## Question 8:

### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| The integer $n$ can be written as $n = 1000a + 100b + 10c + d$ | M1 | Writes $n = abcd$ as sum of multiples of digits |
| As $1000 = 142\times7+6$, $100=14\times7+2$ and $10=7+3$, reducing coefficients modulo 7 gives $n \equiv 6a+2b+3c+d \pmod{7}$ * | A1* | Explains how each coefficient reduces modulo 7 to give the stated answer |

### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $n$ divisible by 9 means $a+b+c+d = 9k$ for some integer $k$ | B1 | Applies the divisibility test for 9 |
| $a+b = c+d \Rightarrow 2(a+b) = 9k$, hence $k$ even | M1 | Uses third fact with second to eliminate $c$ and $d$ and deduce $k$ even |
| But $a+b+c+d$ must be at least 3 and at most 35 (as $b$ smaller than all other numbers) | B1 | Eliminates possibilities $k=0$ or $4$ |
| 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* | Completes the proof with all steps explained |

### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Combining (a) and (b) gives $3 \equiv 2(a+b)+4a+(c+d)+2c \equiv 4a+2c+27 \pmod{7}$ | M1 | Combines results of (a) and (b) to eliminate $b$ and $d$ |
| $\Rightarrow 2c \equiv -4a-24 \equiv -4(a+6) \equiv 3(a-1) \pmod{7}$ $\Rightarrow 8c \equiv 12(a-1)\pmod{7} \Rightarrow c \equiv 5(a-1)\pmod{7}$ * | A1* | Correct completion to given statement. Different approaches possible |

### Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| As $b < a$ and $a+b=9$ we must have $a \geq 5$ | M1 | Begins process of eliminating possibilities for one of the digits |
| If $a=9$: $c \equiv 5\pmod{7} \Rightarrow c=5$, contradicting $c$ even. If $a=8$: $c \equiv 0\pmod{7} \Rightarrow c=0$ or $7$, but 7 not even and $c>b$ so can't be zero. If $a=7$: $c \equiv 2\pmod{7} \Rightarrow c=2$, but $b=2$ for $a=7$ and $b<c$ so not possible. If $a=6$: $c \equiv 4\pmod{7} \Rightarrow c=4$, $d=5$ and $b=3$ which works. If $a=5$: $c \equiv 6\pmod{7} \Rightarrow c=6$, $d=3$ and $b=4$ but then $d<b$, not allowed. | dM1, A1 | Continues eliminating at least two more cases; complete argument leading to one solution only |
| Hence $n = 6345$ (only) | B1 | For $n=6345$ obtained in any way |