# 2015-2016 SoMaS Team Mathematics Challenge

You can browse this year's problems below or view the download/print the PDF copy of problems. You can browse past years problems and solutions here.Problem 1

For \(N=201\), we can reverse the digits to get the number \(\overline{N}=102\). We have \(M=201-102=099\) and \(\overline{M}=990\). This gives \(M+\overline{M}=1089\).

Now let \(N=abc\) be a general three digit number with integers \(a,b,c\in\{0,...,9\}\) and \(a>c\), and let \(\overline{N}=cba\). Show that the three digit number \(M = def := N-\overline{N}\) with \(d,e,f\in\{0,...,9\}\) has the property that \(M+\overline{M}=def+fed=1089\) and generalise the result to base \(n\).

Problem 1 had a good success rate, with all submitted solutions easily proving the base 10 part of the question!
*Calculo Ergo Sum* and *Oscillating Wildly* both generalised the result very well to base \(n\), finding
\( M+\bar M=(n-1)(n+1)^2. \)
This is very good, but the result is more naturally written in base \(n\), namely
\(
M+\bar M= 1\cdot n^3+ 0\cdot n^2+(n-2)\cdot n+(n-1)\cdot n^0.
\)

*Rebecca Pickstock* had a particularly good solution of the base 10 case, very
well written (in prose!) and with illustrating examples. *Kaiwen Zhang* also gave a nice answer,
and *Robit the Hobbit* generalised the problem to several digits and
computed general formulas in base 10 and in base \(n\). This was fantastic,
even if there was a little bit of magic in the proof! ;)

We reproduce the spirit of the solutions that most teams submitted here.

100 lemmings are placed on a levitating 10 metre long steel beam. Each lemming is facing one end of the beam. They all start walking simultaneously in the direction they are facing. If two lemmings bump into each other they both turn and walk in the opposite direction. Are all the lemmings guaranteed to fall off the beam if they all move at a constant speed of 10 metres per hour? Can you find a minimum/maximum number of lemmings that will be on the beam after t minutes?

*Heptadecagon* was worried about animal welfare, but was willing to bet that at least a
couple wouldn't fall from the beam that they were unjustifiable placed on! *Cos(i rule),
Calculo Ergo Sum*, and *Diamond Knights* all observed that we might as well allow the lemmings
to walk through themselves, and we display *Diamond Knights'* diagram below (when we get it) which
speaks a thousand words:

Write down an explicit bijection from the open interval \[I= (0,1):=\{x\in \mathbb{R} \; : \; 0 < x < 1 \} \] to the infinite union of closed intervals \[ \bigcup_{n=1}^{\infty}[2n,2n+1].\]

Solutions from *Calculo Ergo Sum, Cos(i rule), Diamond Knights, Gergely Bodo*, and *Kexixiwazi* all used
the great observation that a monotonic convergent sequence of points in \( (0,1)\) could be
mapped to the set \(\{2,3,4,5,\dots \}\) in the range. All solutions choose different sequences, but the crux was
to note that the complement of the monotonic sequence in \((0,1)\) is a disjoint union of open intervals that can be mapped
directly to the interiors of the interval in the range.

A conglomerated version of their very nice maps is the bijection \(f: (0,1)\rightarrow \bigcup_{n=1}^{\infty}[2n,2n+1]\) defined by \( f(x)=x^{-1} \) is \(x =\frac1k\) for some natural number \(k\), and \(f(x)=k(k+1)x+k\) for \(x\in \Big(\frac1{k+1},\frac1k\Big)\).

*Robit the Hobbit*'s solution, was similar in spirit, but trickier to write down. The *Hobbit*'s idea was to
break the entire real line (which can be put into bijective correspondence with the unit interval) into a
\(\mathbb{Z}\) union a countable union of open intervals and go from there.

What is the largest value of \[\sin(x)\sin(\sqrt{5}y) + \sin(\sqrt{2}y)\sin(\sqrt{6}z) + \sin(\sqrt{3}z)\sin(\sqrt{7}x)\] that you can find, for real values of \(x\), \(y\) and \(z\) between 0 and 100?

We have an upper bound and a lower bound:

*Heptadecagon*assures us it won't be bigger than 3, because the sin function is at most 1. It's clearly a great first observation, and it is nicely offset by...*Cos(i rule)*, who found a maximum at \((x,y,z) = (26.7151, 76.5903, 26.2949)\), where \(f(x,y,z) = 2.996438\).

Those bounds are fairly close together, which is pleasing to see. The setters found \((x,y,z) = (26.7151,34.4247,48.0856)\), which is even closer to 3, namely about \(2.998945\).

