A2 June 2025 Q5
5.
- one capital letter
- 3 distinct single digits
- one of the four symbols @ £ $ &
Determine how many different passwords can be created if
| Scheme | Marks | AO |
|---|---|---|
| Any two out of 26 or \(10 \times 9 \times 8\) or 4 seen multiplied together | M1 | 1.1b |
| \(26 \times 10 \times 9 \times 8 \times 4 = 74880\) | A1 | 1.1b |
| (2) |
Notes
M1: Any two out of 26 or \(10 \times 9 \times 8\) or 4 seen multiplied together
A1: Correct answer
| Scheme | Marks | AO |
|---|---|---|
| \(\text{their } \text{‘}74880\text{’} \times \dfrac{5!}{3!} = 1497600\) | B1ft | 1.1b |
| (1) |
Notes
B1ft: For the correct answer or follow through their answer to (a) multiplied by 20. Note they may start again and use \(26 \times {}^{10}C_3 \times 4 \times 5!\)
| Scheme | Marks | AO |
|---|---|---|
| \(4^{12} \equiv 1 \pmod{13}\) or \(4^{13} \equiv 4 \pmod{13}\) | B1 | 1.2 |
| \(\left(4^{12}\right)^4 \times 4^2 \equiv 1 \times 4^2 \pmod{13}\) or \(\left(4^{13}\right)^3 \times 4^{11} \equiv 4^3 \times 4^{11} \equiv 4^{14} \equiv 4^{13} \times 4^1 \equiv \ldots \pmod{13}\) | M1 | 2.1 |
| \(4^{50} \equiv 3 \pmod{13}\) | A1 | 2.2a |
| (3) |
Notes
B1: Recalls Fermat’s Little Theorem correctly. May be implicit in the working rather than stated separately. Accept if stated in terms of \(a\) and \(p\).
M1: A complete attempt to write \(4^{50}\) as powers of \(4^{12}\) or \(4^{13}\) and completes the process to find the residue.
A1: Deduces the correct residue.
| Scheme | Marks | AO |
|---|---|---|
| The number of different 1-digit odd numbers that do not contain the digit 2 is 5 | B1 | 2.2a |
| The number of different 2-digit odd numbers that do not contain the digit 2 is \(8 \times 5 \{= 40\}\) The number of different 3-digit odd numbers that do not contain the digit 2 is \(8 \times 9 \times 5 \{= 360\}\) | M1 | 1.1b |
| There are 500 odd numbers therefore 500 – (5 + 40 + 360) | dM1 | 2.1 |
| = 95 | A1 | 1.1b |
| (4) | ||
| (10 marks) |
Notes
B1: Deduces that there are 5 different 1-digit odd numbers that do not contain the digit 2. Variation: if they consider 000 to 999 allow B1 for correct option of 9 for first digit. (corrected from the printed mark scheme: the printed note says 3 different 1-digit odd numbers, but there are 5 (1, 3, 5, 7, 9), as in the scheme)
M1: Finds the number of different 2-digit and 3-digit odd numbers that do not contain the digit 2. (In the variation they would do \(9 \times 9 \times 5\) possibilities without 2, giving 405 directly.)
dM1: For 500 – the sum of the number of 1, 2, 3 digits that don’t contain the digit 2. Must have been full method to find all the digits without 2 first.
A1: Correct answer
Alternative 1
| Scheme | Marks | AO |
|---|---|---|
| Identifies that 2 can only appear in the first and/or second position | B1 | 2.2a |
| First digit is 2, second digit 10 possible, third digit 5 possible \(= 10 \times 5 = \ldots\{50\}\) First digit not 2, 9 options (including 0), second digit is 2, third digit 5 possible \(= 9 \times 5 = \ldots\{45\}\) | M1 | 1.1b |
| Therefore total number of digits is “50” + “45” = … | dM1 | 2.1 |
| = 95 | A1 | 1.1b |
| (4) |
B1: Deduces that there are only two positions for 2, first and second. (May be implied)
M1: Considers first digit is 2 and the possible digits for second and third position and considers second digit is 2 and the possible digits for first and third position. Allow if they have double counted some digits for this mark.
dM1: Full method considering the possibilities for 2 in various positions and adds the results. They must subtract any repeats for this mark.
A1: Correct answer
Alternative 2
| Scheme | Marks | AO |
|---|---|---|
| 1 digit number = 0 ways | B1 | 2.2a |
| 2 digit number = 1 x 5 =5 ways 3 digit number starting with 2 =1 x 10 x 5 = 50 ways Other 3 digit number with middle number 2 = 8 x 1 x 5 = 40 ways | M1 | 1.1b |
| Total number = 0 + 5 + 50 + 40 | dM1 | 2.1 |
| = 95 | A1 | 1.1b |
| (4) |
B1: Deduces that there are no single digit numbers
M1: Considers the possible 2 and 3 digit numbers. Allow if they have double counted some digits for this mark or omit some cases as long as at least some cases are considered.
dM1: Adds up all the possible digits following consideration of all possible combinations. They must subtract any repeats for this mark. For 3 digit numbers they may have \(1 \times 10 \times 5\) starting 2, \(9 \times 1 \times 5\) with 2 in middle, but need to note 5 of these are repeated (22x), and so get \(5 + 50 + 45 - 5 = \ldots\)
A1: Correct answer.
Other Alternatives are possible. E.g.
B1: Considers 3 digit numbers from 000 to 999. (May be implicit)
M1: Starting with 2 : 50 possibilities (must be odd), middle 2 but not starting 2 : \(9 \times 1 \times 5\) possibilities.
dM1: So total number of possibilities is 50 + 45 = …
A1: Correct answer.