| Exam Board | OCR |
|---|---|
| Module | Further Additional Pure AS (Further Additional Pure AS) |
| Year | 2017 |
| Session | December |
| Marks | 7 |
| Topic | Number Theory |
| Type | Composite number proofs |
| Difficulty | Challenging +1.8 This is a Further Maths question requiring proof techniques with numbers in arbitrary bases. Part (i) involves converting to base 10 and recognizing a perfect square (n²+1)², which is moderately challenging. Part (ii) requires factorization insight (n²+2n+1 = (n+1)²) to prove compositeness. While the algebra is straightforward once the approach is clear, recognizing the factorization patterns and constructing formal proofs elevates this above standard A-level difficulty. |
| Spec | 4.01a Mathematical induction: construct proofs8.02a Number bases: conversion and arithmetic in base n |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(10201_n = n^4 + 2n^2 + 1 = (n^2 + 1)^2\) | B1 B1 [2] | 2.1 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| For \(n \geq 3\), \(n + 1 \geq 4\) and \(n^2 + n + 1 \geq 13\) so neither factor is equal to 1 and hence \(1221_n\) is composite | B1 M1 M1 A1 E1 [5] | 2.1 3.1a 2.1 1.1 2.3 |
**(i)** $10201_n = n^4 + 2n^2 + 1 = (n^2 + 1)^2$ | B1 B1 [2] | 2.1 1.1
**(ii)** $1221_n = n^3 + 2n^2 + 2n + 1 = (n^2 + 1) + 2n(n+1) = (n+1)(n^2 + n + 1)$
For $n \geq 3$, $n + 1 \geq 4$ and $n^2 + n + 1 \geq 13$ so neither factor is equal to 1 and hence $1221_n$ is composite | B1 M1 M1 A1 E1 [5] | 2.1 3.1a 2.1 1.1 2.3 | Identifying the linear factor $(n+1)$, by use of the factor theorem or otherwise; Attempt at the quadratic factor; Correct; Visible check that neither factor is equal to 15 Given that $n$ is a positive integer greater than 2 , prove that\\
(i) $\quad 10201 _ { n }$ is a square number.\\
(ii) $\quad 1221 _ { n }$ is a composite number.
\hfill \mbox{\textit{OCR Further Additional Pure AS 2017 Q5 [7]}}