Puzzles

Let \(p(n)\) denote the number of people in the correct seat when the table is rotated by \(n\) people. From the question, we know that:

As every guest can be put into the correct place by one rotation, we know that:

As \(p(0)=0\):

This sum has 23 positive integers adding up to 24, so one of the rotations must lead to at least two people being in the correct places.

If the 24 guests sat randomly and one person was in the correct seat, could you still rotate the table so that two people are correctly seated?

If \(n\) is odd, the vertices of a regular \(n\)-gon are balanced.

If \(n\) is even, a balanced set can be constructed as follows:

If \(n=4\), this set is balanced (both the triangles are equilateral):

Draw a triangle with its centre at one of the points, going through the other three points:

For \(n>4\), repeatedly add a two new points on the circle which form an equilateral triangle with the centre. For example, for \(n=6\), this set is balanced:

And for \(n=8\), this set is balanced:

This is true for any base larger than 2. This is because:

Which other base 10 square numbers are always square in high enough bases?

The second hand will always be pointing at one of the 60 graduations. If the minute and hour hand are 120° away from the second hand they must also be pointing at one of the graduations. The minute hand will only be pointing at a graduation at zero seconds past the minute, so the second hand must be pointing at 0. Therefore the hand are either pointing at: hour: 4, minute: 8, second: 0; or hour: 8, minute: 4, second: 0. Neither of these are real times, so it is not possible

If the second hand moves continuously instead of moving every second, will there be a time when the hands of the clock are all 120° apart?

Pick two points (A and B) of the same colour (say red). Consider the points:

If at least one of C, D or E is red, then three equally spaced red points have been found. If C, D and E are blue, then they are three equally spaced blue points.

Will there be four equally spaced points of the same colour?

Yes. To show this, let's suppose there is an arrangement of pins with no monochrome triangle.

Let the middle pin be blue (if it is not blue, then swapping all the colours will make it blue without adding/removing monochrome triangles).

Pins 8, 2 and 6 make a triangle. Therefore one of these pins must be blue. By symmetry, any of these situations is equivalent to 8 being blue.

Pins 9 and 4 must be red to prevent a blue triangle.

Pin 3 must be blue to prevent a red triangle.

Pin 6 must be red to prevent a blue triangle.

But now whichever colour 10 is, it will make a monochrome triangle. Therefore there must be three pins of the same colour which lie on the vertices of an equilateral triangle.

What is the smallest \(n\) such that any mixture of \(n^2\) red and blue pins arranged in a square grid must contain four pins of the same colour which lie on the vertices of a square?

Many more puzzles like this one can be found in my post on the Chalkdust blog.

The maximum you can get is £50, with the taxman getting £28. Here is how to get it:

Can you always beat the taxman if £1 up to £\(n\) are available?