Building ad hoc multiplication tables

Proportion tables are sometimes useful for doing long division. 

For many people, the most difficult part of that standard long division algorithm is estimating which multiple of the divisor to subtract.  Jakow Trachtenberg taught a simple trick for avoiding this difficulty.

In the next posting, we are going to divide 1543 by 21, using the standard long division algorithm and Trachtenberg's trick.

The first step is to invest some time making a table of multiples of `21`, up to the `9 xx 21`.  The `0` and `1` rows are obvious :

`\qquad\qquad\qquad[(0, 0),(1, 21)]`.

The `2` row can be found by doubling the second row :

`\qquad\qquad\qquad[(0, 0),(1, 21),(2,42)]`.

The `3` row can be found by adding the `1` row and the `2` row :

`\qquad\qquad\qquad[(0, 0),(1, 21),(2,42),(3,63)]`.

The `4` row can be found either by adding the `1` row to the `3` row, or else by doubling the `2` row.  One picks whichever is more convenient :

`\qquad\qquad\qquad[(0, 0),(1, 21),(2,42),(3,63),(4,84)]`.

The `5` row can be found either by adding the `2` row and the `3` row, or else by adding the `1` row and the `5` row.  Again, one picks whichever is more convenient :

`\qquad\qquad\qquad[(0, 0),(1, 21),(2,42),(3,63),(4,84),(5,105)]`.

One continues in this way, constructing the next row opportunistically, until at last one has :

`\qquad\qquad\qquad[(0, 0),(1, 21),(2,42),(3,63),(4,84),(5,105),(6,126),(7,147),(8,168),(9,189)]`.

This can be checked by casting out nines, if one knows how to do that.  (If not, it needs to be the subject of yet another post.)

Now we are ready divide anything by `21`.

`\qquad\qquad\qquad[(0, 0,(0)),(1, 21,(3)),(2,42,(6)),(3,63,(0)),(4,84,(3)),(5,105,(6)),(6,126,(0)),(7,147,(3)),(8,168,(6)),(9,189,(0))]`.

This checks out, so we can rely on the table we have made.