What fractions can you find between the square roots of 65 and 67?
Find the polynomial p(x) with integer coefficients such that one solution of the equation p(x)=0 is $1+\sqrt 2+\sqrt 3$.
The squares of any 8 consecutive numbers can be arranged into two sets of four numbers with the same sum. True of false?
$37$ 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.
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.
$1 , 2 , 3 = 1 + 2,$
next comes $4$ ($ 2 + 2$ isn't allowed, because $2$ can only be
$5 = 1 + 4$
$6 = 2 + 4$
$7 = 1 + 2 + 4$
$8$ can't be $4 + 4 or 2 + 2 + 2 + 2$
$9 = 1 + 8$
$10 = 2 + 8$
$11 = 1 + 2 + 8$
$12 = 4 + 8$
$13 = 1 + 4 + 8$
$14 = 2 + 4 + 8$
$15 = 1 + 2 + 4 + 8$
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 $2^7$
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
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
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