| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2023 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Permutations & Arrangements |
| Type | Code/password formation |
| Difficulty | Challenging +1.8 This is a multi-part Further Maths question requiring systematic enumeration and modular arithmetic. Part (i) involves standard permutations with a constraint requiring careful case analysis (letters separated by digits). Part (ii) demands number theory (divisibility by 9, modular arithmetic) combined with logical deduction through multiple constraints. The combination of techniques and the need to work through constraints systematically places this well above average difficulty. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities7.01d Multiplicative principle: arrangements of n distinct objects8.02e Finite (modular) arithmetic: integers modulo n |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(5^4 \times \ldots\) or \(\ldots \times 4 \times 3 \times 2\) or \(\ldots \times {^4}P_3\) | M1 | |
| \(5^4 \times 4 \times 3 \times 2 = 15000\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Structure is \(\_N\_N\_N\_N\_\); three blanks are letters, so \(^5C_3\) choices; new number is \(^5C_3 \times 15000\) or \(10 \times 15000\) | M1 | Correct strategy for new number of codes |
| Increase in number of codes is 135000 | A1ft | Follow through \(9 \times\) their 15000 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Divisible by \(9 \Rightarrow a+b+c+d=9k\); e.g. \(1+3+5+7+9=25\), try \(k=1\) or \(2\); \(1+3+5+7=16>9\) so \(k=2\); \(25-(a+b+c+d)=25-18=7\) missing; or considers only \(1+3+5+9=18\) works | M1 | Full strategy to deduce correct digit set |
| Digits are 1, 3, 5 and 9 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Combination is either 1359 or 9531; \(1359\pmod{13} \equiv 59\pmod{13} \equiv 7\); \(9531\pmod{95} \equiv 31\) | M1 | Checks which combination satisfies final property |
| Combination is 1359 | A1 |
# Question 5:
## Part (i)(a):
| Working | Mark | Guidance |
|---------|------|----------|
| $5^4 \times \ldots$ or $\ldots \times 4 \times 3 \times 2$ or $\ldots \times {^4}P_3$ | M1 | |
| $5^4 \times 4 \times 3 \times 2 = 15000$ | A1 | |
## Part (i)(b):
| Working | Mark | Guidance |
|---------|------|----------|
| Structure is $\_N\_N\_N\_N\_$; three blanks are letters, so $^5C_3$ choices; new number is $^5C_3 \times 15000$ or $10 \times 15000$ | M1 | Correct strategy for new number of codes |
| Increase in number of codes is 135000 | A1ft | Follow through $9 \times$ their 15000 |
## Part (ii)(a):
| Working | Mark | Guidance |
|---------|------|----------|
| Divisible by $9 \Rightarrow a+b+c+d=9k$; e.g. $1+3+5+7+9=25$, try $k=1$ or $2$; $1+3+5+7=16>9$ so $k=2$; $25-(a+b+c+d)=25-18=7$ missing; or considers only $1+3+5+9=18$ works | M1 | Full strategy to deduce correct digit set |
| Digits are 1, 3, 5 and 9 | A1 | |
## Part (ii)(b):
| Working | Mark | Guidance |
|---------|------|----------|
| Combination is either 1359 or 9531; $1359\pmod{13} \equiv 59\pmod{13} \equiv 7$; $9531\pmod{95} \equiv 31$ | M1 | Checks which combination satisfies final property |
| Combination is 1359 | A1 | |
---\begin{enumerate}
\item (i) A security code is made up of 4 numerical digits followed by 3 distinct uppercase letters.
\end{enumerate}
Given that the digits must be from the set $\{ 1,2,3,4,5 \}$ and the letters from the set \{A, B, C, D\}\\
(a) determine the total number of possible codes using this system.
To enable more codes to be generated, the system is adapted so that the 3 letters can appear anywhere in the code but no letter can be next to another letter.\\
(b) Determine the increase in the number of codes using this adapted system.\\
(ii) A combination lock code consists of four distinct digits that can be read as a positive integer, $N = a b c d$, satisfying
\begin{itemize}
\item all the digits are odd
\item $\quad N$ is divisible by 9
\item the digits appear in either ascending or descending order
\item $\quad N \equiv e ( \bmod a b )$ where $a b$ is read as a two-digit number and $e$ is the odd digit that is not used in the code\\
(a) Use the first two properties to determine the four digits used in the code.\\
(b) Hence determine the code on the lock.
\end{itemize}
\hfill \mbox{\textit{Edexcel FP2 2023 Q5 [8]}}