| Exam Board | Edexcel |
|---|---|
| Module | FP2 AS (Further Pure 2 AS) |
| Year | 2020 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Number Theory |
| Type | Euclidean algorithm with linear combination |
| Difficulty | Standard +0.3 This is a standard two-part Euclidean algorithm question requiring mechanical application of the algorithm followed by back-substitution to express the HCF as a linear combination. While it involves multiple steps and careful arithmetic, it's a textbook exercise with no conceptual difficulty or novel insight required—students following the standard procedure will succeed. Slightly above average difficulty due to the computational length and potential for arithmetic errors, but well within routine Further Maths territory. |
| Spec | 8.02d Division algorithm: a = bq + r uniquely8.02i Prime numbers: composites, HCF, coprimality |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(963 = 657 \times 1 + 306\), \(657 = 306 \times 2 + 45\) | M1 | Starts the process by showing 2 correct stages |
| \(306 = 45 \times 6 + 36\), \(45 = 36 \times 1 + 9\), \(36 = 9 \times 4 + 0\) | A1 | Completes the algorithm correctly |
| \(\text{HCF}(963, 657) = 9\) or \(c = 9\) | A1 | Correct HCF |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(9 = 45 - 36 \times 1\), \(9 = 45 - (306 - 45 \times 6)\) | M1 | Starts the reversal process by completing at least 2 stages |
| \(9 = 7 \times 45 - 306 = 7 \times (657 - 2 \times 306) - 306 = 7 \times 657 - 15 \times 306 = 7 \times 657 - 15 \times (963 - 1 \times 657)\) | A1 | Completes the algorithm correctly |
| \(9 = -15 \times 963 + 22 \times 657\) | A1 | Correct values for \(a\) and \(b\) |
## Question 2(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $963 = 657 \times 1 + 306$, $657 = 306 \times 2 + 45$ | M1 | Starts the process by showing 2 correct stages |
| $306 = 45 \times 6 + 36$, $45 = 36 \times 1 + 9$, $36 = 9 \times 4 + 0$ | A1 | Completes the algorithm correctly |
| $\text{HCF}(963, 657) = 9$ or $c = 9$ | A1 | Correct HCF |
## Question 2(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $9 = 45 - 36 \times 1$, $9 = 45 - (306 - 45 \times 6)$ | M1 | Starts the reversal process by completing at least 2 stages |
| $9 = 7 \times 45 - 306 = 7 \times (657 - 2 \times 306) - 306 = 7 \times 657 - 15 \times 306 = 7 \times 657 - 15 \times (963 - 1 \times 657)$ | A1 | Completes the algorithm correctly |
| $9 = -15 \times 963 + 22 \times 657$ | A1 | Correct values for $a$ and $b$ |
---\begin{enumerate}
\item The highest common factor of 963 and 657 is $c$.\\
(a) Use the Euclidean algorithm to find the value of $c$.\\
(b) Hence find integers $a$ and $b$ such that
\end{enumerate}
$$963 a + 657 b = c$$
\hfill \mbox{\textit{Edexcel FP2 AS 2020 Q2 [6]}}