| Exam Board | OCR |
|---|---|
| Module | Further Additional Pure AS (Further Additional Pure AS) |
| Year | 2024 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Number Theory |
| Type | Modular arithmetic properties |
| Difficulty | Challenging +1.8 This is a Further Maths modular arithmetic problem requiring systematic case analysis and use of Fermat's Little Theorem or order calculations. Part (a) involves recognizing n = 3(3k±1) forms and showing 1 + 2^n + 4^n ≡ 0 (mod 73) using 2^36 ≡ 1 (mod 73). Part (b) extends this to n = 9k cases. While conceptually accessible to Further Maths students and more structured than a proof requiring novel insight, it demands careful modular arithmetic manipulation across multiple cases, making it notably harder than standard A-level but not exceptionally difficult for this specification. |
| Spec | 8.02d Division algorithm: a = bq + r uniquely8.02e Finite (modular) arithmetic: integers modulo n8.02l Fermat's little theorem: both forms |
| Answer | Marks | Guidance |
|---|---|---|
| 6 | (a) | 2×𝑃5 5 |
| Answer | Marks |
|---|---|
| 6 | M1 |
| Answer | Marks |
|---|---|
| [1] | 1.1 |
| 1.1 | Must consider a range of possible values of P |
| Answer | Marks | Guidance |
|---|---|---|
| 6 | (b) | i |
| [1] | 2.2a | Must be in context (refers to the number of bacteria) |
| 6 | (b) | ii |
| [1] | 3.4 | (∵ the number of bacteria falls below 10 000) |
| 6 | (c) | i |
| [1] | 3.3 | Other possibilities for f may also be fine. At least |
| Answer | Marks | Guidance |
|---|---|---|
| 6 | (c) | ii |
| [1] | 3.1b | FT their f(n) even from B0 in part (c) provided >80 |
| 6 | (c) | iii |
| Answer | Marks |
|---|---|
| 99 | M1 |
| Answer | Marks |
|---|---|
| [2] | 3.2a |
| 3.5b | i.e. 2x2 – 1008x + 9900 = 0 |
| Answer | Marks | Guidance |
|---|---|---|
| 6 | (d) | 2 P n |
| Answer | Marks |
|---|---|
| n | B1 |
| [1] | 3.5c |
Question 6:
6 | (a) | 2×𝑃5 5
P = + used with P = 266.67
6 5
6 𝑃5
so that P = 88.91 to 2d.p.
6 | M1
A1
[1] | 1.1
1.1 | Must consider a range of possible values of P
6
(ignoring </ at endpoints)
Shown carefully (showing that 𝑃 is monotonic in
6
the interval is not required)
SC1 88.91 (by substituting 266.67 soi)
6 | (b) | i | Number of bacteria initially increases, then declines | B1
[1] | 2.2a | Must be in context (refers to the number of bacteria)
6 | (b) | ii | On the 8th day | B1
[1] | 3.4 | (∵ the number of bacteria falls below 10 000)
6 | (c) | i | f(n) = n | B1
[1] | 3.3 | Other possibilities for f may also be fine. At least
three values need to approximately fit
6 | (c) | ii | Suggest n = 100 (accept n = 99) | B1
[1] | 3.1b | FT their f(n) even from B0 in part (c) provided >80
6 | (c) | iii | 2 x 9 9
Using 1 0 .0 8 = + to create a quadratic eqn. in x = P
99
1 0 0 x
which gives P = 10.02 …
99 | M1
A1
[2] | 3.2a
3.5b | i.e. 2x2 – 1008x + 9900 = 0
It is not necessary to verify that P < 10 (working
98
back another step gives P = 9.98)
98
6 | (d) | 2 P n
P = I N T n + (+1)
n + 1 n + 1 P
n | B1
[1] | 3.5c6 For positive integers $n$, let $f ( n ) = 1 + 2 ^ { n } + 4 ^ { n }$.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Given that $n$ is a multiple of 3 , but not of 9 , use the division algorithm to write down the two possible forms that $n$ can take.
\item Show that when $n$ is a multiple of 3 , but not of 9 , $f ( n )$ is a multiple of 73 .
\end{enumerate}\item Determine the value of $\mathrm { f } ( n )$, modulo 73 , in the case when $n$ is a multiple of 9 .
\end{enumerate}
\hfill \mbox{\textit{OCR Further Additional Pure AS 2024 Q6 [9]}}