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Napier's Location Arithmetic

Stage: 4 Challenge Level: Challenge Level:1

Here is an interactivity for Napier's Location Arithmetic.

Full Screen Version
This text is usually replaced by the Flash movie.

Part One : Pick two numbers as multipliers (factors) to multiply together, say $37$ and $51$. Choose some of the values from $1, 2, 4, 8, 16, 32, 64$ and $128$ to make a sum equal to each of those numbers. For example $37 = 32 + 4 + 1$ and $51 = 32 + 16 + 2 + 1$

Incidentally, could $37$ or $51$ have been made in another way ?

  • Now select the side numbers needed to make the factors you've chosen and press the "press when ready" button to begin.

  • Next click the counters in the grid one by one to see them move towards the bottom line.

  • Finally click the counters remaining in the bottom line to compile an answer (product) for your multiplication question.

Play with the application. Try different numbers.
Why does the process work - why does this method always produce a correct answer?

Continue playing. Perhaps try factor numbers with particular patterns of gaps and counters.

Part Two (quite hard) : Are there multipliers (factors) that produce a product which has a counter in every position along the bottom line?

Part Three (a real challenge) : Which multiplication question requires the most counters to represent both the multipliers (factors) and their product (that means the number of factor counters and product counters together making one single total).
For example $37$ uses $3$ counters at the side, $51$ uses $4$ side counters, and their product uses $9$ counters in the bottom line, altogether a total of $16$ counters ($3 + 4 + 9$).

Sending in solutions : we would love to hear your way of explaining why Napier's Location Arithmetic is a valid method for multiplication, or if you make some progress with parts $2$ or $3$, and can explain what you've done and what you've discovered, that would be wonderful to receive also.

An addition : it's excellent for us to hear that a problem we have offered has stimulated a new question in someone else's mind - thank-you very much Sheldon for this:

" I found this technique fascinating. Is there a way of inverting the process, i.e. Factorising. Starting with particular circles on the bottom line, and finding some process which could create the two factors you started with " ?