BMO2 2018

The second round of the British Mathematical Olympiad was taken yesterday by the 100 or so top scoring eligible participants from the first round, as well as some open entries. Qualifying for BMO2 is worth celebrating in its own right. The goal of the setters is to find the sweet spot of difficult but stimulating for the eligible participants, which ultimately means it’s likely to be the most challenging exam many of the candidates sit while in high school, at least in mathematics.

I know that lots of students view BMO2 as something actively worth preparing for. As with everything, this is a good attitude in moderation. Part of the goal in writing about the questions at such length (and in particular not just presenting direct solutions) is because I think at this level it’s particularly easy to devote more time than needed to preparation, and use it poorly.

All these questions could be solved by able children. In fact, each could be solved by able children in less than an hour. You definitely count as an able child if you qualified or if your teacher allowed you to make an open entry! Others count too naturally. But most candidates won’t in fact solve all the questions, and many won’t solve any. And I think candidates often come up with the wrong reasons why they didn’t solve problems. “I didn’t know the right theorems” is very very rarely the reason. Olympiad problems have standard themes and recurring tropes, but the task is not to look at the problem and decide that it is an example of Olympiad technique #371. The task is actually to have as many ideas as possible, and eliminate the ones that don’t work as quickly as possible.

The best way to realise that an idea works is to solve the problem immediately. For the majority of occasions when we’re not lucky enough for that to happen, the second-best way to realise that an idea works is to see that it makes the problem look a bit more like something familiar. Conversely, the best way to realise that an idea doesn’t work is to observe that if it worked it would solve a stronger but false problem too. (Eg Fermat’s Last Theorem *does* have solutions over the reals…) The second-best way to realise that an idea doesn’t work is to have the confidence that you’ve tried it enough and you’ve only made the problem harder, or less familiar.

Both of these second-best ideas do require a bit of experience, but I will try to explain why none of the ideas I needed for various solutions this year required any knowledge beyond the school syllabus, some similarities to recent BMOs, and a small bit of creativity.

As usual, the caveat that these are not really solutions, and certainly not official solutions, but they are close enough to spoil the problems for anyone who hasn’t tried them by themselves already. Of course, the copyright for the problems is held by BMOS, and reproduced here with permission.

Question One

I wrote this question. Perhaps as a focal point of the renaissance of my interest in geometry, or at least my interest in teaching geometry, I have quite a lot to say about the problem, its solutions, its origin story, the use of directed angles, the non-use of coordinate methods and so on. In an ideal world I would write a book about this sort of thing, but for now, a long and separate post is the answer.

This will be available once I’ve successfully de-flooded my apartment.

Question Two

I also wrote this problem, though I feel it’s only fair to show the version I submitted to the BMO committee. All the credit for the magical statement that appears above lies with them. There is a less magical origin story as well, but hopefully with some interesting combinatorial probability, which is postponed until the end of this post.One quick observation is that in my version Joe / Hatter gets to keep going forever. As we shall see, all the business happens in the first N steps, but a priori one doesn’t know that, and in my version it forces you to strategise slightly differently for Neel / Alice. In the competition version, we know Alice is done as soon as she visits a place for a second time, but not in the original. So in the original we only have to consider ‘avoid one place’ rather than the multiple possibilities now of ‘avoid one place’ or ‘visit a place again’.

But I think the best idea is to get Alice to avoid one particular place c\not\equiv 0 whenever possible. At all times she has two possible options for where to go next, lets say b_k+a_k, b_k-a_k in the language of the original statement. We lose nothing by assuming -N/2 < a_k\le N/2, and certainly it would be ridiculous for Joe / Hatter ever to choose a_k=0. The only time Alice’s strategy doesn’t work is when both of these are congruent to c, which implies N\,|\, 2a_k, and thus we must have N= 2a_k. In other words, Alice’s strategy will always work if N is odd.

I think it’s really worth noticing that the previous argument is weak. We certainly did not show that N must be odd for Alice to win. We showed that Alice can avoid a congruence class modulo an odd integer. We didn’t really need that odd integer to be N for this to work. In particular, if N has an odd factor p (say a prime), then the same argument works to show that we can avoid visiting any site with label congruent to 1 modulo p.

It’s actually very slightly more complicated. In the original argument, we didn’t need to use any property of b_k. But obviously here, if b_k\equiv 1 modulo p and p\,|\,a_k, then certainly b_{k+1}\equiv 1 modulo p. So we have to prove instead that Alice can ensure she never ‘visits 1 modulo p for the first time’. Which is fine, by the same argument.

So, we’ve shown that Neel / Alice wins if N is odd, or has an odd factor. The only values that remain are powers of 2. I should confess that I was genuinely a little surprised that Joe / Hatter wins in the power of 2 case. You can find a construction fairly easily for N=2 and N=4, but I suspected that might be a facet of small numbers. Why? Because it still felt we could avoid a particular site. In order for Alice’s strategy to fail, we have to end up exactly opposite the particular site at exactly the time when the next a_k=N/2, and so maybe we could try to avoid that second site as well, and so on backwards?

But that turned out to be a good example of something that got very complicated quite quickly with little insight. And, as discussed at the beginning, that’s often a sign in a competition problem that your idea isn’t so good. (Obviously, when composing a problem, that’s no guarantee at all. Sometimes things are true but no good ideas work.) So we want other ideas. Note that for N=4, the sequence (2,1,2) works for Joe / Hatter, because that forces Alice / Neel to visit either (0,2,1,3) or (0,2,3,1). In particular, this strategy gave Alice no control on the first step nor the last step, and the consequence is that we force her to visit the evens first, then transfer to an odd, and then force her to visit the other odd.

We might play around with N=8, or we might proceed directly to a general extension. If we have a Joe / Hatter strategy for N, then by doubling all the a_ks, we have a strategy for 2N which visits all the even sites in the first N steps. But then we can move to an odd site eg by taking a_N=1. Just as in the N=4 case, it doesn’t matter which odd site we start from, since if we again double all the a_ks, we will visit all the other odd sites. This gives us an inductive construction of a strategy for powers of two. To check it’s understood, the sequence for N=8 is (4,2,4,1,4,2,4).

Although we don’t use it, note that this strategy takes Alice on a tour of sites described by decreasing order of largest power of two dividing the label of the site.

Question Three

I have a theory that the average marks on Q1, Q2 and Q3 on this year’s paper will be in ascending order rather than, as one might expect, descending order. I think my theory will fail because it’s an unavoidable fact of life that in any exam, candidates normally start at the beginning, and don’t move to the middle until making earlier progress. But I think that’s the only reason my theory will fail.

Like kitchen cleanliness or children’s character flaws, it’s hard to compare one’s own problem proposals with others’ rationally. But I felt that, allowing for general levels of geometry non-preference, Q3 was more approachable than Q2, especially to any candidate who’d prepared by looking at some past papers.

