(i)     By converting the base 7 number to base 10, find the value of $x$, in base 10.

(ii)     Express $x$ as a base 5 number.

Let $y$ be the base 9 number ${({a_n}{a_{n - 1}} \ldots {a_1}{a_0})_9}$. Show that $y$ is exactly divisible by 8 if and only if the sum of its digits, $\sum\limits_{i = 0}^n {{a_i}}$, is also exactly divisible by 8.

Using the method from part (b), find the unit digit when the base 9 number ${(321321321)_9}$ is written as a base 8 number.

(i)     converting to base 10

${(503231)_7} = 5 \times {7^5} + 3 \times {7^3} + 2 \times {7^2} + 3 \times 7 + 1 = 85184$    M1A1A1

so $x = 44$     A1

(ii)     repeated division by 5 gives     (M1)

so base 5 value for $x$ is ${(134)_5}$     A1

Notes: Alternative method is to successively subtract the largest multiple of 25 and then 5.

Follow through if they forget to take the cube root and obtain ${(10211214)_5}$ then award (M1)(A1)A1.

[7 marks]

$9 \equiv 1(\bmod 8)$    A1

${9^i} \equiv {1^i} \equiv 1(\bmod 8)$    $i \in \mathbb{N}$     (M1)(A1)

$y = {a_n}{9^n} + {a_{n - 1}}{9^{n - 1}} +  \ldots  + {a_1}9 + {a_0} \equiv {a_n}{1^n} + {a_{n - 1}}{1^{n - 1}} +  \ldots  + {a_1}1 + {a_0} \equiv$

${a_n} + {a_{n - 1}} +  \ldots  + {a_1} + {a_0} \equiv \sum\limits_{i = 0}^n {{a_i}(\bmod 8)}$    M1A1A1

so $y = 0(\bmod 8)$ and hence divisible by 8 if and only if $\sum\limits_{i = 0}^n {{a_i} \equiv 0(\bmod 8)}$ and hence divisible by 8     R1AG

Note: Accept alternative valid methods eg binomial expansion of ${(8 + 1)^i}$, factorization of $({a^i} - 1)$ if they have sufficient explanation.

[7 marks]

using part (b), ${(321321321)_9} \equiv 3 + 2 + 1 + 3 + 2 + 1 + 3 + 2 + 1 = 18 \equiv 2(\bmod 8)$     M1A1

so the unit digit is 2     A1

[3 marks]