| Exam Board | Edexcel |
|---|---|
| Module | FP2 AS (Further Pure 2 AS) |
| Session | Specimen |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Number Theory |
| Type | Euclidean algorithm with linear combination |
| Difficulty | Moderate -0.3 Part (i) is trivial divisibility checking (even number, digit sum divisible by 3). Part (ii) is a standard textbook application of the extended Euclidean algorithm with small coprime numbers requiring no conceptual insight—just mechanical back-substitution. This is routine for Further Maths students but slightly easier than average A-level difficulty overall due to straightforward execution. |
| Spec | 8.02b Divisibility tests: standard tests for 2, 3, 4, 5, 8, 9, 118.02d Division algorithm: a = bq + r uniquely8.02i Prime numbers: composites, HCF, coprimality |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Adding digits \(8+1+8+4=21\) which is divisible by 3 (or continues to add digits giving \(2+1=3\) which is divisible by 3), so concludes that 8184 is divisible by 3 | M1 | Explains divisibility by 3 rule in context by adding digits |
| 8184 is even, so is divisible by 2 and as divisible by both 3 and 2, so it is divisible by 6 | A1 | Explains divisibility by 2, giving last digit even as reason and makes conclusion that number is divisible by 6 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Starts Euclidean algorithm: \(31=27\times1+4\) and \(27=4\times6+3\) | M1 | Uses Euclidean algorithm showing two stages |
| \(4=3\times1+1\) (so hcf \(=1\)) | A1 | Completes the algorithm. Does not need to state that hcf \(=1\) |
| So \(1=4-3\times1=4-(27-4\times6)\times1=4\times7-27\times1\) | M1 | Starts reversal process, doing two stages and simplifying |
| \((31-27\times1)\times7-27\times1=31\times7-27\times8\), so \(a=-8\) and \(b=7\) | A1cso | Correct completion, giving clear answer following complete solution |
## Question 2:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Adding digits $8+1+8+4=21$ which is divisible by 3 (or continues to add digits giving $2+1=3$ which is divisible by 3), so concludes that 8184 is divisible by 3 | M1 | Explains divisibility by 3 rule in context by adding digits |
| 8184 is even, so is divisible by 2 and as divisible by both 3 and 2, so it is divisible by 6 | A1 | Explains divisibility by 2, giving last digit even as reason and makes conclusion that number is divisible by 6 |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Starts Euclidean algorithm: $31=27\times1+4$ and $27=4\times6+3$ | M1 | Uses Euclidean algorithm showing two stages |
| $4=3\times1+1$ (so hcf $=1$) | A1 | Completes the algorithm. Does not need to state that hcf $=1$ |
| So $1=4-3\times1=4-(27-4\times6)\times1=4\times7-27\times1$ | M1 | Starts reversal process, doing two stages and simplifying |
| $(31-27\times1)\times7-27\times1=31\times7-27\times8$, so $a=-8$ and $b=7$ | A1cso | Correct completion, giving clear answer following complete solution |
---\begin{enumerate}
\item (i) Without performing any division, explain why 8184 is divisible by 6\\
(ii) Use the Euclidean algorithm to find integers $a$ and $b$ such that
\end{enumerate}
$$27 a + 31 b = 1$$
\hfill \mbox{\textit{Edexcel FP2 AS Q2 [6]}}