If Andy Murray wins \(51\%\) of his serves and \(50\%\) of his other shots during Wimbledon 2016, what is the probability that he will win the tournament?

There were three entries, each with something different to say:

*Heptadecagon*is rooting for Murray, but thinks Djokovic is going to win anyway. Many who have watched tennis recently would be forgiven for agreeing.*Kexixiwazi*drew some really helpful diagrams sketching how to go about it explicitly.*Cos(i rule)*recommends a Monte Carlo approach (simulating it randomly repeatedly using a computer, and measuring the observed average), which seems like a very good idea. He calculates the probability as 1.9%. I notice that that's a*lot*better than 1/128, so even a tiny advantage in each point translates to a big advantage in the tournament.

One can compute the probabilities explicitly, but it's not straightforward. The best way to go about it is to break up the game into pieces.

It's easy to see that, in a game in which Murray doesn't serve, he wins with probability \(1/2\). In a game where he does serve, the first problem to solve is what happens with deuce. Writing D for the probability of Murray winning at deuce, A for the probability of Murray winning with an advantage, and B for the probability of Murray winning with an advantage against him, we can form equations \[A = \frac{51}{100} + \frac{49}{100}D, \quad D = \frac{51}{100}A+\frac{49}{100}B, \quad B = \frac{51}{100}D\] saying that if Murray has an advantage, there's a 51% chance of winning and a 49% chance of going back to deuce, and that if he's at deuce he has a 51% chance of getting an advantage and a 49% chance of losing an advantage, and if his opponent has an advantage he has a 51% chance of getting back to deuce.

These simultaneous equations can be solved explicitly, to find that Murray has a probability 2601/5002 of winning at deuce. One can then trace back through the other possible scores in a service game to find the probability of winning a game from the beginning. (The fractions get very complicated very quickly!).

Then, equipped with the probability for a game, one can go on to consider a set, and then a match, and then a tournament.

For which \(n\in \mathbb{N}\) does there exist a regular \(n-\)gon in the plane with all vertices having integer coordinates?

There were some good attempts, with particularly noteworthy intuition from *Calculo ergo sum*,
who gave the bones of an argument that showed \(\sin(2\pi/n)\) times the side length of
the \(n\)-gon was necessarily rational for existence. *Robit the Hobbit*'s solution was the only one that dotted the i's and crossed the
t's by appealing to useful facts like Pick's theorem. *Hobbit*'s solution can be found
here, and contains links to the relevant results/proofs used in the solution.
Heptadecagon directed us to
this page which is well worth a look at.

[Problem 11852 from the *American Mathematical Monthly*] For \(n\in\mathbb{Z}^{+}=\{1,2,3,...\}\),
let \(\nu_n=k\) if \(3^k\) divides \(n\) but \(3^{k+1}\) does not. Let \(x_1=2\), and for \(n\geq2\) let
\[
x_n=4\nu_n+2-\frac{2}{x_{n-1}}.
\]
So that \((x_n)\) begins \(2,1,4,\tfrac32,\tfrac23,3,\dots\). Show that every positive rational
number appears exactly once in the list \((x_n)\).

This is a live problem and you can submit a solution to the journal for publication if you can find one. The judges don't know an answer, but we'd like to see you show us one! You can find the original problem here.

Unfortunately we didn't get a perfect solution for this, though the *Diamond Knights'* made some progress.
We were particularly happy with the distribution of points in the sequence that are (will be) shown below (when we are
given the images):

**Judges:** James Cranch,
Madeleine Jotz Lean,
Jayanta Manoharmayum, and
Fionntan Roukema.

**O.B.S. prize:** (overall best submission) *Cos(i rule) (Kyriacos Georgiades)*

**M.V.S. prize:** (most valuable solution) *Robbit the Hobbit (Rob Nicolaides)* for Question 6.

**Honourable mentions:** *
Calculo ergo sum (Jonny Atkinson),
Diamond Knights (Hanye Zhu, Yifan Jiang, Hanlin Yue),
Gergely Bodo,
KEXIXIWAZI (YiTian Yang (tony), Haoran Xue (Rick), Quan Sun, Kexin LI),
Oscillating Wildly (Vinh Vu,
Sam Lomax, Umang Metha, Tom Galvin, Danny Heard),
Heptadecagon (Tom Bennett), Kaiwen Zhang, Rebecca Pickstock.*

**Best team name:** *Calculo ergo sum (Jonny Atkinson)*

**Best interdisciplinary team:** *Oscillating Wildly (Vinh Vu (Aerospace engineering 2nd year),
Sam Lomax (Mechanical engineering 2nd year), Umang Metha (Aerospace engineering 2nd year),
Tom Galvin (Theoretical physics 2nd year), Danny Heard (Computer science 2nd year)).*