The positive integer N is expressed in base 9 as \({({a_n}{a_{n - 1}} \ldots {a_0})_9}\).

(a)     Show that N is divisible by 3 if the least significant digit, \({a_0}\), is divisible by 3.

(b)     Show that N is divisible by 2 if the sum of its digits is even.

(c)     Without using a conversion to base 10, determine whether or not \({(464860583)_9}\) is divisible by 12.

(a)     let \(N = {a_n}{a_{n - 1}} \ldots {a_1}{a_0} = {a_n} \times {9^n} + {a_{n - 1}} \times {9^{n - 1}} + \ldots  + {a_1} \times 9 + {a_0}\)     M1A1

all terms except the last are divisible by 3 and so therefore is their sum     R1

it follows that N is divisible by 3 if \({a_0}\) is divisible by 3     AG

[3 marks]

(b)     EITHER

consider N in the form

\(N = {a_n} \times ({9^n} - 1) + {a_{n - 1}} \times ({9^{n - 1}} - 1) + \ldots  + {a_1}(9 - 1) + \sum\limits_{i = 1}^n {{a_i}} \)     M1A1

all terms except the last are even so therefore is their sum     R1

it follows that N is even if \(\sum\limits_{i = 0}^n {{a_i}} \) is even     AG

OR

working modulo 2, \({9^k} \equiv 1(\bmod 2)\)     M1A1

hence \(N = {a_n}{a_{n - 1}} \ldots {a_1}{a_0} = {a_n} \times {9^n} + {a_{n - 1}} \times {9^{n - 1}} + \ldots  + {a_1} \times 9 + {a_0} = \sum\limits_{i = 0}^n {{a_i}(\bmod 2)} \)     R1

it follows that N is even if \(\sum\limits_{i = 0}^n {{a_i}} \) is even     AG

[3 marks]

(c)     the number is divisible by 3 because the least significant digit is 3     R1

it is divisible by 2 because the sum of the digits is 44 which is even     R1

dividing the number by 2 gives \({(232430286)_9}\)     M1A1

which is even because the sum of the digits is 30 which is even     R1

N is therefore divisible by a further 2 and is therefore divisible by 12     R1

Note: Accept alternative valid solutions.

[6 marks]

Total [12 marks]

Parts (a) and (b) were generally well answered. Part (c), however, caused problems for many candidates with some candidates even believing that showing divisibility by 2 and 3 was sufficient to prove divisibility by 12. Some candidates stated that the fact that the sum of the digits was 44 (which itself is divisible by 4) indicated divisibility by 4 but this was only accepted if the candidates extended their proof in (b) to cover divisibility by 4.