| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2013 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Proof |
| Type | Divisibility proof for all integers |
| Difficulty | Moderate -0.5 This is a straightforward proof requiring factorisation of a cubic (standard difference of squares technique) followed by showing divisibility by 2 and 3 using consecutive integers. While it requires understanding of proof structure, the factorisation is routine and the divisibility argument is a common textbook exercise, making it slightly easier than average. |
| Spec | 1.01a Proof: structure of mathematical proof and logical steps1.02j Manipulate polynomials: expanding, factorising, division, factor theorem |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(n^3 - n = n(n^2-1)\) | B1 | two correct factors |
| \(= n(n-1)(n+1)\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(n-1\), \(n\) and \(n+1\) are consecutive integers | B1 | |
| so at least one is even, and one is div by 3 | B1 | |
| \([\Rightarrow n^3 - n\) is div by 6\(]\) |
## Question 2:
### Part (i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $n^3 - n = n(n^2-1)$ | B1 | two correct factors |
| $= n(n-1)(n+1)$ | B1 | |
### Part (ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $n-1$, $n$ and $n+1$ are consecutive integers | B1 | |
| so at least one is even, and one is div by 3 | B1 | |
| $[\Rightarrow n^3 - n$ is div by 6$]$ | | |
---2 (i) Factorise fully $n ^ { 3 } - n$.\\
(ii) Hence prove that, if $n$ is an integer, $n ^ { 3 } - n$ is divisible by 6 .
\hfill \mbox{\textit{OCR MEI C3 2013 Q2 [4]}}