| Exam Board | OCR |
|---|---|
| Module | Further Additional Pure (Further Additional Pure) |
| Year | 2020 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Number Theory |
| Type | Modular arithmetic properties |
| Difficulty | Challenging +1.2 This is a structured modular arithmetic question from Further Maths that guides students through Fermat's Little Theorem applications. Parts (a)-(c)(i) are systematic applications of standard results with clear scaffolding. Part (c)(ii) requires recognizing that 30 is not maximal (counterexample: n=2 gives 30, but testing shows no universal N>30 works), but the question structure heavily hints at the approach. Moderately above average due to Further Maths content and proof requirements, but well-scaffolded. |
| Spec | 8.02l Fermat's little theorem: both forms |
| Answer | Marks | Guidance |
|---|---|---|
| 7 | (a) | By “Fermat’s Little Theorem”, since 5 is prime. |
| [1] | 1.2 | i.e. FLT |
| (b) | If n is even, then n5 is also even |
| Answer | Marks |
|---|---|
| Since hcf(2, 5) = 1 it follows that n5 ≡ n (mod 2×5=10) | M1 |
| Answer | Marks |
|---|---|
| [3] | 2.1 |
| Answer | Marks | Guidance |
|---|---|---|
| 2.4 | Conclusion with supporting reason | |
| (c) | (i) | Need to show that n5 ≡ n (mod 3) |
| Answer | Marks |
|---|---|
| i.e. n5 – n is divisible by 30 ∀ n ∊ ℕ | B1 |
| Answer | Marks |
|---|---|
| A1 | 3.1a |
| Answer | Marks |
|---|---|
| 2.4 | Running through all possibilities |
| Answer | Marks |
|---|---|
| Alternative method | B1 |
| Answer | Marks |
|---|---|
| Then n5 = n3.n2 ≡ n.n2 = n3 ≡ n | M1 A1 |
| Because (mod 10), then is divisible by 10 | M1 |
| A1 | Conclusion with supporting reason |
| Answer | Marks | Guidance |
|---|---|---|
| (c) | (ii) | No – since, when n = 2, n5 – n = 30 |
| [1] | 2.3 | Producing a counter-example |
Question 7:
7 | (a) | By “Fermat’s Little Theorem”, since 5 is prime. | B1
[1] | 1.2 | i.e. FLT
(b) | If n is even, then n5 is also even
If n is odd, then n5 is also odd so that n5 ≡ n (mod 2)
Since hcf(2, 5) = 1 it follows that n5 ≡ n (mod 2×5=10) | M1
A1
A1
[3] | 2.1
1.1
2.4 | Conclusion with supporting reason
(c) | (i) | Need to show that n5 ≡ n (mod 3)
mod3, 05 = 0 ≡ 0, 15 = 1 ≡ 1, 25 = 32 ≡ 2
Because (mod 10), then is divisible by 10
Since hcf(35, 10) = 1 it follows that5 n5 ≡ n (mod 3×10=30)
𝑛𝑛 ≡ 𝑛𝑛 𝑛𝑛 −𝑛𝑛
i.e. n5 – n is divisible by 30 ∀ n ∊ ℕ | B1
M1
A1
M1
A1 | 3.1a
2.1
1.1
3.1a
2.4 | Running through all possibilities
All noted correctly
Conclusion with supporting reason
Alternative method | B1
n3 ≡ n (mod 3) by FLT
Then n5 = n3.n2 ≡ n.n2 = n3 ≡ n | M1 A1
Because (mod 10), then is divisible by 10 | M1
A1 | Conclusion with supporting reason
Since hcf(35, 10) = 1 it follows that5 n5 ≡ n (mod 3×10=30)
𝑛𝑛 ≡ 𝑛𝑛 𝑛𝑛 −𝑛𝑛
i.e. n5 – n is divisible by 30 ∀ n ∊ ℕ
[5]
(c) | (ii) | No – since, when n = 2, n5 – n = 30 | B1
[1] | 2.3 | Producing a counter-example
M1
A17 Throughout this question, $n$ is a positive integer.
\begin{enumerate}[label=(\alph*)]
\item Explain why $n ^ { 5 } \equiv n ( \bmod 5 )$.
\item By proving that $n ^ { 5 } \equiv n ( \bmod 2 )$, show that $n ^ { 5 } \equiv n ( \bmod 10 )$.
\item \begin{enumerate}[label=(\roman*)]
\item Prove that $n ^ { 5 } - n$ is divisible by 30 for all positive integers $n$.
\item Is there an integer $N$, greater than 30 , such that $n ^ { 5 } - n$ is divisible by $N$ for all positive integers $n$ ? Justify your answer.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR Further Additional Pure 2020 Q7 [10]}}