| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2019 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Proof |
| Type | Proof involving squares and modular forms |
| Difficulty | Standard +0.8 This requires systematic proof by cases using modular arithmetic (checking n ≡ 0,1,2,...,7 mod 8), which goes beyond routine algebraic manipulation. While the technique is accessible to AS students, the need to recognize the approach and work through all cases methodically makes it moderately challenging for this level. |
| Spec | 1.01a Proof: structure of mathematical proof and logical steps1.01d Proof by contradiction |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| States that if \(n\) is odd, \(n^3\) is odd | M1 | States result of cubing an odd number |
| So \(n^3 + 2\) is odd and therefore cannot be divisible by 8 | A1 | Valid reason why not divisible by 8; e.g. "odd number \(+ 2\) is still odd and odd numbers are not divisible by 8" |
| States that if \(n\) is even, \(n^3\) is a multiple of 8 | M1 | States result of cubing an even number |
| So \(n^3 + 2\) cannot be a multiple of 8; so \(n^3 + 2\) is not divisible by 8 for \(n \in \mathbb{N}\) | A1 | Both valid reasons with concluding statement |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| (If \(n\) is even,) \(n = 2k\) and \(n^3 + 2 = (2k)^3 + 2 = 8k^3 + 2\) | M1 | Correct expression; no need to state "If \(n\) is even, \(n=2k\)" |
| "\(8k^3\) is divisible by 8, \(8k^3 + 2\) isn't" | A1 | |
| (If \(n\) is odd,) \(n = 2k+1\) and \(n^3 + 2 = (2k+1)^3 + 2 = 8k^3 + 12k^2 + 6k + 3\) | M1 | |
| Which is an even number add 3, therefore odd, hence not divisible by 8; so \(n^3 + 2\) is not divisible by 8 for \(n \in \mathbb{N}\) | A1 | Both valid reasons with concluding statement; note \(\frac{8k^3+12k^2+6k+3}{8} = k^3 + \frac{3}{2}k^2 + \frac{3}{4}k + \frac{3}{8}\) "not a whole number" is too vague — A0 |
## Question 15:
### Logical Approach
| Answer | Mark | Guidance |
|--------|------|----------|
| States that if $n$ is odd, $n^3$ is odd | M1 | States result of cubing an odd number |
| So $n^3 + 2$ is odd and therefore cannot be divisible by 8 | A1 | Valid reason why not divisible by 8; e.g. "odd number $+ 2$ is still odd and odd numbers are not divisible by 8" |
| States that if $n$ is even, $n^3$ is a multiple of 8 | M1 | States result of cubing an even number |
| So $n^3 + 2$ cannot be a multiple of 8; so $n^3 + 2$ is not divisible by 8 for $n \in \mathbb{N}$ | A1 | Both valid reasons with concluding statement |
### Algebraic Approach
| Answer | Mark | Guidance |
|--------|------|----------|
| (If $n$ is even,) $n = 2k$ and $n^3 + 2 = (2k)^3 + 2 = 8k^3 + 2$ | M1 | Correct expression; no need to state "If $n$ is even, $n=2k$" |
| "$8k^3$ is divisible by 8, $8k^3 + 2$ isn't" | A1 | |
| (If $n$ is odd,) $n = 2k+1$ and $n^3 + 2 = (2k+1)^3 + 2 = 8k^3 + 12k^2 + 6k + 3$ | M1 | |
| Which is an even number add 3, therefore odd, hence not divisible by 8; so $n^3 + 2$ is not divisible by 8 for $n \in \mathbb{N}$ | A1 | Both valid reasons with concluding statement; note $\frac{8k^3+12k^2+6k+3}{8} = k^3 + \frac{3}{2}k^2 + \frac{3}{4}k + \frac{3}{8}$ "not a whole number" is too vague — A0 |\begin{enumerate}
\item Given $n \in \mathbb { N }$, prove that $n ^ { 3 } + 2$ is not divisible by 8
\end{enumerate}
\hfill \mbox{\textit{Edexcel AS Paper 1 2019 Q15 [4]}}