| Exam Board | OCR |
|---|---|
| Module | Further Additional Pure (Further Additional Pure) |
| Year | 2023 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Number Theory |
| Type | Composite number proofs |
| Difficulty | Challenging +1.8 This is a Further Maths number theory question requiring conversion from base-n notation, algebraic factorization, and HCF analysis. While it involves multiple parts and proof techniques, the factorization of 2n³+n²+2n+1 follows a systematic approach, and the HCF analysis is methodical rather than requiring deep insight. The structured parts guide students through the solution, making it challenging but accessible for Further Maths students. |
| Spec | 8.02a Number bases: conversion and arithmetic in base n8.02i Prime numbers: composites, HCF, coprimality |
| Answer | Marks | Guidance |
|---|---|---|
| 8 | (a) | i |
| Answer | Marks |
|---|---|
| = ( 2 n 3 + 2 n ) + ( n 2 + 1 ) = ( 2 n + 1 ( ) n 2 + 1 ) | M1 |
| Answer | Marks |
|---|---|
| [2] | 1.2 |
| 1.1 | Seen, or implied by correct later algebra |
| Answer | Marks | Guidance |
|---|---|---|
| ii | Since f(n) is the product of two integers greater than 1 | |
| (Allow “not equal to 1”), it is composite | B1 | |
| [1] | 2.4 | Properly justified non-primality of f(n) |
| (b) | i | h |
| i.e. (h | ) ±(n – 2) | |
| h | {(2n + 1) – 2(n – 2)} | |
| i.e. h | 5 and h = 1 or 5 | M1 |
| Answer | Marks |
|---|---|
| A1 | 3.1a |
| Answer | Marks |
|---|---|
| 2.4 | Attempt at any linear combination of their a(n) and |
| Answer | Marks | Guidance |
|---|---|---|
| ALT. h | {4(n2 + 1) – (2n – 1) (2n + 1)} | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| i.e. h | 4𝑛2 +4−(4𝑛2 −1) | A1 |
| i.e. h | 5 | M1 |
| h = 1 or 5 | A1 |
| Answer | Marks |
|---|---|
| ii | From (b) (i), we could look for (n – 2) a multiple of 5 |
| n = 7 | M1 |
| A1 | 3.1a |
| 1.1 | Or solving 2n + 1 0 (mod 5) n 2 (mod 5) |
Question 8:
8 | (a) | i | 2 1 2 1 = 2 n 3 + n 2 + 2 n + 1
n
= ( 2 n 3 + 2 n ) + ( n 2 + 1 ) = ( 2 n + 1 ( ) n 2 + 1 ) | M1
A1
[2] | 1.2
1.1 | Seen, or implied by correct later algebra
Correct factorisation. Noting that a(n),𝑏(𝑛) > 1 is
not required here.
ii | Since f(n) is the product of two integers greater than 1
(Allow “not equal to 1”), it is composite | B1
[1] | 2.4 | Properly justified non-primality of f(n)
(b) | i | h | {n(2n + 1) – 2(n2 + 1)}
i.e. (h |) ±(n – 2)
h | {(2n + 1) – 2(n – 2)}
i.e. h | 5 and h = 1 or 5 | M1
A1
M1
A1 | 3.1a
2.2a
2.1
2.4 | Attempt at any linear combination of their a(n) and
b(n)
Second step of this procedure attempted
AG obtained from fully correct working
ALT. h | {4(n2 + 1) – (2n – 1) (2n + 1)} | M1 | Attempt at any linear combination of their a(n) and
b(n)
i.e. h | 4𝑛2 +4−(4𝑛2 −1) | A1
i.e. h | 5 | M1
h = 1 or 5 | A1
[4]
ii | From (b) (i), we could look for (n – 2) a multiple of 5
n = 7 | M1
A1 | 3.1a
1.1 | Or solving 2n + 1 0 (mod 5) n 2 (mod 5)
and/or 𝑛2 +1 ≡ 0 (mod 5) n 2 (mod 5)
NB n = 12 gives 25 145, etc.
[2]8 Let $f ( n )$ denote the base- $n$ number $2121 _ { n }$ where $n \geqslant 3$.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item For each $n \geqslant 3$, show that $\mathrm { f } ( n )$ can be written as the product of two positive integers greater than $1 , \mathrm { a } ( n )$ and $\mathrm { b } ( n )$, each of which is a function of $n$.
\item Deduce that $\mathrm { f } ( n )$ is always composite.
\end{enumerate}\item Let $h$ be the highest common factor of $\mathrm { a } ( n )$ and $\mathrm { b } ( n )$.
\begin{enumerate}[label=(\roman*)]
\item Prove that $h$ is either 1 or 5 .
\item Find a value of $n$ for which $h = 5$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR Further Additional Pure 2023 Q8 [9]}}