Year: 2007
Paper: 1
Question Number: 1
Course: LFM Stats And Pure
Section: Combinatorics
There were significantly more candidates attempting this paper this year (an increase of nearly 50%), but many found it to be very difficult and only achieved low scores. In particular, the level of algebraic skill required by the questions was often lacking. The examiners' express their concern that this was the case despite a conscious effort to make the paper more accessible than last year's. At this level, the fluent, confident and correct handling of mathematical symbols (and numbers) is necessary and is expected; many good starts to questions soon became unstuck after a simple slip. Graph sketching was usually poor: if future candidates wanted to improve one particular skill, they would be well advised to develop this. There were of course some excellent scripts, full of logical clarity and perceptive insight. It was pleasing to note that the applied questions were more popular this year, and many candidates scored well on at least one of these. It was however surprising how rarely answers to questions such as 5, 9, 10, 11 and 12 began with a diagram. However, the examiners were left with the overall feeling that some candidates had not prepared themselves well for the examination. The use of past papers to ensure adequate preparation is strongly recommended. A student's first exposure to STEP questions can be a daunting, demanding experience; it is a shame if that takes place during a public examination on which so much rides. Further, and fuller, discussion of the solutions to these questions can be found in the Hints and Answers document.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
A positive integer with $2n$ digits (the first of which must not be $0$) is called a \textit{balanced number} if the sum of the first $n$ digits equals the sum of the last $n$ digits.
For example, $1634$ is a $4$-digit balanced number, but $123401$ is not a balanced number.
\begin{questionparts}
\item Show that seventy $4$-digit balanced numbers can be made using the digits $0, 1, 2, 3$ and $4$.
\item Show that $\frac16 {k \left( k+1 \right) \left( 4k+5 \right) }$ $4$-digit balanced numbers can be made using the digits $0$ to $k$.
\textit{You may use the identity }$\displaystyle \sum _{r=0}^{n} r^2 \equiv \tfrac{1}{6} {n \left( n+1 \right)
\left( 2n+1 \right) } \;$.
\end{questionparts}\begin{questionparts}
\item For each number from $1$ to $8$ (4+4), we can count the number of ways it can be achieved in any way, or without including a leading $0$.
\begin{array}{c|c|c|c}
\text{total} & \text{ways with }0 & \text{ways without } 0 & \text{comb}\\ \hline
8 & 1 & 1 & 1\\
7 & 2 & 2 & 4 \\
6 & 3 & 3 & 9 \\
5 & 4 & 4 & 16 \\
4 & 5 & 4 & 20 \\
3 & 4 & 3 & 12 \\
2 & 3 & 2 & 6 \\
1 & 2 & 1 & 2 \\ \hline
&&& 70
\end{array}
\item For $2k$ to $k+1$ there are $1 \times 1 + 2 \times 2 + \cdots i\times i+\cdots + k\times k$ ways to achieve this, (we can choose anything from $k$ to $k-i+1$ for the first digit, and we can never have a $0$.
For $1$ to $k$ we can have $1 \times 2 + 2 \times 3 + \cdots + k \times (k+1)$ since we cannot start with a $0$, but can have anything less than or equal to $i$ for the first digit, and then with the $0$ we can have the same thing starting with $0$.
Hence the answer is:
\begin{align*}
&& S &= \sum_{i=1}^k i^2 + \sum_{i=1}^k i (i+1) \\
&&&= 2\sum_{i=1}^k i^2 + \sum_{i=1}^k i \\
&&&= \frac{1}{3} k(k+1)(2k+1) + \frac12k(k+1) \\
&&&= k(k+1) \left (\frac{2k+1}{3} + \frac{1}{2} \right) \\
&&&= \frac16 k(k+1)(4k+2+3) \\
&&&= \frac16 k(k+1)(4k+5)
\end{align*}
\end{questionparts}
This question required little more than a clear head and some persistence: candidates had either ample or very little of both, and thus most scores were either high or very low. The examiners would like to stress that a solution to a question such as this must be written out methodically and coherently: many answers which began promisingly were soon hopelessly fragmented and incoherent, leaving the candidate unable to regain his or her train of thought. This was especially true when deriving the final expression given on the exam paper. Examiners follow closely a candidate's line of reasoning, and they have to be certain that the candidate has constructed a complete argument, and that he or she has not arrived at a printed result without full justification.