(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]