A2 June 2025 Q7
7.
| Scheme | Marks | AO |
|---|---|---|
| \(\gcd(36, 21) = 3\) | M1 | 1.1b |
| 3 does not divide into 11 therefore no solutions | A1 | 2.4 |
| (2) |
Notes
M1: Finds the gcd of 36 and 21 or in some way considers the hcf of these two numbers.
A1: States that the gcd is 3 and does not divide into 11 therefore no solutions
| Scheme | Marks | AO |
|---|---|---|
| \(12 \times 3y \equiv 12 \times 16 \pmod{40}\) | M1 | 1.1b |
| \(\gcd(12, 40) = 4\) and divides through by 4 | M1 | 1.1b |
| \(3y \equiv 16 \pmod{10} \Rightarrow 3y \equiv 6 \pmod{10}\) \(\gcd(3, 10) = 1\) | dM1 | 2.1 |
| \(y \equiv 2 \pmod{10}\) or \(y \equiv 2, 12, 22, 32 \pmod{40}\) | A1 | 2.2a |
| (4) | ||
| (6 marks) |
Notes
M1: Identifies \(36y\) as \(kay\) and 192 as \(kb\) (may be implied by cancelling).
M1: Finds the gcd of \(k\) and 40. Then divides through by the gcd of \(k\) and 40, Modulus must be reduced to 10.
dM1: Completes the solution by appropriate means. There will be many variations on how they can do this. Accept just \(y = 2\) as an attempt to find the solution. Note the multiplicative inverse of 9 modulo 10 is 9. If a multiplicative inverse is attempted, it must be from a correct method.
A1: Deduces the correct solution
Alternative I
| Scheme | Marks | AO |
|---|---|---|
| \(36y \equiv 192 \equiv 32 \pmod{40}\) | M1 | 1.1b |
| \(\text{hcf}(36, 40) = 4 \Rightarrow 9y \equiv 8 \pmod{10}\) | M1 | 1.1b |
| E.g. \(9y \equiv 18 \pmod{10} \Rightarrow y = 2 \pmod{10}\) | dM1 | 2.1 |
| \(y \equiv 2 \pmod{10}\) or \(y \equiv 2, 12, 22, 32 \pmod{40}\) | A1 | 2.2a |
| (4) |
M1: Reduces RHS modulo 40 to give \(36y \equiv 32 \pmod{40}\)
M1: Finds the gcd of 36 and 40 and divides through by 4. This must include the modulus being reduced to 10.
dM1: Depends on previous M. Finds a value for \(y\) by any appropriate means. There will be many variations on how they can do this. Accept just \(y = 2\) as an attempt to find the solution.
A1: Deduces the correct solution
Note: Candidates to start by finding the hcf of 36 and 40 and divide through to get \(9y \equiv 48 \pmod{10}\) will score M2 at the start of the solution.
(ii) Alt II
| Scheme | Marks | AO |
|---|---|---|
| \(36y \equiv 192 \pmod{40} \Rightarrow 36y = 192 + 40k\) | M1 | 1.1b |
| \(\Rightarrow 9y = 48 + 10k\) | M1 | 1.1b |
| \(\Rightarrow 10y - y = 50 - 2 + 10k \Rightarrow y = 10y - 50 - 10k + 2\) \(\Rightarrow y = 2 + 10(y - 5 - k)\) | dM1 | 2.1 |
| \(y \equiv 2 \pmod{10}\) or \(y \equiv 2, 12, 22, 32 \pmod{40}\) | A1 | 2.2a |
| (4) |
M1: Writes the congruence as a linear equation with \(40k\)
M1: Divides the equation through by 4
dM1: Completes the solution by appropriate means. One example shown in scheme.
A1: Deduces the correct solution.