4
$\begingroup$

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???

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    I assume that you’re talking about what is sometimes called [Russian peasant multiplication](http://en.wikipedia.org/wiki/Ancient_Egyptian_multiplication); you should make this clear in the question.2012-05-22
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    Meh, cross-editing.2012-05-22
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    @Jennifer: First of all, the half of 38 is not 18. And please explain what your question is. Do you not know how to proceed or do you not understand why it works?2012-05-22
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    As Phira says, half of $38$ isn’t $18$; moreover, there’s no remainder there, or when taking half of $18$, but there *is* a remainder when taking half of $9$.2012-05-22
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    Welcome to Math.SE Jennifer.2012-05-22

2 Answers 2