1Recent studies comparing rank in STEP with rank in ﬁrst-year Cambridge mathematics examinations reveal a Spearman correlation coefficient of 0.63, which is very high in comparison with other predictors of university examination results.

2I use the term ‘A-level’ here as a shorthand for a typical school mathematics examination. The particular examinations you take may well be very different in style and format but, even if that is the case, I am sure some of what follows will strike a chord with you.

3It is freely available, so you might like to try it out. You will probably ﬁnd it good fun to use, but quite time-consuming and not really suitable for writing out the solutions of STEP problems.

4These and the other publications mentioned below are obtainable from the STEP web site www.stepmathematics.org.uk. You will also ﬁnd other useful material on the NRICH web site; do google it now if you haven’t already done so.

5Though you must obey instructions in the question: for example, if it says ‘Hence prove ...’, then you must use the previous result in your proof.

6Begin rant: I am very surprised at the scrappy and illegible work that I receive from a few of my students. It seems so disrespectful to expect me to spend ages trying to decipher their work when they could have spent a little more time making it presentable, for example by copying it out neatly or writing more slowly. Why is my time less important than theirs? End rant.

7J. E. Littlewood (1885–1977), distinguished Cambridge mathematician and author of the highly entertaining Mathematicians Miscellany (Cambridge University Press, 1986) used to work seven days a week until an experiment revealed that when he took Sundays off the good ideas had a way of coming on Mondays.

8Another check will reveal that for very small positive $x$, $log$ is positive (since its argument is bigger than 1) whereas the series is negative, so there is clearly something else wrong.

9Which what I normally do; but I am still hoping to heed my own advice in the not too distant future.

10Mathematicians should feel as insulted as engineers by the following joke.

A mathematician, a physicist and an engineer enter a mathematics contest, the ﬁrst task of which is to prove that all odd number are prime. The mathematician has an elegant argument: ‘1’s a prime, 3’s a prime, 5’s a prime, 7’s a prime. Therefore, by mathematical induction, all odd numbers are prime. It’s the physicist’s turn: ‘1’s a prime, 3’s a prime, 5’s a prime, 7’s a prime, 11’s a prime, 13’s a prime, so, to within experimental error, all odd numbers are prime.’ The most straightforward proof is provided by the engineer: ‘1’s a prime, 3’s a prime, 5’s a prime, 7’s a prime, 9’s a prime, 11’s a prime …’.

11There is an old anecdote about the distinguished Professor X. In the middle of a lecture she writes ‘It is obvious that A’, suddenly falls silent and after a few minutes she rubs it out and walks out of the room. The awed students hear her pacing up and down outside. Then after twenty minutes she returns and writes ‘It is obvious that A’ and continues the lecture.

12You can see this easily by sketching a ‘typical’ odd function, or by differentiating the Maclaurin expansion of an odd function.

13It is just the sort of thing that is used in university texts; but I’m not sure that it would be used in STEP papers nowadays.

14I went as far as ${x}_{7}$ to be sure of the pattern: I found that ${x}_{7}=\frac{8{x}_{1}+5}{5{x}_{1}+3}$.

15Fibonacci (short for ﬁlius Bonacci — son of Bonacci) was called the greatest European mathematician of the Middle Ages. He was born in Pisa (Italy) in about 1175 AD. He introduced the series of numbers named after him in his book of 1202 called Liber Abbaci (Book of the Abacus). It was the solution to the problem of the number of pairs of rabbits produced by an initial pair: A pair of rabbits are put in a ﬁeld and, if rabbits take a month to become mature and then produce a new pair every month after that, how many pairs will there be in twelve months time?

16After the ﬁrst edition, a correspondent pointed out that the following argument gives angle $EAB$ more quickly: clearly $A$, $B$ and $E$ all lie on a circle with centre $O$; the angle at $O$ subtended by the chord $AB$ is $\frac{2}{7}\pi$, so the angle at the circumference is $\frac{1}{7}\pi$. Obvious, really — can’t think why I didn’t see it.

17An ellipsoid is roughly the shape of the surface of a rugby ball, or of the giant galaxy ESO 325-G004. A hyperboloid can either be the shape of an inﬁnite radar dish (in fact, a pair of such dishes) or it can be the shape of a power station cooling tower. Our equations in fact represent hyperboloids of the cooling tower type.

