| Exam Board | Edexcel |
|---|---|
| Module | FP2 AS (Further Pure 2 AS) |
| Year | 2024 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Number Theory |
| Type | Linear congruences (single) |
| Difficulty | Standard +0.3 This is a straightforward application of linear congruences with clear real-world context. Part (a) requires translating the problem into modular arithmetic (5x + 6y ≡ 0 (mod 21)), which is direct. Part (b) involves solving x + y = 43 with the congruence constraint, requiring substitution and basic modular arithmetic to find the minimum y. The problem is slightly easier than average as it's well-scaffolded with explicit guidance on the form of the answer and clear constraints. |
| Spec | 8.02e Finite (modular) arithmetic: integers modulo n |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(n \equiv 5x + 6y \pmod{21}\) | B1 | Correct congruence expression |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(x + y = 43\) | B1 | Forms equation in \(x\) and \(y\) from given information |
| \(\Rightarrow n \equiv 5(x+y) + y \equiv 5 \times 43 + y \pmod{21}\) | M1 | Substitutes for \(x\) or \(x+y\) in congruence to form expression in \(y\) only |
| \(\equiv 5 \times 1 + y \equiv 5 + y \pmod{21}\) | A1 | Correct equation in \(y\) (or \(x\)) only |
| \(n \equiv 0 \pmod{21} \Rightarrow 5 + y \equiv 0 \pmod{21} \Rightarrow y \equiv -5 \pmod{21}\) | M1 | Full method to reach residue for \(y\) |
| Minimum for \(y\) is \(16\) | A1 | Interprets situation correctly to give 16 as minimum value |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(x + y = 43\) | B1 | Forms equation in \(x\) and \(y\) |
| \(5x + 6y \pmod{21} = 0 \Rightarrow 5x + 6y = 21N\); solve simultaneously with \(x+y=43\) | M1 | Sets congruence \(= 0\), forms equation with third variable |
| \(y = 21N - 215\) | A1 | Correct equation for \(y\) |
| \(N = 11 \Rightarrow y = 21\times11 - 215 = \ldots\) | M1 | Uses value for third variable to find \(y\) |
| Minimum for \(y\) is \(16\) | A1 | Interprets situation correctly |
# Question 2:
## Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $n \equiv 5x + 6y \pmod{21}$ | B1 | Correct congruence expression |
## Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $x + y = 43$ | B1 | Forms equation in $x$ and $y$ from given information |
| $\Rightarrow n \equiv 5(x+y) + y \equiv 5 \times 43 + y \pmod{21}$ | M1 | Substitutes for $x$ or $x+y$ in congruence to form expression in $y$ only |
| $\equiv 5 \times 1 + y \equiv 5 + y \pmod{21}$ | A1 | Correct equation in $y$ (or $x$) only |
| $n \equiv 0 \pmod{21} \Rightarrow 5 + y \equiv 0 \pmod{21} \Rightarrow y \equiv -5 \pmod{21}$ | M1 | Full method to reach residue for $y$ |
| Minimum for $y$ is $16$ | A1 | Interprets situation correctly to give 16 as minimum value |
### Alternative:
| Answer | Mark | Guidance |
|--------|------|----------|
| $x + y = 43$ | B1 | Forms equation in $x$ and $y$ |
| $5x + 6y \pmod{21} = 0 \Rightarrow 5x + 6y = 21N$; solve simultaneously with $x+y=43$ | M1 | Sets congruence $= 0$, forms equation with third variable |
| $y = 21N - 215$ | A1 | Correct equation for $y$ |
| $N = 11 \Rightarrow y = 21\times11 - 215 = \ldots$ | M1 | Uses value for third variable to find $y$ |
| Minimum for $y$ is $16$ | A1 | Interprets situation correctly |
---\begin{enumerate}
\item Tiles are sold in boxes with 21 tiles in each box.
\end{enumerate}
The tiles are laid out in $x$ rows of 5 tiles and $y$ rows of 6 tiles.\\
All the tiles from a box are used before the next box is opened.\\
When all the rows of tiles have been laid, there are $n$ tiles left in the last opened box.\\
(a) Write down a congruence expression for $n$ in the form
$$a x + b y ( \bmod c )$$
where $a$, $b$ and $c$ are integers.
Given that
\begin{itemize}
\item exactly 43 rows of tiles are laid
\item there are no tiles left in the last opened box\\
(b) use your congruence expression to determine the minimum number of rows of 6 tiles laid.
\end{itemize}
\hfill \mbox{\textit{Edexcel FP2 AS 2024 Q2 [6]}}