Year: 2009
Paper: 1
Question Number: 1
Course: LFM Pure
Section: Proof
There were significantly more candidates attempting this paper again this year (over 900 in total), and the scores were pleasing: fewer than 5% of candidates failed to get at least 20 marks, and the median mark was 48. The pure questions were the most popular as usual; about two-thirds of candidates attempted each of the pure questions, with the exceptions of question 2 (attempted by about 90%) and question 5 (attempted by about one third). The mechanics questions were only marginally more popular than the probability and statistics questions this year; about one quarter of the candidates attempted each of the mechanics questions, while the statistics questions were attempted by about one fifth of the candidates. A significant number of candidates ignored the advice on the front cover and attempted more than six questions. In general, those candidates who submitted answers to eight or more questions did fairly poorly; very few people who tackled nine or more questions gained more than 60 marks overall (as only the best six questions are taken for the final mark). This suggests that a skill lacking in many students attempting STEP is the ability to pick questions effectively. This is not required for A-levels, so must become an important part of STEP preparation. Another "rubric"-type error was failing to follow the instructions in the question. In particular, when a question says "Hence", the candidate must make (significant) use of the preceding result(s) in their answer if they wish to gain any credit. In some questions (such as question 2), many candidates gained no marks for the final part (which was worth 10 marks) as they simply quoted an answer without using any of their earlier work. There were a number of common errors which appeared across the whole paper. These included a noticeable weakness in algebraic manipulations, sometimes indicating a serious lack of understanding of the mathematics involved. As examples, one candidate tried to use the misremembered identity cos β = sin √(1 − β²), while numerous candidates made deductions of the form "if a² + b² = c², then a + b = c" at some point in their work. Fraction manipulations are also notorious in the school classroom; the effects of this weakness were felt here, too. Another common problem was a lack of direction; writing a whole page of algebraic manipulations with no sense of purpose was unlikely to either reach the requested answer or gain the candidate any marks. It is a good idea when faced with a STEP question to ask oneself, "What is the point of this (part of the) question?" or "Why has this (part of the) question been asked?" Thinking about this can be a helpful guide. One aspect of this is evidenced by pages of formulæ and equations with no explanation. It is very good practice to explain why you are doing the calculation you are, and to write sentences in English to achieve this. It also forces one to focus on the purpose of the calculations, and may help avoid some dead ends. Finally, there is a tendency among some candidates when short of time to write what they would do at this point, rather than using the limited time to actually try doing it. Such comments gain no credit; marks are only awarded for making progress in a question. STEP questions do require a greater facility with mathematics and algebraic manipulation than the A-level examinations, as well as a depth of understanding which goes beyond that expected in a typical sixth-form classroom.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
A {\em proper factor} of an integer $N$ is a positive integer, not $1$ or $N$, that divides $N$.
\begin{questionparts}
\item Show that $3^2\times 5^3$ has exactly $10$ proper factors. Determine how many other integers of the form $3^m\times5^n$ (where $m$ and $n$ are integers) have exactly 10 proper factors.
\item Let $N$ be the smallest positive integer that has exactly $426$ proper factors. Determine $N$, giving your answer in terms of its prime factors.
\end{questionparts}\begin{questionparts}
\item All factors of $3^2 \times 5^3$ have factors of the form $3^k \times 5^l$ where $0 \leq k \leq 2$ and $0 \leq l \leq 3$ therefore there are $3$ possible values for $k$ and $4$ possible values for $l$, which gives $3 \times 4 = 12$ factors, which includes $2$ factors we aren't counting, so $10$ proper factors.
By the same argument $3^m \times 5^n$ has $(m+1) \times (n+1) - 2$ proper factors, so we want $(m+1) \times (n+1) = 12$, so we could have
\begin{array}{cccc}
\text{factor} & m+1 & n + 1 & m & n \\
12 = 12 \times 1 & 12 & 1 & 11 & 0 \\
12 = 6 \times 2 & 6& 2 & 5 & 1 \\
12 = 4 \times 3 & 4& 3 & 3 & 2 \\
12 = 3 \times 4 & 3& 4 & 2 & 3 \\
12 = 2 \times 6 & 2& 6 & 1 & 5 \\
12 = 1 \times 12 & 1& 12 & 0 & 11 \\
\end{array}
So we could have $3^{11}, 3^{5} \times 5^1 3^3 \times 5^2, 3^2 \times 5^3, 3^1 \times 5^5, 5^{11}$
\item Suppose $N$ has $426$ proper factors, then it has $428 = 2^2 \times 107$ factors, so it will either factor as $p^{427}$ or $p_1^{106} p_2^{3}$ or $p_1^{106} p_2 p_3$. Clearly the first will be very large, and we should have $p_1 < p_2 < p_3$, so lets consider $2^{106}$ with either $3^3 = 27$ or $3 \times = 15 < 27$. Therefore we should take $2^{106} \times 3 \times 5$
\end{questionparts}
The start of this question was attempted successfully by the majority of candidates, though a significant number failed to provide any justification of why 3² × 5³ has only 10 factors. Disturbingly, many candidates expanded 3² × 5³ = 1125 and then attempted to factorise the latter, rather than noticing that it is much easier to work from the already-factorised form. Few of these candidates made any further progress in the question. Only about half of the candidates were able to make any significant progress on the second half of part (i); this required deducing a general formula for the number of factors. The idea was that by working through the specific example given at the beginning, it would be realised that every factor has the form 3^a × 5^b, and then a simple counting argument would do the job. There were numerous incorrect formulæ used at this point; candidates did not seem to understand that a formula cannot just be pulled from thin air, but requires some justification. Of those candidates who reached a correct formula, several became stuck attempting to solve the equation (m + 1)(n + 1) = 12; expanding the left hand side is likely to lead nowhere (especially in part (ii) of the question). The form mn + m + n = 11 used by some candidates was particularly unhelpful, as it makes it very hard to justify that all solutions have been found. Furthermore, to answer this part successfully, some systematic approach is required, as we want the number of solutions to this equation. Several candidates slipped up at this point by failing to realise that 0 is, in fact, an integer, giving m = 0, n = 11 and vice versa as solutions. Those candidates who understood part (i) generally made good progress on part (ii). However, a sizeable number of candidates failed to write down the most general form for N, often writing things likes N = p^m × q^n or N = 3^m × 5^n. Also, many failed to consider all possible factorisations of 428. Furthermore, they needed to give some justification for the choice of primes and exponents corresponding to each factorisation of 428, and this was often lacking. Nevertheless, there were some very good answers to this question, and those who understood the principles involved generally made very good progress.