Puzzles

Every item bought will cause the pence in the total cost to fall by 1. So to spend £65.76, Susanna must have bought 24 items.

What is the smallest amount Susanna could spend for which we could not tell how many items she bought?

\(T_{n} = \frac{1}{2}n(n+1)\), so:

So, we are looking for \(n\) such that \((n+1)^2+(n+3)^2=(n+5)^2\). This is true when \(n=5\) (\(6^2+8^2=10^2\)).

Find \(n\) such that \(T_{n}+T_{n+1}+T_{n+1}+T_{n+2}=T_{n+2}+T_{n+3}\).

The area of an ellipse is \(\pi ab\) where \(a\) and \(b\) are the distances from the centre of the ellipse to the closest and furthest points on the ellipse.

In the first ellipse, \(a=5\mathrm{cm}\) and \(b=4\mathrm{cm}\), so the area is \(20\pi\mathrm{cm}^2\). In the second ellipse, \(a=5\mathrm{cm}\) and \(b=3\mathrm{cm}\), so the area is \(15\pi\mathrm{cm}^2\). Hence, the first ellipse has the larger area.

How far apart should the pins be placed to give the ellipse with the largest area?

Once the map is folded, it will look like this:

For the final tetrahedron to be regular, the red lengths must be equal. Let each red length be 2 (this will get rid of halves in the upcoming calculations). By drawing a vertical line in we can work out the width and height of the rectangle:

The width of the rectangle is 3 (one and a half red lengths). Using Pythagoras' Theorem in the blue triangle, we find that the height of the rectangle is \(\sqrt{3}\). Therefore, the ratio of the rectangle is \(\sqrt{3}:3\) or \(1:\sqrt{3}\).

If the ratio of the rectangle is \(1:a\), what is the ratio of the lengths of the sides of the tetrahedron?

\(y=x^{x^{x^{x^{...}}}}\) so \(y=x^y=e^{y\ln{x}}\).

By the chain and product rules, \(\frac{dy}{dx}=e^{y\ln{x}}(\frac{dy}{dx}\ln{x}+\frac{y}{x})\).

Rearranging, we get \(\frac{dy}{dx}=\frac{ye^{y\ln{x}}}{x(1-e^{y\ln{x}}\ln{x})}\).

This simplifies to \(\frac{dy}{dx}=\frac{x^{x^{x^{x^{...}}}}x^{x^{x^{x^{...}}}}}{x(1-x^{x^{x^{x^{...}}}}\ln{x})}\).

What would the graph of \(y=x^{x^{x^{x^{...}}}}\) look like?

They are all equal to one third.

The sum of the first \(n\) odd numbers is \(n^2\) (this can be proved by induction). This means that:

What is \(\frac{\mathrm{sum\ of\ the\ first\ }n\mathrm{\ odd\ numbers}}{\mathrm{sum\ of\ the\ first\ }n\mathrm{\ even\ numbers}}\)?

Let A have the equation \(y = mx + c\). B will have the equation \(y = cx + m\).

Therefore, \(mx + c = cx + m\).

Which rearranges to \(x(m - c) = m - c.\)

So \(x = 1\).

Substituting back in, we find \(y=m+c\).

The co-ordinates of the point of intersection are \((1,m+c)\).

Let \(a\), \(b\) and \(c\) be three distinct numbers. What can you say about the points of intersection of the parabolas:

Therefore \(x=\sqrt{2}\).

If \(x^{x^{x^{x^{...}}}}\) is equal to 4, what is the value of \(x\)?