Doodles

Draw a 'doodle' - a closed intersecting curve drawn without taking pencil from paper. What can you prove about the intersections?

Russian Cubes

I want some cubes painted with three blue faces and three red faces. How many different cubes can be painted like that?

Picture Story

Can you see how this picture illustrates the formula for the sum of the first six cube numbers?

Multiplication Square

Age 14 to 16Challenge Level

Well done to Maulik aged 11 who sent in some nice work on this problem. Neil's solution is given below.

For a 2 by 2 square with column headings of x and x+1, and row headings of y and y+1, Neil says that:

For the two by two square you can always express it algebraically like this:

So the diagonal from top right to bottom left is:

$(x+1)y+x(y+1) = xy+y+xy+x = 2xy+x+y$

Lets call that Z.

The diagonal from top left to bottom right is:

$xy+(x+1)(y+1) = xy+xy+x+y+1 = 2xy+x+y+1$

So the first diagonal is Z and the second Z+1 so the diagonal from top left to bottom right is always 1 more than the diagonal from top right to bottom left.

For a 3 by 3 square with column headings of x, x+1 and x+2, and row headings of y, y+1 and y+2, Neil says that:

The 3 by 3 square looks like this:

The diagonal from top right to bottom left is:

$\begin{split}x(y+2)+(x+1)(y+1)+(x+2)y &= xy+2x+xy+x+y+1+xy+2y \\&= 3xy+3x+3y+1\end{split}$

The diagonal from top left to bottom right is:

$\begin{split}xy+(x+1)(y+1)+(x+2)(y+2) &= xy+xy+x+y+1+xy+2x+2y+4 \\&= 3xy+3x+3y+5 \end{split}$

Let's say that $3xy+3x+3y = W$

The diagonal from top right to bottom left is W+1.

The diagonal from top left to bottom right is W+5. So the difference between the diagonals is 4.

More generally

Continuing with the same method:

Note that
1 = 1 ²
4 = 2 ²
10 = 1 ² + 3 ²
20 = 2 ² + 4 ²
35 =1 ² + 3 ² + 5 ²
56 = 2 ² + 4 ² + 6 ²
84 = 1 ² + 3 ² + 5 ² + 7 ²

Neil goes on to point out that the differences are the tetrahedral numbers.
Tetrahedral numbers are the sum of consecutive triangular numbers.
The formula is 1/6n(n+1)(n+2).
The first few tetrahedral numbers are 1, 4, 10, 20, 35, 56, 84, 120, ...
The tetrahedral numbers are found in the fourth diagonal of Pascal's triangle: