Question 7 - A-Level Maths

Edexcel FP2 2022 June — Question 7 8 marks

Exam Board	Edexcel
Module	FP2 (Further Pure Mathematics 2)
Year	2022
Session	June
Marks	8
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Permutations & Arrangements
Type	Digit arrangements forming numbers
Difficulty	Challenging +1.2 This is a two-part question combining straightforward permutation counting with modular arithmetic constraints. Part (i) requires basic counting of arrangements with restrictions (selecting 4 from 5 or 6 distinct values), which is routine. Part (ii) involves applying divisibility rules for 11 and 9, then systematically finding solutions—standard Further Maths material but requires careful case analysis. The multi-step nature and combination of topics elevates it slightly above average difficulty.
Spec	7.01d Multiplicative principle: arrangements of n distinct objects 8.02b Divisibility tests: standard tests for 2, 3, 4, 5, 8, 9, 11 8.02e Finite (modular) arithmetic: integers modulo n

1. The polynomial $\mathrm { F } ( x )$ is a quartic such that

$$\mathrm { F } ( x ) = p x ^ { 4 } + q x ^ { 3 } + 2 x ^ { 2 } + r x + s$$ where $p , q , r$ and $s$ are distinct constants.
Determine the number of possible quartics given that

the constants $p , q , r$ and $s$ belong to the set $\{ - 4 , - 2,1,3,5 \}$
the constants $p , q , r$ and $s$ belong to the set $\{ - 4 , - 2,0,1,3,5 \}$ (ii) A 3-digit positive integer $N = a b c$ has the following properties
- $N$ is divisible by 11
- the sum of the digits of $N$ is even
- $N \equiv 8 \bmod 9$
- Use the first two properties to show that
$$a - b + c = 0$$
Hence determine all possible integers $N$, showing all your working and reasoning.

Show mark scheme Show mark scheme source

Question 7:

Part (i)(a):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
120	B1

Part (i)(b):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
300	B1

Part (ii)(a):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
Divisible by 11 implies $a - b + c = 11p$; due to restrictions on $a$, $b$, $c$ $\Rightarrow a - b + c = 11$ or $0$	M1
$a + b + c$ is even $\Rightarrow (a+b+c) - (2b)$ is even $\Rightarrow a - b + c$ is even; therefore $a - b + c = 0$ cso	A1*	Fully justified conclusion; various valid approaches accepted

Part (ii)(b):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$N = 100a + 10b + c \Rightarrow a + b + c \equiv 8 \pmod{9}$ or $N + 1 \equiv 0 \pmod{9}$ ($N+1$ is a multiple of 9)	B1

Question 7 (ii)(b) continued:

Answer	Marks	Guidance
Working/Answer	Mark	Guidance
$a + b + c \equiv 8 \pmod{9}$ or $a + b + c + 1 \equiv 0 \pmod{9}$	B1	1.1b
Solves $a - b + c = 0$ and $a + b + c \equiv 8 \pmod{9}$; uses $(a+b+c \equiv 8$ or $26)$ to find $b = 4$, forms equation $a + c = \ldots$	M1	2.1
Or: $2a + 2c \equiv 8 \pmod{9} \Rightarrow a + c \equiv 4 \pmod{9}$ leading to $a + c = \ldots$	M1	2.1
Or: $a - b + c = 0$, $b = a+c$ so $2b+1$ is multiple of 9, $2b+1 = 9$ or $18$ leading to $b = \ldots$	M1	2.1
$N = 143, 242, 341, 440$	M1, A1	1.1b, 2.2a

Alternative:

Answer	Marks	Guidance
Working/Answer	Mark	Guidance
$(N) = 11n \equiv 8 \pmod{9}$	B1	1.1b
Finds multiplicative inverse of 11 using $5 \times 9 - 4 \times 11 = 1$; $-4 \times 11n \equiv -4 \times 8 \pmod{9} \Rightarrow n \equiv -32 \pmod{9} \Rightarrow n \equiv 4 \pmod{9}$; so $N = 11(9n+4)$	M1	2.1
$N = 143, 242, 341, 440$	M1	1.1b
All four correct, no extras	A1	2.2a

