Year: 2018
Paper: 2
Question Number: 6
Course: LFM Pure
Section: Proof
The pure questions were again the most popular of the paper, with only two of those questions being attempted by fewer than half of the candidates (none of the other questions was attempted by more than half of the candidates). Good responses were seen to all of the questions, but in many cases, explanations lacked sufficient detail to be awarded full marks. Candidates should ensure that they are demonstrating that the results that they are attempting to apply are valid in the cases being considered. In several of the questions, later parts involve finding solutions to situations that are similar to earlier parts of the question. In general candidates struggled to recognise these similarities and therefore spent a lot of time repeating work that had already been done, rather than simply observing what the result must be.
Difficulty Rating: 1600.0
Difficulty Comparisons: 0
Banger Rating: 1484.7
Banger Comparisons: 1
\begin{questionparts}
\item Find all pairs of positive integers $(n,p)$, where $p$ is a prime number, that satisfy
\[
n!+ 5 =p
\,.
\]
\item In this part of the question you may use the following two theorems:
\begin{enumerate}
\item
For $n\ge 7$, $1! \times 3! \times \cdots \times (2n-1)! > (4n)!\,$.
\item For every positive integer $n$, there is a prime number between $2n$ and $4n$.
\end{enumerate}
Find all pairs of positive integers $(n,m)$ that satisfy
\[
1! \times 3! \times \cdots \times (2n-1)! = m! \,.
\]
\end{questionparts}\begin{questionparts}
\item Let $n! + 5 = p$. If $n \geq 5$ then $5$ divides the LHS and so must also divide the RHS. Since $n!+5 > 5$ this means the RHS cannot be prime. Therefore consider $n = 1, 2, 3, 4$.
\begin{align*}
n = 1: && 1! + 5 = 6 &&\text{ nope} \\
n=2: && 2! + 5 = 7 && \checkmark \\
n=3: && 3! + 5 = 11 && \checkmark \\
n=4: && 4! + 5 = 29 && \checkmark
\end{align*}
Therefore the solutions are $(2,7), (3,11), (4,29)$.
\item Suppose $1! \times 3! \times \cdots \times (2n-1)! = m!$. If $n \geq 7$ then $m! > (4n)!$ (by the first theorem) in particular $m > 4n$. Therefore (by the second theorem) the RHS is divisible by some prime which cannot divide the LHS. Therefore consider $n = 1,2,3,4,5,6$
\begin{align*}
n = 1: && 1! = 1 = 1! && \checkmark \\
n = 2: && 1! \times 3! = 6 = 3! && \checkmark \\
n = 3: && 1! \times 3! \times 5! = 6! && \checkmark \\
n = 4: && 1! \times 3! \times 5! \times 7! = 6! \times 7! = 10! && \checkmark \\
n = 5: && 1! \times 3! \times 5! \times 7! \times 9! = 10! 9! > 11! && \text{would need a factor of } 11\text{ so no} \\
n = 6: && 1! \times 3! \times 5! \times 7! \times 9! \times 11! = 10! 11! 9! > 13! && \text{would need a factor of } 13\text{ so no} \\
\end{align*}
Therefore all solutions are $(1,1), (2,3), (3,6), (4,10)$
\end{questionparts}
Solutions on this question either scored very well or very poorly, depending on the quality of explanation provided by candidates in their solutions. In the weakest cases the only marks that were awarded were for finding some of the particular cases. In the first part of the question candidates were generally able to find the cases that satisfied the equation, but many of the explanations that there are no solutions if ≥ 5 were not sufficiently well produced to receive full marks. In the second part of the question many students just restated the theorems without explaining the reasoning that followed from them. The cases for small values of were generally found, but some candidates struggled to find the pair (4,10). Some candidates also did not attempt to explain why the cases = 5 and = 6 did not produce solutions. There were some attempts to calculate the values of large factorials in this question. Candidates should be aware that such an approach will not be the correct method with which to tackle the questions.