Consider an integer $a$ with $(n + 1)$ digits written as $a = {10^n}{a_n} + {10^{n - 1}}{a_{n - 1}} +  \ldots  + 10{a_1} + {a_0}$, where $0 \leqslant {a_i} \leqslant 9$ for $0 \leqslant i \leqslant n$, and ${a_n} \ne 0$.

(a)     Show that $a \equiv ({a_n} + {a_{n - 1}} +  \ldots  + {a_0}) ({\text{mod9}})$.

(b)     Hence find all pairs of values of the single digits $x$ and $y$ such that the number $a = 476x212y$ is a multiple of $45$.

(c)     (i)     If $b = 34\,390$ in base 10, show that $b$ is $52\,151$ written in base 9.

(ii)     Hence find ${b^2}$ in base 9, by performing all the calculations without changing base.

(a)     $10 \equiv 1(\bmod 9) \Rightarrow {10^i} \equiv 1({\text{mod9}}),{\text{ }}i = 1,{\text{ }} ... ,{\text{ }}n$     M1A1

$\Rightarrow {10^i}{a_i} \equiv {a_i}({\text{mod9}}),{\text{ }}i = 1,{\text{ }}n$     M1

Note: Allow $i = 0$ but do not penalize its omission.

$\Rightarrow ({10^n}{a_n} + {10^{n - 1}}{a_{n - 1}} +  \ldots  + {a_0}) \equiv ({a_n} + {a_{n - 1}} +  \ldots  + {a_0})({\text{mod9}})$     AG

[3 marks]

(b)     $4 + 7 + 6 + x + 2 + 1 + 2 + y = 9k,{\text{ }}k \in \mathbb{Z}$     (M1)

$\Rightarrow (22 + x + y) \equiv 0(\bmod 9),{\text{ }} \Rightarrow (x + y) \equiv 5({\text{mod9}})$

$\Rightarrow x + y = 5$ or $14$     A1

if $5$ divides $a$, then $y = 0$ or $5$     M1

so $y = 0 \Rightarrow x = 5,{\text{ }}\left( {ie{\text{ }}(x,{\text{ }}y) = (5,{\text{ }}0)} \right)$     A1

$y = 5 \Rightarrow x = 0$ or $x = 9,{\text{ }}\left( {ie{\text{ }}(x,{\text{ }}y) = (0,{\text{ }}5){\text{ or }}(x,{\text{ }}y) = (9,{\text{ }}5)} \right)$     A1A1

[6 marks]

(c)     (i)    
     (M1)A1

$b = {(52\,151)_9}$     AG

(ii)    
     M1A3

Note: M1 for attempt, A1 for two correct lines of multiplication, A2 for two correct lines of multiplication and a correct addition, A3 for all correct.

[6 marks]

Surprisingly few good answers. Part (a) had a number of correct solutions, but there were also many that seemed to be a memorised solution, not properly expressed – and consequently wrong. In part (b) many failed to understand the question, not registering that x and y were digits rather than numbers. Part (c)(i) was generally well answered, although there were a number of longer methods applied, and few managed to do (c)(ii).