A2 June 2019 Q4
4.
| Scheme | Marks | AO |
|---|---|---|
| \(6^{13-1} \equiv 1\ (\mathrm{mod}\ 13)\) or \(6^{13} \equiv 6\ (\mathrm{mod}\ 13)\) | B1 | 1.2 |
| Attempts \(542 = 45 \times 12 + 2\) or \(542 = 41 \times 13 + 9\) (seen or implied) | M1 | 1.1b |
| \(6^{542} = \left(6^{12}\right)^{45} \times 6^2\) or \(6^{542} = \left(6^{13}\right)^{41} \times 6^9\) | A1 | 1.1b |
| \(\equiv 1 \times 6^2 \equiv \ldots\ (\mathrm{mod}\ 13)\) or \(\equiv 6^{41} \times 6^9 \equiv \left(6^{13}\right)^3 \times 6^2 \times 6^9 \equiv 6^3 \times 6^2 \times 6^9 \equiv 6^{13} \times 6 \equiv 6^2 \equiv \ldots\ (\mathrm{mod}\ 13)\) | M1 | 1.1b |
| \(\equiv 10\ (\mathrm{mod}\ 13)\) | A1 | 1.1b |
| (5) |
Notes
B1: Recalls Fermat’s Little Theorem correctly. May be implied in their work.
M1: Attempts 542 in the form \(12a + b\). Score if an attempt at \(\left(6^{12}\right)^{45}\) or similar is seen with attempt to combine with another term. Allow M1 for attempt \(\text{“}542 = \text{their } 12\text{”} \times a + b\)
A1: Uses their \(a\) and \(b\) to write \(6^{542}\) correctly in terms of \(6^{12}\) or \(6^{13}\) (must be one of these two)
M1: Completes the process to find the residue (more convoluted roots are possible but look for a complete process to reach a residue using their attempt at Fermat’s Little Theorem at least once).
A1: Correct residue. Allow if the “45” was incorrect so long as the remainder was 2.
| Scheme | Marks | AO |
|---|---|---|
| \(7! = 5040\) | B1 | 1.1b |
| (1) |
Notes
B1: Correct value. Accept as 7! for this part but must be evaluated in the remaining parts.
| Scheme | Marks | AO |
|---|---|---|
| \(4! \times 4! = 576\) | M1 A1 | 3.1b 1.1b |
| (2) |
Notes
M1: Evidence that the 4 students have been considered as one unit among many and sees the problem as permutations of 4 items. Score for \(4! \times k\) where \(k \ne 1\)
A1: Correct value
| Scheme | Marks | AO |
|---|---|---|
| \(5! \times 2! = 240\) | M1 A1 | 3.1b 1.1b |
| (2) |
Notes
M1: Realises that the other 5 students can sit in any position – evidenced by sight of 5! (in (c))
A1: Correct value
| Scheme | Marks | AO |
|---|---|---|
| \(7! - 6! \times 2! = 3600\) or \(5! \times (2 \times 5 + 2 \times 4 + 2 \times 4 + 4) = 3600\) | M1 A1 | 3.1b 1.1b |
| (2) | ||
| (12 marks) |
Notes
M1: A correct strategy applied. E.g. interprets the situation as the answer to part (ii)(a) minus the ways that they can sit together So score for \(7! - \ldots\) or \(5040 - \ldots\) where \(\ldots\) is not zero.
Alternatively, considers the different positions Devindra can sit for each different position for Charles, and with 5! positions for the rest. Look for 5!\(\times\)(sum of 7 terms) oe.
A1: Correct value