A2 October 2020 Q8
8. The four digit number \(n = abcd\) satisfies the following properties:
(1) \(n \equiv 3\ (\mathrm{mod}\ 7)\)
(2) \(n\) is divisible by 9
(3) the first two digits have the same sum as the last two digits
(4) the digit \(b\) is smaller than any other digit
(5) the digit \(c\) is even
| Scheme | Marks | AO |
|---|---|---|
| The integer \(n\) can be written as \(n = 1000a + 100b + 10c + d\) | M1 | 1.1b |
| As \(1000 = 142 \times 7 + 6\), \(100 = 14 \times 7 + 2\) and \(10 = 7 + 3\), reducing coefficients modulo 7 gives \(n \equiv 6a + 2b + 3c + d\ (\mathrm{mod}\ 7)\) * | A1* | 2.4 |
| (2) |
Notes
Allow credit for any part for relevant work seen throughout.
M1: Writes \(n = abcd\) as sum of multiples of the digits, as shown in scheme.
A1*: Explains how each coefficient reduces modulo 7 to give the stated answer.
| Scheme | Marks | AO |
|---|---|---|
| \(n\) divisible by 9 means \(a + b + c + d = 9k\) for some integer \(k\) | B1 | 1.1b |
| \(a + b = c + d \Rightarrow 2(a + b) = 9k\), hence \(k\) even | M1 | 3.1a |
| But \(a + b + c + d\) must be at least 3 and at most 35 (as \(b\) smaller than all other numbers). | B1 | 2.1 |
| So since \(k\) must be even, only possibility is \(k = 2\) to keep \(a + b + c + d\) in range 0-36, hence \(a + b = 9\) * | A1* | 2.2a |
| (4) |
Notes
B1: Applies the divisibility test for 9, either as in scheme or stating \(a + b + c + d\) is a multiple of 9 (or similar). \(a + b + c + d \equiv 0\ (\mathrm{mod}\ 9)\) is fine.
M1: Uses the third fact in combination with the second to eliminate \(c\) and \(d\) and deduce \(k\) even. Alternatively, \(a + b + c + d \equiv 2(a + b) \equiv 0\ (\mathrm{mod}\ 9) \Rightarrow 10(a + b) \equiv 5 \times 0\ (\mathrm{mod}\ 9) \Rightarrow a + b \equiv 0\ (\mathrm{mod}\ 9)\)
B1: Eliminates the possibilities that \(k = 0\) or 4 using the fourth fact. Alt: reasons as \(b \lt a\) then \(a + b\) cannot be 0 or 18.
A1*: Completes the proof, with all steps explained. Allow if the previous B has not been earned.
| Scheme | Marks | AO |
|---|---|---|
| Combining (a) and (b) gives \(3 \equiv 2(a + b) + 4a + (c + d) + 2c \equiv 4a + 2c + 27\ (\mathrm{mod}\ 7)\) | M1 | 3.1a |
| \(\Rightarrow 2c \equiv -4a - 24 \equiv -4(a + 6) \equiv 3(a - 1)\ (\mathrm{mod}\ 7)\) \(\Rightarrow 8c \equiv 12(a - 1)\ (\mathrm{mod}\ 7) \Rightarrow c \equiv 5(a - 1)\ (\mathrm{mod}\ 7)\) * | A1* | 2.1 |
| (2) |
Notes
M1: Combines the results of (a) and (b) to eliminate \(b\) and \(d\) from the result in (a).
A1*: Correct completion to the given statement. There will be different ways to achieve this, one example is shown. Check work carefully!
| Scheme | Marks | AO |
|---|---|---|
| As \(b \lt a\) and \(a + b = 9\) we must have \(a \geqslant 5\) | M1 | 3.1a |
| If \(a = 9\) then \(c \equiv 5\ (\mathrm{mod}\ 7) \Rightarrow c = 5\), contradicting \(c\) is even. If \(a = 8\) then \(c \equiv 0\ (\mathrm{mod}\ 7) \Rightarrow c = 0\) or \(7\) but 7 not even, and \(c \gt b\) so can’t be zero. If \(a = 7\) then \(c \equiv 2\ (\mathrm{mod}\ 7) \Rightarrow c = 2\), but also \(b = 2\) for \(a = 7\), and \(b \lt c\) so not possible. If \(a = 6\) then \(c \equiv 4\ (\mathrm{mod}\ 7) \Rightarrow c = 4, d = 5\) and \(b = 3\) which works. If \(a = 5\) then \(c \equiv 6\ (\mathrm{mod}\ 7) \Rightarrow c = 6\), \(d = 3\) and \(b = 4\) but then \(d \lt b\), not allowed. | dM1 A1 | 1.1b 1.1b |
| Hence \(n = 6345\) (only) | B1 | 2.2a |
| (4) | ||
| (12 marks) |
Notes
M1: There may be many approaches here. Scored for beginning a process of eliminating possibilities for one of the digits, e.g. deducing \(a\) is at least 5 as shown. (May be done by individually eliminating the lower values.)
dM1: Continues the process eliminating at least two more cases.
A1: Complete argument, leading to one solution only (and no possibility of others). Need not be expressed as eloquently as above, but should be clear all cases have been considered.
B1: For \(n = 6345\) obtained in any way.
(corrected from the printed mark scheme: the first line is printed as \(\text{“}a\,..\,5\text{”}\), a garbled symbol for \(a \geqslant 5\))