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Do Unto Caesar

At the beginning of the night three poker players; Alan, Bernie and Craig had money in the ratios 7 : 6 : 5. At the end of the night the ratio was 6 : 5 : 4. One of them won $1 200. What were the assets of the players at the beginning of the evening?

Plutarch's Boxes

According to Plutarch, the Greeks found all the rectangles with integer sides, whose areas are equal to their perimeters. Can you find them? What rectangular boxes, with integer sides, have their surface areas equal to their volumes?


Using some or all of the operations of addition, subtraction, multiplication and division and using the digits 3, 3, 8 and 8 each once and only once make an expression equal to 24.

How Long Is the Cantor Set?

Age 11 to 14 Challenge Level:

Gregory from Magnus C of E School in Nottingham and Luke from St Patrick's School reasoned correctly. Here is Luke's explanation:

For this question I found a pattern in it:
For example $C_1$ has length of 1 and $C_2$ has length of $\frac{2}{3}$
and there is a pattern in this because 1 $\times$2 = 2, so there is your numerator,
and to get your denominator you multiply 1 by 3, so there you have your $\frac{2}{3}$.

So to put this into easier words, it simply means multiply the numerator by 2 the whole way and multiply the denominator by 3.
So there is my strategy.

So the answer to $C_3$ is $\frac{4}{9}$
and the answer to $C_4$ is $\frac{8}{27}$
and finally $C_5$ is $\frac{16}{81}$.
$C_n$ is $\left(\frac{2}{3}\right)^{n-1}$
and when n $\rightarrow$ infinity, $ C_n \rightarrow$ 0

David from Gordonstoun School got the same result and added:
In a geometric progression,
$a$ = $ 1^{st}$ term
$r$ = constant factor
$n$ = number of terms

Any value in a geometric progression is $ ar^{n-1}$
In this case
$a$ = 1
$r$ = $\frac{2}{3}$

$r$ = $\frac{2}{3}$ (which is less than 1),
so the higher its power, the closer the result is to zero.
(Any positive number smaller than 1, to the power of infinity, tends to zero.
Therefore $ r^{n}$ tends to zero)

So as $n$ tends to infinity,
$ar^{n-1}$ tends to $a \times 0 = 0$
So the Cantor Set's length is zero.

Liam from Wilbarston School reasoned in a similar way:

The length of $C_{n+1}$ is simply two thirds of the length of $C_n$,
as $C_{n+1}$ is purely $C_n$ with the middle thirds removed.

Now taking $L_n$ to be the length of $C_n$:
$L_2$ = $\frac{2}{3}$,
$L_3$ = $\frac{4}{9}$,
$L_4$ = $\frac{8}{27}$ etc. etc.

It's obvious that $L_n$ = $\left(\frac{2}{3}\right)^{n-1}$.
So as n tends to infinity, $L_n$ gets increasingly smaller, i.e. tends to zero.

Therefore the length of the Cantor set is zero. In fact, the Cantor set is a set of points, because endpoints of line segments will never be removed, only middle thirds.

And as Euclid said, 'A point is that which has no part', i.e. a point has zero length, zero width and zero height.

Well done to you all.