| Exam Board | OCR |
|---|---|
| Module | Further Additional Pure (Further Additional Pure) |
| Year | 2018 |
| Session | September |
| Marks | 5 |
| Topic | Number Theory |
| Type | Composite number proofs |
| Difficulty | Standard +0.8 This is a Further Maths question requiring students to convert from base-n notation to polynomial form, perform algebraic factorization, and make a number theory argument about compositeness. While the individual steps are accessible (polynomial manipulation and factorization), the question requires insight to connect base representation to divisibility, and the final part demands understanding that a non-trivial factorization implies compositeness. The multi-step reasoning and conceptual sophistication place it moderately above average difficulty. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem4.02j Cubic/quartic equations: conjugate pairs and factor theorem8.02a Number bases: conversion and arithmetic in base n8.02d Division algorithm: a = bq + r uniquely |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(10001_n = n^3 + n + 1\) | B1 | |
| (ii) \(n^3 + n + 1 = (n^2 + n + 1)(n^2 - n^2 + 1)\) | M1, A1 | Attempt at long division or equivalent; Correct cubic factor |
| (iii) For \(n \geq 2\), \(n^2 + n + 1 \geq 7\) and \(n^3 - n^2 + 1 \geq 5\) and both factors \(> 1 \Rightarrow N\) composite | B1, E1 | Statement that both factors \(> 1 \Rightarrow N\) composite; Supporting working (any convincing justification) |
**(i)** $10001_n = n^3 + n + 1$ | B1 |
**(ii)** $n^3 + n + 1 = (n^2 + n + 1)(n^2 - n^2 + 1)$ | M1, A1 | Attempt at long division or equivalent; Correct cubic factor
**(iii)** For $n \geq 2$, $n^2 + n + 1 \geq 7$ and $n^3 - n^2 + 1 \geq 5$ and both factors $> 1 \Rightarrow N$ composite | B1, E1 | Statement that both factors $> 1 \Rightarrow N$ composite; Supporting working (any convincing justification)\begin{enumerate}[label=(\roman*)]
\item Write the number $100011_n$, where $n \geq 2$, as a polynomial in $n$. [1]
\item Show that $n^2 + n + 1$ is a factor of this expression. [2]
\item Hence show that $100011_n$ is composite in any number base $n \geq 2$. [2]
\end{enumerate}
\hfill \mbox{\textit{OCR Further Additional Pure 2018 Q1 [5]}}