$37$ expressed as $32 + 4 + 1$ multiplies $51$ expressed as $32 + 16 + 2 + 1$
The yellow discs mark the answers to each part of that calculation, for example the disc with an orange dot is $4$ lots of $32$ ( $4$ from the '$37$' multiplying $32$ from the '$51$' ) .
In the red diagonal line there are two yellow discs (products) to include : $4$ lots of $16$ and $32$ lots of $2$, these answers are the same ($64$), in fact everything in the same diagonal has the same value.
And, here's the important bit, two discs in the same diagonal are worth one disc in the next diagonal up.
Look again at the interactivity doing the calculation to see this in action gathering up and combining the bits which together make the answer.Well done Patrick from Woodbridge School and Tom from Silverdale Schoolfor illuminating what's going on here.
In fact any number can only be made in one way.
To see why this is, remember that it must be a sum of powers of two and that any power can only be used once.Count up from one and watch what happens :
$1 , 2 , 3 = 1 + 2,$
next comes $4$ ($ 2 + 2$ isn't allowed, because $2$ can only be used once)
$5 = 1 + 4$
$6 = 2 + 4$
$7 = 1 + 2 + 4$
$8$ can't be $4 + 4 or 2 + 2 + 2 + 2$
$9 = 1 + 8$
$10 = 2 + 8$
$11 = 1 + 2 + 8$
$12 = 4 + 8$
$13 = 1 + 4 + 8$
$14 = 2 + 4 + 8$
$15 = 1 + 2 + 4 + 8$When you add the extra one each time if there's a one already that makes a two, and if there was already a two the pair of twos now make a four, and so on until there's a gap, a power of $2$ that wasn't already present.
Napier's Location Arithmetic works because it is based on binary numbers. $37$ and $51$ cannot be made in any other way because in binary they are $10011$ and $11011$, and any different set of ones and zeros would result in a different number.
Part Two : Are there multipliers (factors) that produce a product which has a counter in every position along the bottom line?
It is impossible to fill every space on the bottom line as the total of a counter on each of the bottom lines is $65535$ (adding up the powers of $2$) and the maximum number that can be achieved is $255 \times 255 = 65025$ as the total of $2^0$ through to $2^7$ is $255$.
Therefore this is the largest number that can be displayed and so it is not possible to have a counter on each of the bottom squares.
The maximum number of squares on the bottom line that can be displayed is actually $13$ as after this to fill any new square another square must always be sacrificed.
There is no power a $2$ that can be taken from $65535$ (which is what would happen if all the bottom squares where filled) that is divisible by two numbers $255$ or under.
The largest total number of counters possible is $24$ for $255\times 255$.
This is also the largest number that can be represented.
Removing one $255$ from this number changes a one to a two and loses the $256$, removing two $225$s adds a two and loses the $512$ (totalling one counter loss in total).
This continues adding a power of $4$ while losing another power of four and so all of these total $24$ counters (apart from $255 \times254$ which totals $23$).
Now if a multiple of $253$ is removed from $255 \times253$ giving $254 \times 253$ then a $1024$ and $1$ is lost and a $512, 256$ and $4$ gained, giving a zero net sum.
This works in the same way for all the others and so does not increase the number of counters in total.
This trend continues with either counters being lost or none gained and so the largest possible total is $24$ for $255 \times255$.Thank-you very much Patrick and Tom