| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2024 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Number Theory |
| Type | Fermat's Little Theorem |
| Difficulty | Standard +0.3 This is a direct application of Fermat's Little Theorem with minimal steps: recognize that 23 is prime, apply a^22 ≡ 1 (mod 23), decompose 80 = 22×3 + 14, then compute 21^14 mod 23. While it's Further Maths content, it's a straightforward textbook exercise requiring only theorem recall and basic modular arithmetic, making it easier than average overall. |
| Spec | 8.02l Fermat's little theorem: both forms |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(21^{22} \equiv 1 \pmod{23}\) or \(21^{23} \equiv 21 \pmod{23}\) or \(21^{23} \equiv -2 \pmod{23}\) | B1 | Accurately recalls Fermat's Little Theorem. Allow if stated correctly in general form \(a^{p-1} \equiv 1 \pmod{p}\). May be implied. |
| \(80 = 3 \times 22 + 14 \Rightarrow 21^{80} \equiv 21^{14} \pmod{23}\) or \(80 = 3 \times 23 + 11 \Rightarrow 21^{80} \equiv -8 \times 21^{11} \pmod{23}\) | M1 | Attempts to write the index in a suitable form and then applies the theorem; look for \(80 = 22k + r\) or equivalent leading to a reduced index. |
| Full reduction method, e.g. \(\equiv \left((-2)^4\right)^3 \times (-2)^2 \equiv 16^3 \times 4 \equiv (-7)^2 \times -7 \times 4 \equiv 49 \times -28 \equiv 3 \times 18 \equiv 54\) \(\equiv (-2)^{14} \pmod{23} \equiv 16384 \pmod{23}\) \(\equiv (-2)^{14} \equiv (4)^7 \pmod{23}\) | dM1 | For a full method to reduce to a least residue. Many valid splitting approaches accepted. Condone a single sign slip e.g. using \(21^2 \equiv 19 \pmod{23}\) instead of \(21^2 \equiv -19 \pmod{23}\) as long as there is a complete method and only one sign slip. |
| \(\equiv 8 \pmod{23}\) | A1 | Correct answer. |
| (4 marks) |
## Question 1:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $21^{22} \equiv 1 \pmod{23}$ or $21^{23} \equiv 21 \pmod{23}$ or $21^{23} \equiv -2 \pmod{23}$ | **B1** | Accurately recalls Fermat's Little Theorem. Allow if stated correctly in general form $a^{p-1} \equiv 1 \pmod{p}$. May be implied. |
| $80 = 3 \times 22 + 14 \Rightarrow 21^{80} \equiv 21^{14} \pmod{23}$ or $80 = 3 \times 23 + 11 \Rightarrow 21^{80} \equiv -8 \times 21^{11} \pmod{23}$ | **M1** | Attempts to write the index in a suitable form and then applies the theorem; look for $80 = 22k + r$ or equivalent leading to a reduced index. |
| Full reduction method, e.g. $\equiv \left((-2)^4\right)^3 \times (-2)^2 \equiv 16^3 \times 4 \equiv (-7)^2 \times -7 \times 4 \equiv 49 \times -28 \equiv 3 \times 18 \equiv 54$ $\equiv (-2)^{14} \pmod{23} \equiv 16384 \pmod{23}$ $\equiv (-2)^{14} \equiv (4)^7 \pmod{23}$ | **dM1** | For a full method to reduce to a least residue. Many valid splitting approaches accepted. Condone a single sign slip e.g. using $21^2 \equiv 19 \pmod{23}$ instead of $21^2 \equiv -19 \pmod{23}$ as long as there is a complete method and only one sign slip. |
| $\equiv 8 \pmod{23}$ | **A1** | Correct answer. |
| **(4 marks)** | | |\begin{enumerate}
\item In this question you must show detailed reasoning.
\end{enumerate}
Use Fermat's Little Theorem to determine the least positive residue of\\
$21 { } ^ { 80 } ( \bmod 23 )$\\
(4)
\hfill \mbox{\textit{Edexcel FP2 2024 Q1 [4]}}