Trial and error:

Answer	Marks	Guidance
Working/Answer	Mark	Guidance
Identifying $a + c = b$	B1	—
One correct value (implies B1)	M1	—
Two correct values	M1	—
All four correct values of $N$, no extras	A1	—

## Question 7:

### Part (i)(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| 120 | B1 | |

### Part (i)(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| 300 | B1 | |

### Part (ii)(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Divisible by 11 implies $a - b + c = 11p$; due to restrictions on $a$, $b$, $c$ $\Rightarrow a - b + c = 11$ or $0$ | M1 | |
| $a + b + c$ is even $\Rightarrow (a+b+c) - (2b)$ is even $\Rightarrow a - b + c$ is even; therefore $a - b + c = 0$ cso | A1* | Fully justified conclusion; various valid approaches accepted |

### Part (ii)(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $N = 100a + 10b + c \Rightarrow a + b + c \equiv 8 \pmod{9}$ or $N + 1 \equiv 0 \pmod{9}$ ($N+1$ is a multiple of 9) | B1 | |

# Question 7 (ii)(b) continued:

| Working/Answer | Mark | Guidance |
|---|---|---|
| $a + b + c \equiv 8 \pmod{9}$ or $a + b + c + 1 \equiv 0 \pmod{9}$ | B1 | 1.1b |
| Solves $a - b + c = 0$ and $a + b + c \equiv 8 \pmod{9}$; uses $(a+b+c \equiv 8$ or $26)$ to find $b = 4$, forms equation $a + c = \ldots$ | M1 | 2.1 |
| Or: $2a + 2c \equiv 8 \pmod{9} \Rightarrow a + c \equiv 4 \pmod{9}$ leading to $a + c = \ldots$ | M1 | 2.1 |
| Or: $a - b + c = 0$, $b = a+c$ so $2b+1$ is multiple of 9, $2b+1 = 9$ or $18$ leading to $b = \ldots$ | M1 | 2.1 |
| $N = 143, 242, 341, 440$ | M1, A1 | 1.1b, 2.2a |

**Alternative:**

| Working/Answer | Mark | Guidance |
|---|---|---|
| $(N) = 11n \equiv 8 \pmod{9}$ | B1 | 1.1b |
| Finds multiplicative inverse of 11 using $5 \times 9 - 4 \times 11 = 1$; $-4 \times 11n \equiv -4 \times 8 \pmod{9} \Rightarrow n \equiv -32 \pmod{9} \Rightarrow n \equiv 4 \pmod{9}$; so $N = 11(9n+4)$ | M1 | 2.1 |
| $N = 143, 242, 341, 440$ | M1 | 1.1b |
| All four correct, no extras | A1 | 2.2a |

**Trial and error:**

| Working/Answer | Mark | Guidance |
|---|---|---|
| Identifying $a + c = b$ | B1 | — |
| One correct value (implies B1) | M1 | — |
| Two correct values | M1 | — |
| All four correct values of $N$, no extras | A1 | — |

---

Show LaTeX source

\begin{enumerate}
  \item (i) The polynomial $\mathrm { F } ( x )$ is a quartic such that
\end{enumerate}

$$\mathrm { F } ( x ) = p x ^ { 4 } + q x ^ { 3 } + 2 x ^ { 2 } + r x + s$$

where $p , q , r$ and $s$ are distinct constants.\\
Determine the number of possible quartics given that\\
(a) the constants $p , q , r$ and $s$ belong to the set $\{ - 4 , - 2,1,3,5 \}$\\
(b) the constants $p , q , r$ and $s$ belong to the set $\{ - 4 , - 2,0,1,3,5 \}$\\
(ii) A 3-digit positive integer $N = a b c$ has the following properties

\begin{itemize}
  \item $N$ is divisible by 11
  \item the sum of the digits of $N$ is even
  \item $N \equiv 8 \bmod 9$\\
(a) Use the first two properties to show that
\end{itemize}

$$a - b + c = 0$$

(b) Hence determine all possible integers $N$, showing all your working and reasoning.

\hfill \mbox{\textit{Edexcel FP2 2022 Q7 [8]}}

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