Russian Peasant Method for multiplication What exactly happens with the remainder in this algorithm? I don't understand why it is "dropped". 
Example:
$$\begin{array}{c}
\text{Half}&&\text{Double}&\text{Remainder}\\ \hline
38&\times&15&1\\
18&\times&30&1\\
9&\times&60\\
4&\times&120\\
1&\times&480
\end{array}$$
What is happening with the $1$'s???
 A: This is a method for multiplication?
Well, in the ideal case, we can precisely halve one side, and double the other:
$4\times120 = 2\times240 = 1\times480$
However, in some cases, we cannot exactly halve, because we have an odd number to halve. We can deal with this by having a remainder:
$9\times60 = 4\times120 + 60$. So, we have an additional 60 that is not dealt with - so we'll ignore it for now, keep going, and add 60 to the final result later on.
I'm not exactly clear about your method, but I assume it would be better as follows:
$$\begin{array}{c}
\text{Half}&&\text{Double}&\text{Remainder}\\ \hline
38&\times&15&\\
19&\times&30&\\
9&\times&60&30\\
4&\times&120&60\\
1&\times&480
\end{array}$$
So, the result will be $38 \times 15 = 480 + 60 + 30 = 570$
This algorithm would seem to be quicker if we halved the smaller number. But, this leads to two problems - 38 is harder to double than 15 (which would double to a multiple of 10), and because 15 is just one below 16 (a power of 2), meaning we get a lot of nasty remainders:
$$\begin{array}{c}
\text{Half}&&\text{Double}&\text{Remainder}\\ \hline
15&\times&38&\\
7&\times&76&38\\
3&\times&152&76\\
1&\times&304&152\\
\end{array}$$
It seems like we did one less step of work. But, the result is $15 \times 38 = 308+154+76+38 = 570$. Yuck! This way around doesn't make it simple to calculate by hand.