I’m in no way a number theorist, but I know three or four common themes when one is asked to prove that a certain sequence contains no squares, or almost no squares. [3a]

  • Number theoretic properties of the sequence of squares. Squares cannot be 3 modulo 4 for example. They also cannot be 2 modulo 4, and thus they also cannot be 2^{k-1} modulo 2^k for any even k. This first observation was essentially the body of most solutions to Q4 of BMO1 2016, among many others.
  • Soft properties of the sequence of squares. The sequence of squares grows quadratically. Sometimes we can show a quadratic sequence will have no overlap with some other sequence for basic reasons. This is especially common if the second sequence is also quadratic or similar. For example, the expression n^2+3n-4 is typically not a square because

(n+1)^2 = n^2+2n+1 < n^2 + 3n - 4 < n^2+4n+4 = (n+2)^2,

  • when n is large. In fact the right hand inequality is always true, and the left hand inequality is true for n\ge 6, which doesn’t leave too many cases to check (and n=5 does actually give a square). This type of argument has been quite common on BMO recently, directly on Q1 of BMO1 2011 and also Q3 of BMO1 2016. An example in a more abstract setting is Q3 of Balkan MO 2007, which I greatly enjoyed at the time…
  • Number theoretic properties of the definition of a square. A square is the product of an integer with itself, and so if we want the product of two or more integers to be a square, then this imposes conditions on the shared factors of the two integers. I’ll cite some examples shortly.
  • Huge theorems. Some old paper which I encountered as a child asked us to find all solutions to x^2-1=2^y. Or similar – I can’t find it now – but Q2 of BMO2 2006 is close enough to the sensible approach to the problem. I think it’s more helpful to think about this as proving that a particular sequence rarely includes powers of two than that a particular sequence rarely includes squares. But either way, one could in principle use the Catalan conjecture, which controls all non-trivial solutions to a^p - b^q=1. Fortunately, the Catalan conjecture was proved, by Mihailescu (readable blog about it), between the paper being set, and me attempting it a few years later. I’m being flippant. This is not a standard trope in solving these questions. For very obvious reasons. If it can be killed by direct reference to a known theorem, it won’t be set.

Anyway, those references (and more to follow) are to illuminate why I thought this question was not too hard. Indeed, I feel one can make substantial meta-progress in your head. The given information is interesting, but for the purpose of this question is just a black box. By subtracting the expression for m from the expression for 2m, we can derive an expression for the required sum. It’ll be a quartic in m, because the leading terms won’t cancel.

This leaves all three of the methods above very accessible. Unfortunately m=0 would be a square were it not excluded specifically, so a modular arithmetic approach is unlikely to work directly. Bounding between two quadratics is entirely plausible, as is factorising and comparing number theoretic properties of the factors. I thought the second one seemed more promising, but either way, having two potentially good ideas based only on recent BMO problems before even writing anything down is a good opening.

We do have to calculate the sum, and I make it \frac{1}{4}m^2(5m+3)(3m+1). Now I’m not so sure how to bound this between two quadratics, because the leading coefficient is 15/4, which is not the square of a rational. But the factor analysis approach is definitely on.

Let’s review this generally. Throughout, suppose m,n are positive integers.

Claim 1: if mn is a square, then m and n are squares too.

Claim 2: if mn is a square, then m=n.

Both of these claims are false. However, a version of Claim 1 is true.

Claim 1′: if mn is a square, and m,n are coprime, then each is a square.

Even though this isn’t a named theorem, it is true, and well-known and can be used without proof. One way to prove it is to write m,n as products of primes, and show that since the primes are disjoint, the exponents must all be even. Most other methods will be equivalent to this, maybe with less notation.

What is good about Claim 1′ is that more complicated versions are true for for essentially similar reasons. For example

Claim 3: if mn is 6k^2, and m,n are coprime, then either one is a square and the other is six times a square; or one is two times a square, and the other is three times a square.

Claim 4: if mn is a square, and the greatest common divisor (m,n) is either 5 or 1, then either each is a square, or each is five times a square.

I cited some examples of the other methods I proposed. Here are some examples of this sort of thing in recent BMOs:

  • Q4 of BMO2 2016. Even the statement is suggestive. There are more complicated routes, but showing that (2p-u-v)(2p+u+v) is a square is one way to proceed, and then Claim 4 directly applies after checking a gcd.
  • Q2 of BMO1 2014 is similar, but it is much more explicit that this is the correct approach. Expose p^2 then use a (correct) version of Claim 2.
  • Q1 of BMO2 2009. Show that a and b must each be a square times 41 for rationality reasons.
  • Q6 of BMO1 2006. After sensible focused substitutions, obtain 3n^2=q(q-1). Rather than try to ‘solve’ this, extract the key properties along the lines of Claim 3, eliminate one of the cases by modular arithmetic, and return to the required statement.
  • Q3 of BMO2 2010 requires the student to reproduce the essentials of the arguments above in the case of a particular degree six polynomial with a tractable factorisation, along with some mild square-sandwiching or bounding arguments as discussed earlier.

In conclusion, I’m trying to say that if I claim I am confident I can find all integers m such that \frac14 m^2(5m+3)(3m+1) is a square, this is not based on complicated adult experience, but rather on recent problems at a similar sensible level. And I still don’t think it counts as Olympiad technique #371 – thinking about divisibility of factors is a good thing to do when talking about integers, and so it’s just a natural entry point into problems about squares. Plenty of problems might have this sort of thing as a starting point or an ending point.

For this problem we need a different ending point. To be brief, the factors (5m+3) and (3m+1) cannot both be squares because 5m+3 is never a square. So since the gcd of these factors is 1, 2 or 4, the only other option is that they are both squares times 2. And because -1 is not a square modulo 3, so 1 is not a (square times 2) modulo 3, and we are done. Note that this was a literal example of the first technique for proving something is not a square, proposed all the way back at the start of this section.

Footnotes

[3a] – some common themes for proving that sequences do include squares might be comparison with Pell’s Equations, or comparison with the explicit construction of solutions to Pythagoras’s equation.

Question Four

An example of an absorbing function is f(x)=\lfloor x\rfloor. One challenge is thinking of many other examples. This one is fine, but it’s true under replacing 2018 by 1 in the statement, and so it doesn’t really capture the richness of the situation.

Notation: the pre-image of a function is the language used to describe the inverse of a function which doesn’t have a uniquely-defined inverse. That is, if f is not injective, and multiple arguments have the output. We write f^{-1}(y)=\{x: f(x)=y\}. In particular, this is a set of values, not necessarily a single value. We also use \mathbb{Z} to denote the integers. We can apply pre-images to sets as well. So for example f^{-1}(\mathbb{Z})=\{x : f(x)\in \mathbb{Z}\}.

This question is tricky, and I will be surprised to see many full solutions from the eligible candidates. It rewards the sort of organisation and clear-thinking that is easier said than done in a time-pressured contest environment. There are also many many possible things to consider, and so is particularly challenging in the short timeframe of BMO2 as opposed to, for example, appearing as the middle question on a 4.5 hour international-level paper.

At a meta-level we are being asked to confirm or deny the existence of absorbing functions where f^{-1}(\mathbb{Z}) is small in some sense, firstly when actually having finite size, secondly when, although infinite, being a small sort of infinite, namely spread out in a sparse, well-ordered way (you might say countable if familiar with that language). The general idea is presumably that it’s hard to be absorbing if the pre-image of the integers is small, and so it’s reasonable to assume that it’s too hard if this is finite; but perhaps not quite too hard if it’s merely countable. So (no, yes) is a sensible guess at the answer to the question, though (no, no) might also fit, maybe with a harder argument for the second no.

Ok, instead of trying a) or b), just play with the configuration. Let A=f^{-1}(\mathbb{Z}). We will use this frequently. In the picture below, f maps the real line on top to the real line below. If two reals get mapped to the same image, then whether or not the image is an integer, the whole (closed) interval bounded by the two reals also gets mapped to the same image. This is because f is weakly increasing.

This means that A consists of various intervals (which include single points). But in both a) and b) we know that A is ‘small’, and so it cannot contain any intervals of positive length. So in fact A is a set of separated real values. In the case of a) it’s a finite set.

Do we want to try and iterate this, and look at f^{-1}(A)? Well maybe, but we don’t know much about about pre-images of A, only about pre-images of \mathbb{Z}.

But note that the pre-image of the pre-image of the … of the pre-image [2017 times] of A must be the whole real line, so at some point, some value has a pre-image that is an interval. So if we’re guessing that the answer to b) is yes, then we need to give a construction.

