A2 June 2022 Q7
7.
where \(p\), \(q\), \(r\) and \(s\) are distinct constants.
Determine the number of possible quartics given that
- \(N\) is divisible by 11
- the sum of the digits of \(N\) is even
- \(N \equiv 8 \bmod 9\)
| Scheme | Marks | AO |
|---|---|---|
| 120 | B1 | 1.1b |
| (1) |
Notes
B1: See scheme
| Scheme | Marks | AO |
|---|---|---|
| 300 | B1 | 1.1b |
| (1) |
Notes
B1: See scheme
| Scheme | Marks | AO |
|---|---|---|
| As divisible by 11 implies \(a - b + c = 11p\) where \(p\) is an integer Due to the restrictions on \(a\), \(b\) and \(c\) \(\Rightarrow a - b + c = 11\) or 0 | M1 | 2.4 |
| For example \(a + b + c\) is even \(\Rightarrow (a + b + c) - (2b)\) is even \(\Rightarrow a - b + c\) is even or Sum of digits is even implies that \(a + b + c = 2q\) \((a - b + c = (a + b + c) - 2b = 2q - 2b = 2n)\) Alternative approach 1 \(a + b + c = 2q\) where \(q \in \mathbb{Z}\) either \(a + b + c = 2q\) and \(a - b + c = 0 \Rightarrow 2b = 2q \Rightarrow b = q\) valid \(a + b + c = 2q\) and \(a - b + c = 11 \Rightarrow 2b = 2q - 11 \Rightarrow b = q - 5.5\) not valid Or \(a + b + c = 2q\) and \(a - b + c = 0 \Rightarrow 2a + 2c = 2q\) valid \(a + b + c = 2q\) and \(a - b + c = 11 \Rightarrow 2a + 2c = 2q + 11\) not valid as \(2(a + c)\) even or This approach may be in words Alternative approach 2 If sum \(a - b + c\) is odd/11 then either one or all three numbers odd. This would mean sum \(a + b + c\) would be odd/not even Alternative approach 3 If \(a - b + c = 11\) this implies \(a + b + c \equiv 1 \bmod 2\) (contradiction) Conclusion therefore \(a - b + c = 0\)* cso | A1* | 2.1 |
| (2) |
Notes
M1: Uses the divisibility rule for 11 and the restrictions on the values \(a\), \(b\) and \(c\) leading to \(a - b + c = 0\) or 11 only
A1*: Uses the information that the sum of the digits is even and that \(a - b + c = 0\) or 11 from the divisibility rule for 11 to show that \(a - b + c = 0\) cso
| Scheme | Marks | AO |
|---|---|---|
| \(N = 100a + 10b + c \Rightarrow a + b + c \equiv 8 \bmod 9\) o.e or \(N + 1 \equiv 0 \bmod 9\) (\(N + 1\) is a multiple of 9) o.e. \(\Rightarrow a + b + c + 1 \equiv 0 \bmod 9\) (\(a + b + c + 1\) is a multiple of 9) o.e | B1 | 1.1b |
| Solves \(a - b + c = 0\) and \(a + b + c \equiv 8 \bmod 9\) either \((a + b + c \equiv 8\) or \(26)\) to find a value for \(b\) (= 4) and uses this value to form an equation \(a + c = \ldots (4)\) Or \(2a + 2c \equiv 8 \bmod 9 \Rightarrow a + c \equiv 4 \bmod 9\) leading to \(a + c = \ldots (4)\) or \(2a + 2c \equiv 8 \bmod 9 \Rightarrow a + c \equiv 4 \bmod 9 \Rightarrow b \equiv 4 \bmod 9\) leading to \(b = \ldots (4)\) Or \(a - b + c = 0,\ b = a + c\) so \(2b + 1\) is a multiple of 9, \(2b + 1 = 9\) or 18 leading to \(b = \ldots\) (4) | M1 | 2.1 |
| \(N = 143,\ 242,\ 341,\ 440\) | M1 A1 | 1.1b 2.2a |
| (4) | ||
| (8 marks) |
Notes
B1: States \(a + b + c \equiv 8 \bmod 9\) or \(a + b + c + 1 \equiv 0 \bmod 9\)
M1: Uses their equations to find a value for \(a + c = \ldots\) or a value for \(b\)
M1: Uses their values to find at least two correct values for \(N\)
A1: Deduces all four correct values of \(N\) and no extra values
Trial and error
B1: For identifying \(a + c = b\)
M1: For one correct value, this implies B1
M1: For two correct values
A1: Deduces all four correct values of \(N\) and no extra values
Alternative
| Scheme | Marks | AO |
|---|---|---|
| \((N) = 11n \equiv 8 \bmod 9\) | B1 | 1.1b |
| Finds multiplicative inverse of 11 using \(5 \times 9 - 4 \times 11 = 1\) \(-4 \times 11n \equiv -4 \times 8 \bmod 9 \Rightarrow n \equiv -32 \bmod 9 \Rightarrow n \equiv 4 \bmod 9\) So \(N = 11(9n + 4)\) | M1 | 2.1 |
| \(N = 143,\ 242,\ 341,\ 440\) | M1 | 1.1b |
| A1 | 2.2a | |
| (4) |
B1: States \((N) = 11n \equiv 8 \bmod 9\)
M1: Finds the multiplicative inverse of 11 and uses this to find an expression for \(N\)
M1: Uses their expression for \(N\) to find at least two values for \(N\)
A1: Deduces all four correct values of \(N\) and no extra values