Year: 2016
Paper: 1
Question Number: 7
Course: LFM Pure
Section: Proof
This year, more than 2000 candidates signed up to sit this paper, though just under 2000 actually sat it. This figure is about the same as the entry figure for 2015, though the number of candidates opting to sit STEP I has risen significantly over recent years; for instance, it was around 1500 in 2013. There is no doubt that the purpose of the STEPs is to learn which students can genuinely use their mathematical knowledge, skills and techniques in an arena that demands of them a level of performance that exceeds anything they will have encountered within the standard A-level (or equivalent) assessments. The ability to work at an extended piece of mathematical work, often with the minimum of specific guidance, allied to the need for both determination and the ability to "make connections" at speed and under considerable time pressure, are characteristics that only follow from careful preparation, and there is a great benefit to be had from an early encounter with, and subsequent prolonged exposure to, these kinds of questions. It is not always easy to say what level of preparation has been undertaken by candidates, but the minimum expected requirement is the ability to undertake routine A-level-standard tasks and procedures with speed and accuracy. At the top end of the scale, almost 100 candidates produced a three-figure score to the paper, which is a phenomenal achievement; and around 250 others scored a mark of 70+, which is also exceptionally impressive. At the other end of the scale, over 400 candidates failed to reach a total of 40 marks out of the 120 available. For STEP I, the most approachable questions are always set as Qs.1 & 2 on the paper, with Q1 in particular intended to afford every candidate the opportunity to get something done successfully. So it is perfectly reasonable for a candidate, upon opening the paper, to make an immediate start at the first and/or second question(s) before looking around to decide which of the remaining 10 or 11 questions they feel they can tackle. It is very important that candidates spend a few minutes – possibly at the beginning – reading through the questions to decide which six they intend to work, since they will ultimately only be credited with their best six question marks. Many students spend time attempting seven, eight, or more questions and find themselves giving up too easily on a question the moment the going gets tough, and this is a great pity, since they are not allowing themselves thinking time, either on the paper as a whole or on individual questions. The other side to the notion of strategy is that most candidates clearly believe that they need to attempt (at least) six questions when, in fact, four questions (almost) completely done would guarantee a Grade 1 (Distinction), especially if their score on these first four questions were then to be supplemented by a couple of early attempts at the starting parts of a couple more questions (for the first five or six marks); attempts which need not take longer than, say, ten minutes of their time. It is thus perfectly reasonable to suggest to candidates, in their preparations, that they can spend more than 30 minutes on a question, but only IF they think they are going to finish it off satisfactorily, although it might be best if they were advised to spend absolutely no more than 40-45 minutes on any single question; if they haven't finished by then, it really is time to move on. Curve-sketching skills are usually a common weakness, but were only tested on this paper in Q3. The other common area of weakness – algebra – was tested relatively frequently, and proved to be as testing as usual. Calculus skills were generally "okay" although the integration of first-order differential equations by the separation of variables, as appearing repeatedly in Q4, was found challenging by many of the candidates who attempted this question. The most noticeable deficiency, however, was in the widespread inability to construct an argument, particularly in Qs. 5, 7 & 8. Vectors are often poorly handled, and this year proved no exception.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
The set $S$ consists of all the positive integers that leave a remainder of 1 upon division by 4.
The set $T$ consists of all the positive integers that leave a remainder of 3 upon division by 4.
\begin{questionparts}
\item
Describe in words the sets
$S \cup T$ and $S \cap T$.
\item Prove that the product of any two integers in $S$ is also in $S$.
Determine whether the product of any two integers in $T$
is also in $T$.
\item
Given an integer in $T$ that is not a prime number,
prove that at least one of its prime factors is in $T$.
\item
For any set $X$ of positive integers,
an integer in $X$ (other than 1) is said to be
\textit{$X$-prime} if it cannot be expressed as the product of
two or more integers \textit{all in $X$} (and all different from 1).
\begin{itemize}
\item[\bf (a)] Show that every integer in $T$ is
either $T$-prime or is the product of an odd number of
$T$-prime integers.
\item[\bf (b)] Find an
example of an integer in $S$ that can be expressed as the product
of $S$-prime integers in two distinct ways. [Note:
$s_1s_2$ and $s_2s_1$ are not counted as distinct ways of
expressing the product of $s_1$ and $s_2$.]
\end{itemize}
\end{questionparts}\begin{questionparts}
\item $S \cup T$ is the set of odd positive integers. $S \cap T$ is the empty set.
\item Suppose we have two integers in $S$, say $4n+1$ and $4m+1$, so $(4n+1)(4m+1) = 16nm + 4(n+m) + 1 = 4(4nm + n+m) + 1$ which clearly is in $S$.
The product of any two integers in $T$ is in $S$, since $(4n+3)(4m+3) = 16nm + 12(n+m) + 9 = 4(4nm+3(n+m) + 2) + 1$
\item Suppose $p \in T$ is not prime, then it must have prime factors. They must all be odd, so they are in $T$ or $S$. If they are all in $S$ then $p$ must also be in $S$ (as the product of numbers in $S$) but this is a contraction. Therefore $p$ must have a prime factor in $T$.
\item \begin{itemize}
Suppose $n \in T$ is the product of two or more integers in $T$. Since the product of integers in $T$ is in $S$, we can see that (by pairing them off) a product of an even number of integers in $T$ is also in $S$). Therefore $n$ is a product of an odd number of integers in $T$.
\item $3, 7, 11, 19$ are all primes in $T$. Notice therefore that any product of two of them must be $S$-prime. But then $(3 \times 7) \times (11 \times 19)$ and $(3 \times 11) \times (7 \times 19)$ are distinct factorisations into $S$-prime factors.
\end{itemize}
\end{questionparts}
This was a rather splendid reasoning question, and many candidates responded very well to its demands. Almost half of all candidates made an attempt at it, achieving a mean score of over 7 out of 20. Whilst many attempts went little beyond the first two parts, many candidates actually did well on the question as a whole. Part (iii) required a mixture of a proof by contradiction based on at least an informal understanding of the method of proof by induction, so it was no surprise that many gave up at this stage. The "fun" part was (v), in which candidates had to choose some numbers that demonstrated the required properties; unfortunately, it is clear this sort of request is not always well answered.