\mathbb{R} \stackrel{f}\longrightarrow ?? \stackrel{f}\longrightarrow\quad\ldots\quad \stackrel{f}\longrightarrow ??\stackrel{f}\longrightarrow A \stackrel{f}\longrightarrow f(A)\subset \mathbb{Z}.

If you play around for a bit, it seems very unlikely to be absorbing if the integers don’t get mapped to the integers. You can try to prove this, but at the moment we’re just aiming for a construction, so let’s assume f(\mathbb{Z})\subset \mathbb{Z}. It would be convenient if f(n)=n for all n\in \mathbb{Z}, but we already know that this won’t work because then the pre-image of the pre-image of the… of \mathbb{Z} is always \mathbb{Z}, but we need it to be \mathbb{R}.

The ideal situation would be if A= \mathbb{Z}\cup \{\ldots, a'_{-1},a'_0,a'_1,\ldots\}, where the pre-image of \{\ldots, a'_{-1},a_0,a'_1,\ldots\} is pretty much everything.

Informally, we are specifically banned from mapping intervals directly onto an integer. So have an intermediate set, and try to map almost everything (except the integers and the set itself) onto that set, so and map that set into the integers.

At this point, you really just have to have the right idea and finish it. Many things will work, but this seems the easiest to me. Let the set A consist of the integers and the (integers plus 1/2). And for x\in A, f(x)=2x. This is what f looks like so far.

Here the black crosses are integers, and the purple crosses are (integers plus 1/2). But now we need to make as many reals as possible in the top row map to a purple cross (which is allowed, because purple crosses aren’t integers), but we need also to preserve the weakly increasing property. Fortunately, we can exactly do that. Each cross of either colour in the top row maps to a black cross in the middle row (ie an integer), so we can map the open interval between crosses in the top row to a purple cross in the middle row. As shown in red:

Note that this is consistent. The fact that I haven’t drawn in the red cones into the bottom row is only because I didn’t use the bottom row to motivate doing this. I’ve shown a consistent definition of f that maps all the reals onto the integers in two steps. If it’s an integer to begin with, that was great; if it was an (integer plus 1/2) to begin with then it becomes an integer in one step and stays an integer; and otherwise it first maps to an (integer plus 1/2), and then to an integer in the second step.

To check you’ve understood, try to write down a standalone definition of this function.

I’ve therefore solved part b) with the alternative condition \ldots a_{-1}<a_0<a_1<a_2<\ldots which isn’t exactly as required. It requires one small and simple idea to convert to a solution to the actual statement. See if you can find it yourself!

I think part a) is harder, not because the solution will look more complicated, but because there are so many potential partial results you could try to prove, because there are so many sets you could consider. To name a few: the image of f, the image of f intersected with \mathbb{Z}, the image of \mathbb{Z}, the 2018-composition image f^{2018}(\mathbb{R}), the 2018-composition image f^{2018}(\mathbb{Z}) and so on and so forth. You might have good insight into the wrong things.

For me, the crucial observation (which you can see from the figure in the b) construction) is that when composing an increasing function with itself, the ‘trajectories’ are either increasing or decreasing. That is, if x\le f(x) (respectively, x\ge f(x)), then x\le f(x)\le f^2(x)\le f^3(x)\le\ldots (respectively x\ge f(x)\ge f^2(x)\ge \ldots). Again, you can think of this as Olympiad technique #371 if you insist, but I don’t think that’s helpful. There are lots of things one could try to say here, and this turns out to be natural, true and useful, but you can’t know it’s useful until you play with it.

Anyway, we’re playing with part a), and we know that f^k(x) is an integer for all large enough k, and that f^{k+1}(x) is also an integer, so f^k(x) is one of a finite set of integers because of the condition on A. But we’ve seen the sequence x,f(x),f^2(x),\ldots is weakly increasing or weakly decreasing, and so if we also know it’s eventually bounded (because eventually it’s in this finite set) then it must eventually be constant. And this constant is one of the integers, say n. But unless we started from n, this means that f(n)=n, but also f(x)=n for some other real value x. And so exactly as at the very very beginning, that’s bad, because then the whole interval [x,n] gets mapped to n, which is a contradiction.

Question Two – Origin story

The origin story for Q2 started in a talk I heard by Renan Gross at Weizmann, who referenced some of the history of Scenery Reconstruction. Roughly speaking, we colour the integers (say with two colours), and then let loose a random walker, who tells us the sequence of colours she observes during her walk, but no other information about the walk itself.

How much information can we recover about the colouring? Obviously, the best we can hope for is to recover the colouring, up to translations and reflection, since for every possible random walk trajectory, the exact reflection is equally probable, and we are given no information about the starting point.

Since lots of the transitions between recoverable and unrecoverable depend on the periodicity of the colouring, a reasonable toy model is to do it on a cycle. Note that the Strong Law of Large Numbers tells us that we almost surely recover the number of black sites and white sites from an the infinite trajectory of the random walk. Of course it’s possible that there are only two black vertices, and they are adjacent, and the walker oscillates between them, thus seeing BBBBBB… But this is extremely unlikely. You could think of this in Bayesian terms as strongly increasing the prior on the whole cycle being black, but I think initially it’s best to do this as an infinite-time, SLLN problem not as finite time WLLN/CLT reweightings of anything.

But what more? It’s clear that the lengths of all black substrings should follow some mixed geometric-ish distribution, and this distribution will almost surely wash out as the empirical distribution in an SLLN sense. But it’s tricky to justify why such a mixed geometric-ish distribution should be uniquely determined by the lengths of black arcs in the cycle. But it does definitely feel like we should have enough information to reconstruct the colouring up to reflection/rotation with probability one. For example, analogously to the number of black vertices and the number of white vertices, we should be able to recover the number of adjacent black vertices, the number of adjacent white vertices, and the number of black-white adjacent vertices, and so on.

Anyway, this can be done, and it follows as a consequence of various authors’ work answering some more general conjectures of Benjamini and, separately, of den Hollander and Keane. Douglas Howard [DH] shows a handful of generalisations of this, as do Benjamini and Kesten [BK]. Most of this work is focused on sceneries on \mathbb{Z}, but periodic sceneries are often used as a basis, and of course, the only difference between periodic sceneries on \mathbb{Z} and sceneries on the N-cycle are whether you know the period in advance. [BK] show that ‘almost all’ sceneries are distinguishable in a particular sense, in response to which Lindenstrauss [L99] exhibits a large family of sceneries which are not distinguishable. A readable but technical review is [ML].

So Renan’s talk was about the similar problem (and generalisations) on the hypercube [GG]. Rather than paraphrase the main differences badly, you can read his own excellent blog post about the work.

On the train back to Haifa from Rehovot, I was thinking a bit about the cycle case, and what happens if you generalise the random walk with varying jump lengths, or indeed introduce a demon walker, whose goal is to make it as hard as possible for the reviewer to deduce the colouring. One way this can certainly happen is if the walker can avoid visiting some particular site, as then how could one possibly deduce the colour of the never-visited site? And so we get to the statement posed.

References

[BK] – Benjamini, Kesten, 1996 – Distinguishing sceneries by observing the scenery along a random walk path

[dH] – den Hollander, 1988 – Mixing properties for random walk in random scenery

[DH] – Douglas Howard, 1996 – Detecting defects in periodic scenery by random walks on Z

[GG] – Grupel, Gross, 2017 – Indistinguishable sceneries on the Boolean hypercube

[L99] – Lindenstrauss, 1999 – Indistinguishable sceneries

[ML] – Matzinger, Lember, 2003 – Scenery reconstruction: an overview [link]

 

Advertisements

BMO1 2017 – Questions 5 and 6

The first round of the British Mathematical Olympiad was sat yesterday. The questions can be found here and video solutions here. My comments on the first four questions are in the previous post.