18Now I look at it again, it seems to me that the last sentence of my solution (‘Therefore the inequality holds ...’) is a bit suspect without sketches to show that the inequality actually does hold.

19The term rider seems a bit old-fashioned now. It referred to the ﬁnal part at the end of an examination question in the days when examination questions often consisted of a relatively straightforward ﬁrst part, perhaps the proof of a theorem, followed by a more tricky part extending or applying the ﬁrst part.

20I mentioned this in a talk I gave to sixth formers, and later saw irritatingly on www.thestudentroom.co.uk a comment to the effect that ‘even Dr Siklos can’t do this STEP question’. Of course I could do it; I wanted to understand it, as I hope you do too.

21M. Aigler and G.M. Ziegler (Springer, 1999). The idea behind the proof was given by Christian Goldbach (1690–1764), who is best known for his conjecture that there are an inﬁnite number of twin primes, that is, prime numbers that are two apart such as 17 and 19.

22See The Man Who Loved Only Numbers by Paul Hoffman published by Little Brown and Company in 1999 — a wonderfully readable and interesting book.

23The theorem says that the equation ${x}^{n}+{y}^{n}={z}^{n}\phantom{\rule{0.3em}{0ex}}$, where $x\phantom{\rule{0.3em}{0ex}}$, $y\phantom{\rule{0.3em}{0ex}}$, $z$ and $n$ are positive integers, can only hold if $n=1$ or $2$

24https://www.desmos.com/calculator is brilliant: you can use a slider to change the value of $a$; but please don’t.

25I should come clean at this point: I just differentiated $\left(\ast \right)$ on autopilot, without considering whether I could simplify the task by rewriting the formula as suggested in my hint on the previous page. It was hard work. Serves me right for not following my own general advice.

26For large $n$ we are just adding up a lot of sin curves and it doesn’t much matter what the exact value of $\pi$ is (so we can take it to be a variable); for the exact result, the exact value of $\pi$ does matter.

27Joseph Fourier, 1768– 1830, was a French mathematician and administrator. He is best known for his work on the theory of heat ﬂow. He is generally credited with the discovery of the greenhouse effect. He spent some time in Egypt with Napoleon, and contributed much to the modern study of Egyptology.

28Suppose $p\sqrt{2}=q$. Then $2{p}^{2}={q}^{2}$. Writing $p$ and $q$ as the product of primes gives the contradiction since there are an odd number of powers of 2 one side of this equation and an even number on the other.

29The expansion sums to 1;  I don’t know how you can see that directly. I thought it would come from using the classical formula for the root of a cubic:

$-{\left[\frac{3}{2}\left(\phantom{\rule{0.3em}{0ex}}1+i\sqrt{\frac{13}{243}}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\right)\right]}^{\frac{1}{3}}-{\left[\frac{3}{2}\left(\phantom{\rule{0.3em}{0ex}}1-i\sqrt{\frac{13}{243}}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\right)\right]}^{\frac{1}{3}}\phantom{\rule{0.3em}{0ex}},$

expanding each bracket binomially but it doesn’t seem to. The roots are obtained from this complicated expression by noticing that $-{\left[\frac{3}{2}\left(\phantom{\rule{0.3em}{0ex}}1±i\sqrt{\frac{13}{243}}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\right)\right]}^{\frac{1}{3}}=\frac{1}{2}\left(\phantom{\rule{0.3em}{0ex}}1±i\sqrt{\frac{13}{3}}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\right)$.

30Note that this equation cannot in general be solved by spotting roots. In fact, by translation and scaling, any cubic equation can be reduced to this form, so our series solution can be used to ﬁnd a solution of any cubic ewquation.

31Now located in Poland. Originally, the ﬁrm made chocolate apples which, being apple-shaped, would not have worked well for this question (the centre of mass of an orange segment was hard enough for me). The chocolate lemon, introduced in 1974, would not have been good either but luckily it was discontinued rapidly after it turned out that no one wanted to eat it.

32Draw a diagram! I would, but there isn’t enough room on the page.

33Think of fences and posts: $k$ posts here but only $k-1$ fences.