Using mathematical induction or otherwise, prove that the number \({(1020)_n}\) , that is the number \(1020\) written with base \(n\) , is divisible by \(3\) for all values of \(n\) greater than \(2\).

Explain briefly why the case \(n = 2\) has to be excluded.

\({(1020)_n} = {n^3} + 2n\)     (R1)

so we are required to prove that \({n^3} + 2n\) is divisible by \(3\) for \(n \ge 3\)     (R1)

EITHER

when \(n = 3\) , \({n^3} + 2n = 33\) which is divisible by \(3\) so the result is true for \(n = 3\)     A1

assume the result is true for \(n = k\) , i.e. \({k^3} + 2k\) is divisible by \(3\)     M1

for \(n = k + 1\) ,

\({(k + 1)^3} + 2(k + 1) = {k^3} + 3{k^2} + 3k + 1 + 2k + 2\)     M1

\( = ({k^3} + 2k) + 3({k^2} + k + 1)\)     A1

the second term is clearly divisible by \(3\) and the first term is divisible by \(3\) by hypothesis     A1

therefore true for \(n = k \Rightarrow \)  true for \(n = k + 1\) and since shown true for \(n = 3\) ,

the result is proved by induction     R1

Note: Award the final R1 only if the two M1 marks have been awarded.

OR

there are three cases to consider, let \(N\) be a positive integer

case 1: \(n = 3N\) , in this case

\(n({n^2} + 2) = 3N(9{N^2} + 2)\) which is divisible by \(3\)     M1A1

case 2: \(n = 3N + 1\) , in this case,

\(n({n^2} + 2) = (3N + 1)(9{N^2} + 6N + 3)\) which is divisible by \(3\)     M1A1

case 3: \(n = 3N + 2\) , in this case,

\(n({n^2} + 2) = (3N + 2)(9{N^2} + 12N + 6)\) which is divisible by \(3\)     M1A1

this proves the required result for all \(n > 2\)

[8 marks]

numbers to base \(2\) do not use the digit \(2\) or equivalent     R1

[1 mark]

A significant number of candidates struggled with this question, but a number of wholly correct answers were seen. The majority of candidates tried to use a method of proof by induction although a number got lost in trying to establish the result for \(k + 1\) . Again the presentation and explanation of some of the solutions was poor. Some excellent solutions were seen using other methods including one which noted simply that all positive integers \(n\) are either 0 or \( \pm 1\) modulo \(3\) so that \(n({n^2} + |2) \equiv 0\) modulo \(3\) for all \(n\).

A significant number of candidates struggled with this question, but a number of wholly correct answers were seen. The majority of candidates tried to use a method of proof by induction although a number got lost in trying to establish the result for \(k + 1\) . Again the presentation and explanation of some of the solutions was poor. Some excellent solutions were seen using other methods including one which noted simply that all positive integers \(n\) are either 0 or \( \pm 1\) modulo \(3\) so that \(n({n^2} + |2) \equiv 0\) modulo \(3\) for all \(n\).