Overall, I didn’t think any of the questions on this paper were unusually difficult by the standard of BMO1, but I found everything slightly more time-consuming than typical. I thought Question 5 was a great problem, and I tried lots of things unsuccessfully, first, and so wanted to discuss it in slightly more technical language. For Question 6 I made a decisive mistake, which I’ll explain, and which cost a lot of time. But in general, my point is that the back end of the paper was a little fiddlier than normal, and required longer written solutions, and perhaps many students might have had less time than expected to attack them anyway after details earlier in the paper.

Question Five

As I said before, I thought this question was quite challenging. Not because the solution is particularly exotic or complicated, but because there were so many possible things that might have worked. In my opinion it would not have been out of place at the start of an IMO paper, because it’s perfectly possible to have enough good ideas that eliminating the ones that don’t work takes an hour, or hours. Even though it slightly spoils the flow of the solution, I’m particularly trying to emphasise the tangents that didn’t work, mostly for reassurance to anyone who spent a long time struggling.

I was thinking about this question in terms of a 2Nx2N board, where N is even, and for the given question equal to 100. I spent a while thinking that the bound was 8N-4, corresponding to taking the middle two rows and the middle two columns, but not the 2×2 square which is their intersection. If you think of a comb as a ‘handle’ of 1xN cells, with an extra N/2 alternating cells (say, ‘teeth’) bolted on, then it’s clear this construction works because there’s never space to fit in a handle, let alone the teeth.

I couldn’t prove that this was optimal though. A standard way to prove a given bound K was optimal would be to produce a tiling on the board with K combs, where every cell is included in exactly one comb. But this is clearly not possible in this situation, since the number of cells in a comb (which is 150) does not divide the total number of cells on the board.

Indeed, the general observation that if you take a comb and a copy of the comb rotated by 180, the teeth of the second comb can mesh perfectly with the teeth of the first comb to generate a 3xN unit. I wasted a moderate amount of time pursuing this route.

[Note, it will be obvious in a minute why I’m writing ‘shaded’ instead of ‘coloured’.]

But in motivating the construction, I was merely trying to shade cells so that they intersected every possible 1xN handle, and maybe I could prove that it was optimal for this. In fact, I can’t prove it’s optimal because it isn’t optimal – indeed it’s clear that a handle through one of the middle rows intersects plenty of shaded cells, not just one. However, with this smaller problem in mind, it didn’t take long to come up with an alternative proposal, namely splitting the board into equal quarters, and shading the diagonals of each quarter, as shown.

It seems clear that you can’t fit in a 1xN handle, and any sensible tiling with 1xN handles contains exactly one shaded cell, so this shading (with 4N shaded cells) is optimal. But is it optimal for a comb itself?

Consider a shading which works, so that all combs include a shaded cell. It’s clear that a comb is contained within a 2xN block, and in such a 2xN block, there are four possible combs, as shown.

You need to cover all these combs with some shading somewhere. But if you put the shaded cell on a tooth of comb A, then you haven’t covered comb B. And if you put the shaded cell on the handle of comb A, then you haven’t covered one of comb C and comb D. You can phrase this via a colouring argument too. If you use four colours with period 2×2, as shown

then any comb involves exactly three colours, and so one of them misses out the colour of the shaded cell. (I hope it’s clear what I mean, even with the confusing distinction between ‘shaded’ and ‘coloured’ cells.)

Certainly we have shown that any 2xN block must include at least two shaded cells. And that’s pretty much it. We have a tiling with 2N copies of a 2xN block, and with at least two shaded cells in each, that adds to at least 4N shaded cells overall.

Looking back on the method, we can identify another way to waste time. Tiling a board, eg a chessboard with dominos is a classic motif, which often relies on clever colouring. So it’s perhaps lucky that I didn’t spot this colouring observation earlier. Because the argument described really does use the local properties of how the combs denoted A-D overlap. An attempt at a global argument might start as follows: we can identify 2N combs which don’t use colour 1, and tile this subset of the grid with them, so we need to shade at least 2N cells from colours {2,3,4}. Similarly for sets of colours {1,3,4}, {1,2,4}, and {1,2,3}. But if we reduce the problem to this, then using roughly 2N/3 of each colour fits this global requirement, leading to a bound of 8N/3, which isn’t strong enough. [1]

Question Six

A word of warning. Sometimes it’s useful to generalise in problems. In Q5, I was thinking in terms of N, and the only property of N I used was that it’s even. In Q4, we ignored 2017 and came back to it at the end, using only the fact that it’s odd. By contrast, in Q2, the values did turn out to be important for matching the proof bounds with a construction.

You have to guess whether 300 is important or not here. Let’s see.

I have a natural first question to ask myself about the setup, but some notation is useful. Let a_1,a_2,\ldots,a_{300} be the ordering of the cards. We require that \frac{a_1+\ldots+a_n}{n} is an integer for every 1\le n\le 300. Maybe the values of these integers will be important, so hold that thought, but for now, replace with the divisibility statement that n | a_1+\ldots+a_n.

I don’t think it’s worth playing with small examples until I have a better idea whether the answer is 5 or 295. So the natural first question is: “what does it mean to have (a_1,\ldots,a_{n-1}) such that you can’t pick a suitable a_n?”

It means that there is no integer k in \{1,\ldots,300\}\backslash\{a_1,\ldots,a_{n-1}\} such that n\,\big|\,(a_1+\ldots+a_{n-1})+k, which for now we write as

k\equiv -(a_1+\ldots+a_{n-1})\,\mod n.

Consider the congruence class of -(a_1+\ldots+a_{n-1}) modulo n. There are either \lfloor \frac{300}{n}\rfloor or \lceil \frac{300}{n}\rceil integers under consideration in this congruence class. If no such k exists, then all of the relevant integers in this congruence class must appear amongst \{a_1,\ldots,a_{n-1}\}. At this stage, we’re trying to get a feel for when this could happen, so lower bounds on n are most relevant. Therefore, if we get stuck when trying to find a_n, we have

\lfloor \frac{300}{n} \rfloor\text{ or }\lceil \frac{300}{n}\rceil \le n-1, (*)

which is summarised more succinctly as

\lfloor \frac{300}{n} \rfloor \le n-1. (**)

[Note, with this sort of bounding argument, I find it helpful to add intermediate steps like (*) in rough. The chance of getting the wrong direction, or the wrong choice of \pm 1 is quite high here. Of course, you don’t need to include the middle step in a final write-up.]

We can check that (**) is false when n\le 17 and true when n\ge 18. Indeed, both versions of (*) are true when n\ge 18.

So we know the minimum failure length is at least 17. But is there a failing sequence of length 17? At a meta-level, it feels like there should be. That was a very natural bounding argument for 17 (which recall corresponds to n=18), and it’s easy to believe that might be part of an official solution. If we achieve equality throughout the argument, that’s most of the way to a construction as well. It won’t be so easy to turn this argument into a construction for n\ge 19 because there won’t be equality anywhere.

We have to hope there is a construction for n=18. What follows is a description of a process to derive (or fail to derive) such a construction. In a solution, one would not need to give this backstory.

Anyway, in such a construction, let \alpha\in\{1,2,\ldots,18\} describe the congruence class modulo 18 which is exhausted by \{a_1,\ldots,a_{17}\}. I’m going to hope that \alpha=18 because then the calculations will be easier since everything’s a multiple of 18. We haven’t yet used the fact that for a problem, we need \alpha\equiv-(a_1+\ldots+a_{n-1}). We definitely have to use that. There are 16 multiples of 18 (ie relevant integers in the congruence class), so exactly one of the terms so far, say a_j, is not a multiple of 18. But then

0 \equiv 0+\ldots+0+a_j+0+\ldots+0,

which can’t happen. With a bit of experimentation, we find a similar problem making a construction using the other congruence classes with 16 elements, namely \alpha\in \{13,14,\ldots,18\}.

