(a)     Using Fermat’s little theorem, show that, in base 10, the last digit of n is always equal to the last digit of ${n^5}$ .

(b)     Show that this result is also true in base 30.

(a)     using Fermat’s little theorem ${n^5} \equiv n(\bmod 5)$     (M1)

${n^5} - n \equiv 0(\bmod 5)$     A1

now ${n^5} - n = n({n^4} - 1)$     (M1)

$= n({n^2} - 1)({n^2} + 1)$

$= n(n - 1)(n + 1)({n^2} + 1)$     A1

hence one of the first two factors must be even     R1

i.e. ${n^5} - n \equiv 0(\bmod 2)$

thus ${n^5} - n$ is divisible by 5 and 2

hence it is divisible by 10     R1

in base 10, since ${n^5} - n$ is divisible by 10, then ${n^5} - n$ must end in zero and hence ${n^5}$ and n must end with the same digit     R1

[7 marks]

(b)     consider ${n^5} - n = n(n - 1)(n + 1)({n^2} + 1)$

this is divisible by 3 since the first three factors are consecutive integers     R1

hence ${n^5} - n$ is divisible by 3, 5 and 2 and therefore divisible by 30 

in base 30, since ${n^5} - n$ is divisible by 30, then ${n^5} - n$ must end in zero and hence ${n^5}$ and n must end with the same digit     R1

[2 marks]

Total [9 marks]

There were very few fully correct answers. If Fermat‟s little theorem was known, it was not well applied.