| Exam Board | Edexcel |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2023 |
| Session | October |
| Marks | 3 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Proof |
| Type | Proof by exhaustion with table |
| Difficulty | Easy -1.2 This is a straightforward proof by exhaustion requiring students to substitute c = 3a + 1 into a + b + c = 15 to get b = 14 - 4a, then test small positive integer values of a (only a = 1, 2, 3 work). The algebraic setup is simple, the cases are few, and checking divisibility by 4 is routine arithmetic. This is easier than average A-level work as it requires minimal problem-solving and is highly scaffolded with a table. |
| Spec | 1.01a Proof: structure of mathematical proof and logical steps |
| \(a\) | \(b\) | \(c\) | \(a b c\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| One correct set: \((a,b,c) \in \{(1,10,4),(2,6,7),(3,2,10)\}\) | B1 | Any correct set of values for \(a\), \(b\), \(c\); ignore attempt at product \(abc\) |
| Two correct rows with correct \(abc\) calculations | M1 | Condone extra/incorrect rows |
| All three rows correct with \(abc\) values (40, 84, 60) all multiples of 4, plus minimal statement confirming divisibility by 4 | A1 | Must state awareness that 40, 84, 60 are multiples of 4; \(\frac{40}{4}=10\) alone insufficient without "to give whole numbers"; no incorrect calculations e.g. \(\frac{60}{4}=16\) gives A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Solves \(a+b+c=15\) with \(c=3a+1\) to get \(4a+b=14\), states \(b\) is even; OR uses \(c=3a+1\) to state parity relationship between \(a\) and \(c\) | B1 | Either approach acceptable |
| Attempts to generalise; states \(abc = odd \times even \times even\) or \(abc = even \times even \times odd\) = multiple of 4 | M1 | One of the two parity possibilities |
| Fully correct with both cases and no incorrect calculations; \(even \times even\) argued via different variables e.g. \(2m \times 2n = 4mn\) | A1 | \(2n \times 2n = 4n^2\) is A0 |
# Question 1:
| Answer/Working | Marks | Guidance |
|---|---|---|
| One correct set: $(a,b,c) \in \{(1,10,4),(2,6,7),(3,2,10)\}$ | B1 | Any correct set of values for $a$, $b$, $c$; ignore attempt at product $abc$ |
| Two correct rows with correct $abc$ calculations | M1 | Condone extra/incorrect rows |
| All three rows correct with $abc$ values (40, 84, 60) all multiples of 4, plus minimal statement confirming divisibility by 4 | A1 | Must state awareness that 40, 84, 60 are multiples of 4; $\frac{40}{4}=10$ alone insufficient without "to give whole numbers"; no incorrect calculations e.g. $\frac{60}{4}=16$ gives A0 |
**Alt 1 (Deduction):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| Solves $a+b+c=15$ with $c=3a+1$ to get $4a+b=14$, states $b$ is even; OR uses $c=3a+1$ to state parity relationship between $a$ and $c$ | B1 | Either approach acceptable |
| Attempts to generalise; states $abc = odd \times even \times even$ or $abc = even \times even \times odd$ = multiple of 4 | M1 | One of the two parity possibilities |
| Fully correct with both cases and no incorrect calculations; $even \times even$ argued via different variables e.g. $2m \times 2n = 4mn$ | A1 | $2n \times 2n = 4n^2$ is A0 |
---\begin{enumerate}
\item Given that $a , b$ and $c$ are integers greater than 0 such that
\end{enumerate}
\begin{itemize}
\item $c = 3 a + 1$
\item $a + b + c = 15$\\
prove, by exhaustion, that the product $a b c$ is always a multiple of 4\\
You may use the table below to illustrate your answer.
\end{itemize}
You may not need to use all rows of this table.
\begin{center}
\begin{tabular}{|l|l|l|l|}
\hline
$a$ & $b$ & $c$ & $a b c$ \\
\hline
& & & \\
\hline
& & & \\
\hline
& & & \\
\hline
& & & \\
\hline
& & & \\
\hline
& & & \\
\hline
\end{tabular}
\end{center}
\hfill \mbox{\textit{Edexcel P2 2023 Q1 [3]}}