So we have to tackle a different class. If \alpha\le 12 then our sequence must be

\alpha,18+\alpha,2\times 18 +\alpha, \ldots, 16\times 18 + \alpha,

in some order. In fact, let’s add extra notation, so our sequence is

(a_1,\ldots,a_{17}) = (18\lambda_1+ \alpha,\ldots,18\lambda_{17}+\alpha),

where (\lambda_1,\ldots,\lambda_{17}) is a permutation of {0,…,16}. And so we require

k \,\big|\, 18(\lambda_1+\ldots+\lambda_k) + k\alpha, (%)

for 1\le k\le 17. But clearly we can lop off that k\alpha, and could ignore the 18. Can we find a permutation \lambda such that

k \,\big|\, \lambda_1+\ldots+\lambda_k.

This was where I wasted a long time. I played around with lots of examples and kept getting stuck. Building it up one term at a time, I would typically get stuck around k=9,10. And I had some observations that in all the attempted constructions, the values of \frac{\lambda_1+\ldots+\lambda_k}{k} were around 8 and 9 too when I got stuck.

I became convinced this subproblem wasn’t possible, and decided that would be enough to show that n=18 wasn’t a possible failure length. I was trying to show the subproblem via a parity argument (how must the a_is alternate odd/even to ensure all the even partial sums are even) but this wasn’t a problem. Then I came up with a valid argument. We must have

\lambda_1+\ldots+\lambda_{17}=136= 16\times 8 + 8\quad\text{and}\quad 16\,\big|\,\lambda_1+\ldots+\lambda_{16},

which means \lambda_1+\ldots+\lambda_{16} must be 128 = 15×8 + 8, ie \lambda_{17}=8. But then we also have 15\,\big|\, \lambda_1+\ldots+\lambda_{15}, which forces $latex\lambda_{16}=8$ also. Which isn’t possible.

If this then hadn’t wasted enough time, I then tried to come up with a construction for n=19, for which there are lots more variables, and took a lot more time, and seemed to be suffering from similar problems, just in a more complicated way. So I became convinced I must have made a mistake, because I was forced down routes that were way too complicated for a 3.5 hour exam. Then I found it…

What did I do wrong? I’ll just say directly. I threw away the 18 after (%). This made the statement stronger. (And in fact false.) Suppose instead I’d thrown away a factor of 9 (or no factors at all, but it’s the residual 2 that’s important). Then I would be trying to solve

k\,\big|\,2(\lambda_1+\ldots+\lambda_k).

And now if you experiment, you will notice that taking \lambda_1=0,\lambda_2=1,\lambda_3=2,\ldots seems to work fine. And of course, we can confirm this, using the triangle number formula for the second time in the paper!

This had wasted a lot of time, but once that thought is present, we’re done, because we can go straight back and exhibit the sequence

(a_1,\ldots,a_{17}) = (1, 18+1,2\times 18 +1,\ldots, 16\times 18 +1).

Then the sum so far is congruent to -1 modulo 18, but we have exhausted all the available integers which would allow the sum of the first 18 terms to be a multiple of 18. This confirms that the answer to the question as stated is 17.

At the start, I said that we should be cautious about generalising. In the end, this was wise advice. We definitely used the fact that 18 was even in the stage I over-reduced the first time. We also used the fact that there was at least one value of \alpha with an ‘extra’ member of the congruence class. So I’m pretty sure this proof wouldn’t have worked with 288 = 16×18 cards.

Footnotes

[1] – If shading were a weighted (or continuous or whatever you prefer) property, ie that each cell has a quantity of shading given by a non-negative real number, and we merely demand that the total shading per comb is at least one, then the bound 8N/3 is in fact correct for the total shading. We could look at a 2xN block, and give 1/3 shading to one cell of each colour in the block. Alternatively, we could be very straightforward and apply 2/3N shading to every cell in the grid. The fact that shading has to be (in this language) zero or one, imposes meaningful extra constraints which involve the shape of the comb.

BMO1 2017 – Questions 1-4

The first round of the British Mathematical Olympiad was sat yesterday. The questions can be found here. I recorded some thoughts on the questions while I was in Cyprus, hence the nice Mediterranean sunset above. I hope this might be useful to current or future contestants, as a supplement to the concise official solutions available. It goes without saying that while these commentaries may be interesting at a general level, they will be much more educational to students who have at least digested and played around with the questions, so consider trying the paper first. Video solutions are available here. These have more in common with this blog post than the official solutions, though inevitably some of the methods are slightly different, and the written word has some merits and demerits over the spoken word for clarity and brevity.

The copyright for these questions lies with BMOS, and are reproduced here with permission. Any errors or omissions are obviously my own.

I found the paper overall quite a bit harder than in recent years, or at least harder to finish quickly. I’ve therefore postponed discussion of the final two problems to a second post, to follow shortly.

Question One

A recurring theme of Q1 from BMO1 in recent years has been: “it’s possible to do this problem by a long, and extremely careful direct calculation, but additional insight into the setup makes life substantially easier.”

This is the best example yet. It really is possible to evaluate Helen’s sum and Phil’s sum, and compare them directly. But it’s easy to make a mistake in recording all the remainders when the divisor is small, and it’s easy to make a mistake in summation when the divisor is large, and so it really is better to have a think for alternative approaches. Making a mistake in a very calculation-heavy approach is generally penalised heavily. And this makes sense intellectually, since the only way for someone to fix an erroneous calculation is to repeat it themselves, whereas small conceptual or calculation errors in a less onerous solution are more easily isolated and fixed by a reader. Of course, it also makes sense to discourage such attempts, which aren’t really related to enriching mathematics, which is the whole point of the exercise!

Considering small divisors (or even smaller versions of 365 and 366) is sometimes helpful, but here I think a ‘typical’ divisor is more useful. But first, some notation will make any informal observation much easier to turn into a formal statement. Corresponding to Helen and Phil, let h(n) be the remainder when n is divided by 365, and p(n) the remainder when n is divided by 366. I would urge students to avoid the use of ‘mod’ in this question, partly because working modulo many different bases is annoying notationally, partly because the sum is not taken modulo anything, and partly because the temptation to use mod incorrectly as an operator is huge here [1].

Anyway, a typical value might be n=68, and we observe that 68 x 5 + 25 = 365, and so h(68)=25 and p(68)=26. Indeed, for most values of n, we will have p(n)=h(n)+1. This is useful because

p(1)+p(2)+\ldots+p(366) - \left(h(1)+h(2)+\ldots+h(365)\right)

= \left(p(1)-h(1)\right) + \ldots+\left(p(365)-h(365)\right) + p(366),

and now we know that most of the bracketed terms are equal to one. We just need to handle the rest. The only time it doesn’t hold that p(n)=h(n)+1 is when 366 is actually a multiple of n. In this case, p(n)=0 and h(n)=n-1. We know that 366 = 2 x 3 x 61, and so its divisors are 1, 2, 3, 6, 61, 122, 183.

Then, in the big expression above, seven of the 365 bracketed terms are not equal to 1. So 358 of them are equal to one. The remaining ones are equal to 0, -1, -2, -5, -60, -121, -182 respectively. There are shortcuts to calculate the sum of these, but it’s probably safer to do it by hand, obtaining -371. Overall, since p(366)=0, we have

p(1)+p(2)+\ldots+p(366) - \left(h(1)+h(2)+\ldots+h(365)\right)

= -371 + 358 + 0 = -13.

