(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.