In laser physics, a “white cell” is a mirror system that acts as a delay line for the laser beam. The beam enters the cell, bounces around on the mirrors, and eventually works its way back out.

The specific white cell we will be considering is an ellipse with the equation $4x^2+y^2=100$.

The section corresponding to $-0.01\leq x\leq 0.01$ at the top is missing, allowing the light to enter and exit through the hole.

The light beam in this problem starts at the point (0.0, 10.1) just outside the white cell, and the beam first impacts the mirror at (1.4, -9.6).

Each time the laser beam hits the surface of the ellipse, it follows the usual law of reflection “angle of incidence equals angle of reflection.” That is, both the incident and reflected beams make the same angle with the normal line at the point of incidence.

In the figure on the left, the red line shows the first two points of contact between the laser beam and the wall of the white cell; the blue line shows the line tangent to the ellipse at the point of incidence of the first bounce.

The slope $m$ of the tangent line at any point $(x,y)$ of the given ellipse is: $m=-4x/y$.

The normal line is perpendicular to this tangent line at the point of incidence.

The animation on the right shows the first 10 reflections of the beam.

How many times does the beam hit the internal surface of the white cell before exiting?

The 5-digit number, $16807=7^5$, is also a fifth power. Similarly, the 9-digit number, $134\,217\,728=8^9$, is a ninth power.

How many $n$-digit positive integers exist which are also an $n$th power?

The sequence 1, 1, 1, 3, 5, 9, 17, 31, 57, 105, 193, 355, 653, 1201 … is defined by $T_1=1, T_2=1, T_3=1$ and $T_n=T_{n-1}+T_{n-2}+T_{n-3}$.

It can be shown that 27 does not divide any terms of this sequence. In fact, 27 is the first odd number with this property.

Find the 124th odd number that does not divide any number of this sequence.

Consider the fraction, $n/d$, where $n$ and $d$ are positive integers. If $n<d$ and $HCF(n,d)=1$, it is called a reduced proper fraction.

If we list the set of reduced proper fractions for $d\leq 8$ in ascending order of size, we get: \[\frac{1}{8},\frac{1}{7},\frac{1}{6},\frac{1}{5},\frac{1}{4},\frac{2}{7},\frac{1}{3},\mathbf{\frac{3}{8}},\mathbf{\frac{2}{5}}, \mathbf{\frac{3}{7}},\frac{1}{2},\frac{4}{7},\frac{3}{5},\frac{5}{8},\frac{2}{3},\frac{5}{7},\frac{3}{4},\frac{4}{5},\frac{5}{6},\frac{6}{7},\frac{7}{8}\]

It can be seen that there are 3 fractions between $1/3$ and $1/2$.

How many fractions lie between $1/3$ and $1/2$ in the sorted set of reduced proper fractions for $d\leq12000$?

Consider the fraction, $n/d$, where $n$ and $d$ are positive integers. If $n<d$ and $HCF(n,d)=1$, it is called a reduced proper fraction.

If we list the set of reduced proper fractions for $d\leq 8$ in ascending order of size, we get: \[\frac{1}{8},\frac{1}{7},\frac{1}{6},\frac{1}{5},\frac{1}{4},\frac{2}{7},\frac{1}{3},\frac{3}{8},\mathbf{\frac{2}{5}}, \frac{3}{7},\frac{1}{2},\frac{4}{7},\frac{3}{5},\frac{5}{8},\frac{2}{3},\frac{5}{7},\frac{3}{4},\frac{4}{5},\frac{5}{6},\frac{6}{7},\frac{7}{8}\]

It can be seen that there are 21 elements in this set.

How many elements would be contained in the set of reduced proper functions for $d\leq 1\,000\,000$?

Taking three different letters from the 26 letters of the alphabet, character strings of length three can be formed.

Examples are ‘abc’, ‘hat’ and ‘zyx’.

When we study these three examples we see that for ‘abc’ two characters come lexicographically after its neighbour to the left. For ‘hat’ there is exactly one character that comes lexicographically after its neighbour to the left. For ‘zyx’ there are zero characters that come lexicographically after its neighbour to the left.

In all there are 10400 strings of length 3 for which exactly one character comes lexicographically after its neighbour to the left.

We now consider strings of $n\leq 26$ different characters from the alphabet. For every $n$, $p(n)$ is the number of strings of length of $n$ for which exactly one character comes lexicographically after its neighbour to the left.

What is the maximum value of $p(n)$?

A printing shop runs 16 batches (jobs) every week and each batch requires a sheet of special colour-proofing paper of size A5.

Every Monday morning, the supervisor opens a new envelope, containing a large sheet of the special paper with size A1.

The supervisor proceeds to cut it in half, thus getting two sheets of size A2. Then one of the sheets is cut in half to get two sheets of size A3 and so on until an A5-size sheet is obtained, which is needed for the first batch of the week.

All the unused sheets are placed back in the envelope.

At the beginning of each subsequent batch, the supervisor takes from the envelope one sheet of paper at random. If it is of size A5, then it is used. If it is larger, then the ‘cut-in-half’ procedure is repeated until an A5-size sheet is obtained, and any remaining sheets are always placed back in the envelope.

Excluding the first and last batch of the week, find the expected number of times (during each week) that the supervisor finds a single sheet of paper in the envelope.

Give your answer rounded to six decimal places using the format x.xxxxxx .

Using a combination of grey square tiles and oblong tiles chosen from: red tiles (measuring two units), green tiles (measuring three tiles), and blue tiles (measuring four tiles), it is possible to tile a row measuring five units in length in exactly fifteen different ways.

How many ways can a row measuring fifty units in length be tiled?

This is related to #116 - Red, green or blue tiles

A row of five grey square tiles is to have a number of its replaced with coloured oblong tiles chosen from red (length two), green (length three), or blue (length four).

If red tiles are chosen there are exactly seven ways this can be done.

If green tiles are chosen there are three ways.

And if blue tiles are chosen there are two ways.

Assuming that colours cannot be mixed there are 7 + 3 + 2 = 12 ways of replacing the grey tiles in a row measuring five units in length.

How many different ways can the grey tiles in a row measuring fifty units in length be replaced if colours cannot be mixed and at least one coloured tile must be used?

This is related to #117 - Red, green, and blue tiles

This is a much more difficult version of #114 - Counting block combinations I

A row measuring $n$ units in length has red blocks with a minimum length of $m$ units placed on it, such that any two red blocks (which are allowed to be different lengths) are separated by at least one black square.

Let the fill-count function, $F(m,n)$, represent the number of ways that a row can be filled.

For example, $F(3,29) = 673135$ and $F(3, 30) = 1089155$.

That is, for $m=3$, it can be seen that $n=30$ is the smallest value for which the fill-count function first exceeds one-million.

In the same way, for $m=10$, it can be verified that $F(10, 56) = 880711$ and $F(10, 57) = 1148904$, so $n=57$ is the least value for which the fill-count function first exceeds one million.

For $m=50$, find the least value of $n$ for which the fill-count function first exceeds one million.