So, possibly counter-intuitively, Helen has the larger sum, with difference 13, and we didn’t have to do a giant calculation…

Question Two

Suppose each person chooses which days to go swimming ‘at random’, without worrying about how to define this. Is this likely to generate a maximum or minimum value of n? I hope it’s intuitively clear that this probably won’t generate an extreme value. By picking at random we are throwing away lots of opportunity to force valuable overlaps or non-overlaps. In other words, we should start thinking about ways to set up the swimming itinerary with lots of symmetry and structure, and probably we’ll eventually get a maximum or a minimum. At a more general level, with a problem like this, one can start playing around with proof methods immediately, or one can start by constructing lots of symmetric and extreme-looking examples, and see what happens. I favour the latter approach, at least initially. You have to trust that at least one of the extreme examples will be guess-able.

The most obvious extreme example is that everyone swims on the first 75 days, and no-one swims on the final 25 days. This leads to n=75. But we’re clearly ‘wasting’ opportunities in both directions, because there are never exactly five people swimming. I tried a few more things, and found myself simultaneously attacking maximum and minimum, which is clearly bad, so focused on minimum. Just as a starting point, let’s aim for something small, say n=4. The obstacle is that if you demand at most four swimmers on 96 days, then even with six swimmers on the remaining four days, you don’t end up with enough swimming having taken place!

Maybe you move straight from this observation to a proof, or maybe you move straight to a construction. Either way, I think it’s worth saying that the proof and the construction come together. My construction is that everyone swims on the first 25 days, then on days 26-50 everyone except A and B swim, on days 51-75 everyone except C and D swim, and on days 76-100 everyone except E and F swim. This exactly adds up. And if you went for the proof first, you might have argued that the total number of swim days is 6×75 = 450, but is at most 4n + 6(100-n). This leads immediately to n\ge 25, and I just gave the construction. Note that if you came from this proof first, you can find the construction because your proof shows that to be exact you need 25 days with six swimmers, and 75 days with four swimmers, and it’s natural to try to make this split evenly. Anyway, this clears up the minimum.

[Less experienced contestants might wonder why I was worried about generating a construction despite having a proof. Remember we are trying to find the minimum. I could equally have a proof for n\ge 10 which would be totally totally valid. But this wouldn’t show that the minimum was n=10, because that isn’t in fact possible (as we’ve seen), hence it’s the construction that confirms that n=25 is the true minimum.]

It’s tempting to go back to the drawing board for the maximum, but it’s always worth checking whether you can directly adjust the proof you’ve already given. And here you can! We argued that

450\le 4n + 6(100-n)

to prove the minimum. But equally, we know that on the n days we have at least five swimmers, and on the remaining days, we have between zero and four swimmers, so

450 \ge 5n + 0\times (100-n), (*)

which gives n\le 90. If we have a construction that attains this bound then we are done. Why have I phrased (*) with the slightly childish multiple of zero? Because it’s a reminder that for a construction to attain this bound, we really do need the 90 days to have exactly five swimmers, and the remaining ten days to have no swimmers. So it’s clear what to do. Split the first 90 days into five groups of 15 days. One swimmer skips each group. No-one swims in the final ten days, perhaps because of a jellyfish infestation. So we’re done, and 25\le n\le 90.

At a general level, it’s worth noting that in the story presented, we found an example for the minimum which we turned into a proof, and then a proof for the maximum, which we then analysed to produce a construction.

Note that similar bounding arguments would apply if we fiddled with the numbers 5, 75 and 100. But constructions matching the bounds might not then be possible because the splits wouldn’t work so nicely. This would make everything more complicated, but probably not more interesting.

Question Three

It’s understandable that lots of students attempting this paper might feel ill-at-ease with conventional Euclidean geometry problems. A good first rule of thumb here, as in many settings, is “don’t panic!”, and a more specific second rule of thumb is “even if you think you can calculate, try to find geometric insight first.”

Here, it really does look like you can calculate. A configuration based on a given isosceles triangle and a length condition and a perpendicular line is open to several coordinate approaches, and certainly some sensible trigonometry. It’s also very open to organised labelling of the diagram. You have three equal lengths, and a right-angle, as shown.

The key step is this. Drop the perpendicular from A to BC, and call its foot D. That alone really is the key step, as it reduces both parts of the question to an easy comparison. It’s clear that the line AD splits the triangle into two congruent parts, and thus equal areas and perimeters. So it is enough to show that triangle BMN has the same area as triangle ABD, and that their outer-perimeters (ie the part of its perimeter which is also the perimeter of ABC) are the same.

But they’re congruent, so both of these statements are true, and the problem is solved.

My solution could be as short as two or three lines, so for the purposes of this post all that remains is to justify why you might think of the key step. Here are a few possible entry routes:

  • You might notice that line AD induces the required property for triangle ABD.
  • You might try to find a triangle congruent to AMN, and come up with D that way.
  • There’s already a perpendicular in the question so experimenting with another one is natural, especially since the perpendicular from A has straightforward properties.
  • AMN is a right angle, and so constructing D gives a cyclic quadrilateral. We didn’t use that directly in the proof above, but constructing cyclic quadrilaterals is usually a good idea.
  • If you were trying a calculation approach, you probably introduced the length AD, or at least the midpoint D as an intermediate step.

On the video, Mary Teresa proposes a number of elegant synthetic solutions with a few more steps. You might find it a useful exercise to try to come up with some motivating reasons like the bullet points above to justify her suggestion to reflect A in M as a first step.

Question Four

BMO1 2017Q4

I wasn’t paying enough attention initially, and I calculated a_2=0\text{ or }2. This made life much much more complicated. As with IMO 2017 Q1, if trying to deduce general behaviour from small examples, it’s essential to calculate the small examples correctly!

Once you engage your brain properly, you find that a_2=0 \text{ or }3, and of course a_2=0 is not allowed, since it must be positive. So a_2=3, and a similar calculation suggests a_3=1\text{ or }6. It’s clear that the set of values for a_{k+1} depends only on a_k, so if you take a_3=1, then you’re back to the situation you started with at the beginning. If you choose to continue the exploration with a_3=6, you will find a_4=2\text{ or }10, at which point you must be triggered by the possibility that triangle numbers play a role here.

As so often with a play-around with small values, you need to turn a useful observation into a concrete statement, which could then be applied to the problem statement. It looks like in any legal sequence, every term will be a triangle number, so we only need to clarify which triangle number. An example of a suitable statement might be:

Claim: If a_n=T_k=\frac{k(k+1)}{2}, the k-th triangle number, then a_{n+1}=T_{k-1}\text{ or }T_{k+1}.

There are three stages. 1) Checking the claim is true; 2) checking the claim is maximally relevant; 3) proving it. In this case, proving it is the easiest bit. It’s a quick exercise, and I’m omitting it. Of course, we can’t prove any statement which isn’t true, and here we need to make some quick adjustment to account for the case k=1, for which we are forced to take a_{n+1}=T_{k+1}.

The second stage really concerns the question “but what if a_n\ne T_k?” While there are deductions one could make, the key is that if a_1 is a triangle number, the claim we’ve just made shows that a_n is always a triangle number, so this question is irrelevant. Indeed the claim further shows that a_{2017}\le T_{2017}, and also that a_{2017}=T_k for some odd value of k. To be fully rigorous you should probably describe a sequence which attains each odd value of k, but this is really an exercise in notation [2], and it’s very obvious they are all attainable.

In any case, the set of possible values is \{T_1,T_3,\ldots,T_{2017}\}, which has size 1009.

Final two questions

These are discussed in a subsequent post.

