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I'm a bit embarrassed to ask this, since it seems like such a trivial question, but I'm wondering why we put $i$ zeroes in the $i$th iteration of the long multiplication (pencil-and-paper) algorithm.

I suspect it has something to do with us moving up one number place, but I just can't make the connection more explicit in my mind. My closest formulation is that for the first iteration, we're working with a multiplier digit in the $b^0$ place in a system with base $b$, and then $b^1$ for then next rightmost multiplier digit, and so on.

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As an example consider $$ 325\times 326 $$ When we do the algorithm we are using the distributive property i.e. $$ 325\times 326=325(300+20+6)=325(6)+325(20)+325(300)\tag{1}. $$ For the first step of the algorithm, we compute $325\times 6$. Then in the algorithm we add a zero to the next line and concatenate to the $0$ the result of $325 \times 2$. But this corresponds exactly to computing $325(20)$. Lastly we would add two zeroes to the next line and concatenate to the zeros the result of $325\times 3$. But this would correspond exactly to $325\times 300$. Finally the algorithm says to add up the numbers in the rows, exactly what we have in (1).

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  • $\begingroup$ "Then in the algorithm we add a zero to the next line and concatenate to the 0 the result of 325×2. But this corresponds exactly to computing 325(20)." Ah, there it is. Thank you! $\endgroup$ – AleksandrH Jul 1 '18 at 15:26

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