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Station  11F

Earlier we computed 19×519 \times 5 and got this picture for the answer 9595.

I11S11F - Image01

If we were given this picture of 9595 first and was told that it came from a multiplication problem with one of the factors being 55, could we deduce what the other factor must have been? That is, can we use the picture to compute 95÷595 \div 5?

Since 5=4+15 = 4 + 1 we will need to slide counters on this picture so that two copies of the same pattern appear in the shaded two rows.

I11S11F - Image02

Slide the leftmost dot up to the top shaded row and we see it “completes” the 1616 column. Let’s not touch the counters in that column ever again.

I11S11F - Image03

We are now left with a smaller division problem: dividing 8+4+2+18+4+2+1 (that is, 1515) by 55.

Slide its leftmost dot up to the top shaded row. This completes the 22s column and let’s never touch the counters in that column again.

I11S11F - Image04

This leaves us with a smaller division problem to contend with: 4+14+1 divided by 55. Slide its leftmost dot up to the top shaded row to complete the 11s column.

I11S11F - Image05

We see that we have now created the picture of 19×519 \times 5 and so 95÷5=1995 \div 5 = 19.

This loosely illustrates the general principle for doing division on Napier’s checkerboard:

Represent the dividend by dots in the bottom row and the divisor by shaded rows.

Slide the leftmost dot to the top shaded row.

Complete the leftmost column of dots possible in some way you can (you might need to unexplode some dots) and when done never touch those dots again. What is left is a smaller division problem and repeat this procedure for the leftmost dot of that problem.

The procedure described here is loose as our computation 95÷5=1995 \div 5 = 19 ran into no difficulties.

Let’s try 250÷13250 \div 13 for something more involved. Here’s its setup.

I11S11F - Image06

Slide the leftmost dot to the highest shaded row. Doing so shows we need to work with the 1616s column, but it is not complete.

I11S11F - Image07

We can complete it by sliding the current leftmost dot into that column. (That’s convenient!)

I11S11F - Image08

Now we have a smaller division problem to work on. Slide the leftmost dot up to the highest shaded row.

I11S11F - Image09

What’s the leftmost column we can complete right now without ever touching those dost of the 1616s column? We see that there is no means complete the 88s column. (What dot can we slide into its top?)

There is no means to complete the 44s column either. (How do we slide a dot into that 4×44 \times 4 cell?)

So let’s work on the 22s column. I can see by sliding the dot in the 88s column and performing a (horizontal) unexplosion from the 44s column we can fill up the 22s column.

I11S11F - Image10

The 22s column is a bit overloaded. Let’s unexplode one of the dots the top pair (horizontally).

I11S11F - Image11

All the action is now left in the 11s column. What can we do to make that column complete? (Remember, dots in completed columns are never to be touched again.) Let’s unexplode downwards a number of times.

I11S11F - Image12

This does complete the 11s column, but with three ones too many.

If we had three less dots —247247 instead of 250250— then we would have, right now, a picture of 19×1319 \times 13 showing that 247÷13=19247 \div 13 = 19. So it must be then that 250÷13250 \div 13 has a remainder of three and so

250÷13=19+313250 \div 13 = 19 + \dfrac{3}{13}.

Question: Compute 256÷10256 \div 10 via Napier’s method.

Question: Is it possible to do polynomial division with Napier’s checkerboard? (Can one compute 11x\dfrac{1}{1-x}?)

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