Footnotes

[1] – mod n is not an operator, meaning you shouldn’t think of it as ‘sending integers to other integers’, or ‘taking any integer, to an integer in {0,1,…,n-1}’. Statements like 19 mod 5 = 4 are useful at the very start of an introduction to modular arithmetic, but why choose 4? Sometimes it’s more useful to consider -1 instead, and we want statements like a^p\equiv a modulo p to make sense even when a\ge p. 19 = 4 modulo 5 doesn’t place any greater emphasis on the 4 than the 19. This makes it more like a conventional equals sign, which is of course appropriate.

[2] – Taking a_n=T_n for $1\le n\le k$, and thereafter a_n=T_k if k is odd, and $a_n=T_{k+1}$ if k is even will certainly work, as will many other examples, some perhaps easier to describe than this one, though make sure you don’t accidentally try to use T_0!

EGMO 2016 Paper II

Continuing from yesterday’s account of Paper I, this is a discussion of my thoughts about Paper II of EGMO 2016, happening at the moment in Busteni, Romania. This is not an attempt to describe official solutions, but rather to describe the thought process (well, a thought process) of someone tackling each question. I hope it might be interesting or useful, but for students, it will probably be more useful after at least some engagement with the problems. These are excellent problems, and reading any summary of solutions means you miss the chance to hunt for them yourself.

In actual news, you can follow the scoreboard as it is updated from Romania here. Well done to the UK team on an excellent performance, and hope everyone has enjoyed all aspects of the competition!

Question 4

Circles \omega_1,\omega_2 with the same radius meet at two points X_1,X_2. Circle \omega is externally tangent to \omega_1 at T_1, and internally tangent to \omega_2 at T_2. Prove that lines X_1T_1,X_2T_2 meet on \omega.

Thought 1: I’m not the biggest fan of geometry ever, but I thought this looked like a nice problem, because it’s only really about circles, so I figured it probably wouldn’t require anything too exotic.

Thought 2: I bet lots of people try inversion. But the equal radius condition means I’m probably happy with the circles as they are. I hope lots of people don’t try to place the diagram in some co-ordinate system, even if it possible to do it sensibly (eg by making \omega the reference circle).

Thought 3: The labelling of X_1,X_2 is unrelated to the rest of the indexing. So the intersection of X_1T_2,X_2T_1 should also lie on \omega, and possibly has some relationship (antipodal?) to the point I actually need to find out. But I can’t think of any reason why it’s easier to prove two points lie on a circle than just one, so let’s leave this as a thought rather than an idea.

Idea 1: I drew a terrible diagram on the back of a draft of my abstract, and for once, this was actually kind of helpful. Forget about radii being equal, one of them wasn’t even a circle. Anyway, while drawing in the later points, I was struggling to make it look convincingly like all the lengths which were supposed to be equal were in fact equal. So the idea was: almost all the segments in the diagram (once I’ve defined the circle centres O_{\omega_1} etc) have one of two lengths (the radii of \omega_1,\omega – red and green-ish in the diagram below), and with this in mind I can forget about the circles. We’ve got a rhombus, which is even better than a parallelogram, which is itself a really useful thing to have in a configuration. Another consequence of the proliferation of equal lengths is that almost all triangles are isosceles, and we know that similarity of isosceles triangles is particularly easy, because you only have to match up one angle.

Idea 2: How to prove it? We have to prove that two lines and a circle concur. This is where I actually need to stop and think for a moment: I could define the point where the lines meet and try to show it’s on the circle, or intersect one line with the circle, and show it’s on the other line. Idea 1 basically says I’m doing the problem using lengths, so I should choose the way that fits best with lengths.

20160414_093104

If I define the point P where X_2T_2 meets the circle (this was easier to draw in my diagram), then I know PO_\omega=T_2 O_\omega and so on. Then there were loads of isosceles triangles, and some of them were similar, which led to more parallel lines, and from this you could reverse the construction in the other direction to show that P also lay on the other line.

Question 5

Let k, n be integers such that k\ge 2,\, k\le n\le 2k-1. Place rectangular k x 1 or 1 x k tiles on an n x n chessboard in the natural way with no overlap until no further tile can be placed. Determine the minimum number of tiles that such an arrangement may contain.

Idea 1: It took me a while to parse the question. Minimum over what? I rephrased it in my head as: “to show the answer is eg n+5, I need to show that whenever you place n+4 tiles legally, you can’t add another. I also need to show that you can place n+5 such that you can’t add another.” This made life a lot easier.

Thought 1: What goes wrong if you take n=2k and beyond? Well, you can have two horizontal tiles on a given row. I’m not really sure how this affects the answer, but the fact that there is still space constraint for n<2k is something I should definitely use.

Diversion: I spent a while thinking that the answer was 4 and it was a very easy question. I spent a bit more time thinking that the answer was n, and it was a quite easy question, then realised that neither my construction nor my argument worked.

Thought 2: can I do the cases n=k,or 2k-1 or k+1? The answers were yes, unsure, and yes. The answer to k+1, which I now felt confident was actually four, was helpful, as it gave me a construction for k+2, …, 2k-1 that seemed good, even though it was clearly not optimal for 2k-1. Therefore, currently my potential answer has three regimes, which seemed unlikely, but this seemed a good moment to start trying to prove it was optimal. From now on, I’m assuming I have a configuration from which you can’t add another block.

20160414_100244

Idea 2: About this diagram, note that once I’ve filled out the top-left (k+1)x(k+1) sub-board in this way, there are still lots of ways to complete it, but I do have to have (n-k-1) horizontal and (n-k-1) vertical tiles roughly where I’ve put them. Why? Because I can’t ‘squeeze in’ a vertical tile underneath the blue bit, and I can’t squeeze in a horizontal tile to the right of the blue bit. Indeed, whenever I have a vertical block, there must be vertical blocks either to the left or to the right (*) (or possibly both if we’re near the middle). We need to make this precise, but before doing that, I looked back at where the vertical blocks were in the proposed optimum, and it turns out that all but (k-1) columns include a vertical block, and these (k-1) columns are next to each other.

This feels like a great idea for two reasons: 1) we’ve used the fact that n<2k at (*). 2) this feels very pigeonhole principle-ish. If we had fewer tiles, we’d probably have either at least k columns or least k rows without a vertical (or, respectively, horizontal) tile. Say k columns don’t include a vertical tile – so long as they are next to each other (which I think I know) we can probably include a horizontal tile somewhere in there.

So what’s left to do? Check the previous sentence actually works (maybe it’s full of horizontal tiles already?), and check the numerics of the pigeonhole bound. Also work out how the case n=2k-1 fits, but it seems like I’ve had some (/most) of the good ideas, so I stopped here.

Question 6

EGMO2016 Q6

I don’t actually want to say very much about this, because I didn’t finish off all the details. I want to talk briefly in quite vague terms about what to do if you think this problem looks scary. I thought it looked a bit scary because it looked similar to two number theoretic things I remember: 1) primes in arithmetic progressions. This is very technical in general, but I can remember how to do 3 mod 4 fairly easily, and 1 mod 4 with one extra idea; 2) if a square-free number can be written as a sum of two squares, this controls its factors modulo 4.

Vague Ideas: It seemed unlikely that this would involve copying a technical argument, so I thought about why I shouldn’t be scared. I think I shouldn’t be scared of the non-existence part. Often when I want to show there are no integer solutions to an equation, I consider showing there are no solutions modulo some base, and maybe this will be exactly what I do here. I’ll need to convert this statement about divisibility into an equation (hopefully) and check that n\equiv 3,4 modulo 7 doesn’t work.

For the existence of infinitely many solutions, maybe I’d use Chinese Remainder Theorem [1], or I’ll reduce it to something that I know has lots of solutions (eg Pythagoras), or maybe I can describe some explicit solutions?

