends on multiplying and dividing by two. To see the connection more clearly, let's investigate the problem 12*13" quoted from the link. Hope this helps.
A: You could actually use a modulus other than 2 to make this work, but the process will not be necessarily convenient.  Below is an example using base-10 representation, but doing doubling and halving, and after that, an example using base-10 representation, but doing tripling and taking thirds.  The difference with taking thirds, is that sometimes, you have to handle the case where you have a modulus of 0, 1, and 2 distinctly, whereas with the 'pure' form, you only need to worry about the cases with modulus 0 and 1.
Example with base 10 using doubling and halving:   
43 * 20 = 86 * 10 = 172 * 5 = 172 * (1 + 4) = 172 + 172 * 4 = 172 + 344 * 2 =
172 + 688 * 1 = 172 + 688 = 860

Example with base 10 representation and tripling/taking thirds:
43 * 20 = 43 * (2 + 18) = 43 * 2 + 43 * 18 = 86 + 43 * 18 = 86 + 129 * 6 =
86 + 387 * 2 = 86 + 774 = 860

Notice that since 20 mod 3 = 2, we keep track of 43 * 2, then, after tripling and taking thirds a couple times, we add in 387 * 2.  Sometimes, the remainder mod 3 will be 1, so you have to handle that case as well.
