Puzzles

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 \(x\)?

By Euler's formula, \(i=e^{i\frac{\pi}{2}}\). This means that:

It is notable that this is a real number

What is \(-i^{-i}\)?

Let \(y = \arccos(x)\)

This means that \(x = \cos(y)\)

This gives us this triangle, as cosine is adjacent ÷ hypotenuse.

Call the angle at the top of the triangle \(z\).

Sine is opposite ÷ hypotenuse, so \(\sin(z) = x\)

So \(z = \arcsin(x)\)

Angles in triangle add to \(\pi\), so \(z = \frac{\pi}{2}-y\)

Hence, \(\arcsin(x)= \frac{\pi}{2}-y\)

Therefore \(\arccos(x) + \arcsin(x) = y + \frac{\pi}{2}-y\) \(=\frac{\pi}{2}\)

What are the values of \(\mathrm{arcsec}(x) + \mathrm{arccosec}(x)\) and \(\arctan(x) + \mathrm{arccot}(x)\)?

There are two ways to make 20 by multiplying three digits: \(2\times 2\times 5\) and \(1\times 4\times 5\). Listing all the possible orderings of these, we have:

145

154

415

451

514

541

225

252

522

Therefore, there are 9 different three digit numbers where the product of the digits is 20.

How many 4 digit numbers are there where the product of the digits is 20?

5 digit?

\(n\) digit?

By walking in a straight line, the people will follow a great circle. They begin by travelling parallel to each other, so the 1km they are apart at the start is the furthest they will be apart. They will be 1km apart again after walking half way around the world and will collide at one quarter and three quarters of the way around.

Hence, each person will walk one quarter of the circumference of the Earth before colliding, which will be 10,018km.

Another way to see that this is the answer is to assume that the people stand on the equator and walk North. They will meet at the North Pole, one quarter of the way round.

If two people stand 10km apart and walk in the same direction, how far will the have to walk until they collide due to the curvature of the Earth? (diameter of Earth = 12,742km)

CG is equal to AF as they are both diagonals in a rectangle.

AF is a radius of the triangle, so AF = 10.

Therefore CG = 10.

Calculate the lengths of CI and CJ.

Calculate the lengths of HG.