Actual Idea 1: We’ve got n^2+m | n^4, but this is a very inefficient statement, since the RHS is a lot larger than the LHS, so to be useful I should subtract off a large multiple of the LHS. Difference of two squares is a good thing to try always, or I could do it manually. Either way, I get n^2+m | m^2 which is genuinely useful, given I know m=1,2, …, 2n, because the RHS is now comparable in size to the LHS, so I’ve narrowed it down from roughly n^2 possibilities to just three:

n^2+m=m^2,\quad 2(n^2+m)=m^2,\quad 3(n^2+m)=m^2. (*)

I’m going to stop now, because we’ve turned it into a completely different problem, which might be hard, but at least in principle this is solvable. I hope we aren’t actually scared of (*), since it looks like some problems we have solved before. I could handle one of these in a couple of lines, then struggled a bit more with the other pair. I dealt with one by recourse to some theory, and the final one by recourse to some theory after a lot of rearranging which I almost certainly got wrong, but I think I made an even number of mistakes rather than an odd number because I got the correct solution set modulo 7. Anyway, getting to (*) felt like the majority of the ideas, and certainly removed the fear factor of the Q6 label, so to fit the purpose of this discussion I’ll stop here.

[1] During one lunch in Lancaster, we were discussing why Chinese Remainder Theorem is called this. The claim was that an ancient Chinese general wanted to know the size of their army but it was too big to count, so had them arrange themselves into columns of various sizes, and counted the remainders. The general’s views on the efficiency of this algorithm remain lost in the mists of time.

Modular arithmetic – Beyond the Definitions

Modular arithmetic is a relatively simple idea to define. The natural motivation is to consider a clock. The display of a standard analogue clock makes no distinction between 4am, 4pm, and 4pm next Thursday. This is a direct visualisation of the integers modulo 12. Instead of counting in the usual way, where each successive integer is different to all those considered previously, here, every time we get to a multiple of 12, we reset our count back to zero. As a result, this procedure is often referred to as `clock arithmetic’.

A common problem for good students, for example those starting the UKMT’s Senior Mentoring Scheme, is that the idea of modular arithmetic seems very simple, but it’s hard to work out how it might be useful in application to problems. My claim is that the language of modular arithmetic is often the best way to discuss divisibility properties in problems about whole numbers. In particular, the behaviour of powers (ie m^n and so forth) is nice in this context, and the notation of modular arithmetic is the only sensible way to approach it. Anyway, let’s begin with a quick review of the definitions.

Definitions

We are interested in divisibility by some fixed integer n\geq 2, and the remainders given after we divide by n. Given an integer M, we can write this as:

M=kn+a,\quad\text{ where }a=0,1,\ldots,n-1,

and this decomposition is unique. We then say that M is congruent to a modulo n. Note that working modulo n, means that we are interested in remainders after division by n (rather than a or k or M or anything else). This has the feeling of a function or algorithm applied to M. We get told what M is, then work out the remainder after division by n, and say that this is `M mod n‘.

This is fine, but it very much worth bearing in mind a slightly different interpretation. Working modulo n is a way of saying that we aren’t interested in the exact value of an integer, only where it lies on the n-clock. In particular, this means we are viewing lots of integers as the same. The `sameness’ is actually more important in lots of arguments than the position on the n-clock.

More formally, we say that a\equiv b or a is congruent to b modulo n, if they have the same remainder after division by n. Another way of writing this is that

a\equiv b\quad \iff \quad n|a-b.

Sets of integers which are equivalent are called congruence classes. For example \{\ldots,-4,-1,2,5,8,\ldots\} is a congruence class modulo 3. Note that under the first definition, we consider all elements here to be congruent to 2, but in a particular question it may be more useful to consider elements congruent to -1, or some combination.

These definitions are equivalent, but it can be more useful to apply this second definition for proving things, rather than writing out b=kn+a or whatever all the time.

Addition and Multiplication

Everything has been set up in terms of addition, so it is easy to see that addition works well on congruence classes. That is:

\text{If }a\equiv b,c\equiv d,\quad\text{then }a+c\equiv b+d.

We could argue via a clock argument, but the second definition works very well here:

\text{We have }n|a-b,n|c-d,\quad\text{and so }n|(a+c)-(b+d),\text{ exactly as required.}

We want to show that a similar result happens for multiplication. But this holds as well:

\text{If }a\equiv b,c\equiv d,\quad\text{then }n|c(b-a)\text{ and }n|b(c-d).

\Rightarrow n|ac-bd,\text{ that is }ac\equiv bd.

Let’s just recap what this means. If we want to know what 157\times 32 is modulo 9, it suffices to say that 157\equiv 4 and 32\equiv 5, and so 157\times 32\equiv 4\times 5\equiv 2. In a more abstract setting, hopefully the following makes sense:

\text{ODD}\times\text{ODD}=\text{ODD};\quad \text{EVEN}\times\text{EVEN}=\text{EVEN};\quad \text{ODD}\times\text{EVEN}=\text{EVEN}.

This is exactly the statement of the result in the case n=2, where the congruence classes are the odd integers and the even integers.

Powers

What happens if we try to extend to powers? Is it the case that

\text{if }a\equiv b,c\equiv d,\quad\text{then }a^c\equiv b^d? Continue reading

Senior Mentoring – Solving equations in integers

At school, we learn how to solve equations like 4x+3=-5. Sometimes the answer is an integer (that is, a whole number), sometimes it isn’t. If we change that -5 to a -6, the solution to the equation above won’t be an integer any more. What is more, if we change the 5 to \sqrt{5}, the solution won’t even be rational (that is, a normal fraction that looks like \frac{p}{q}). But, the key point is that in the introduction to the chapter on linear equations, or whatever they call it, in a textbook, there will be an explanation of how to solve this generic class of problems, which will work whether the numbers in the equation are integers, fractions, or even complex numbers!

Later, we move on to quadratic equations. Spotting factorisations is a good way to get started, then you could learn the quadratic formula to use in those cases where the factors are harder to spot. Alternatively, there’s the method of ‘completing the square’, which takes more steps, but means you can do the arithmetic in clear stages. Then, if you think about why the quadratic formula works, you realise that you are just completing the square for all quadratics (using a, b, c in place of any numerical coefficients). In other words, though there are a few different ways of approaching the problem, the sensible methods are essentially the same. And then you might get problems about logarithms or matrices for which you use your new knowledge about these objects to turn the question into a quadratic equation, say. At all times, there’s an implicit assumption that any question can be turned into an equation which you ‘know how to solve’.

So when you first come across a problem which asks you to find integer solutions to an equation, as have appeared a few times on this year’s Senior Mentoring scheme, it is hard to know where to start. You could try to solve it as if it wasn’t about integers, then select the integer solutions at the end. But this could be difficult, or might not even make sense. You might have a condition that x is odd: what does this mean if x isn’t even an integer? You also often start making sensible-looking substitutions: x is odd, so write x=2n+1 for some n which is also an integer, then work with n. But this can sometimes cause you to end with a huge number of variables which you can’t really relate to, and horribly complicated expressions.

As with so much of maths, experience is very useful. If you’ve seen solutions to twenty moderately tricky integer equations, you’ll have more ideas about how to think about starting another problem. So here are a few tricks that might come in handy.

Smallest solution: Are you stuck trying to prove there aren’t any solutions? Do all your substitutions just give more complicated versions of the original equation? What about considering the smallest solution? You might have to be careful about what you mean by ‘small’, but suppose you consider the smallest solution x, and end up finding a smaller solution x'<x. What could this mean? Well it can only mean that there isn’t a smallest solution! In other words, there are no solutions. You should think about why the fact that we are working with integers means that if there is any solution there must be a smallest one. Then see if you can find all solutions in positive (>0) integers to x^2-2y^2=0. Continue reading