Find the values of \(f(3)\), \(f(4)\) and \(f(5)\).

Use the Euclidean algorithm to find

(i)     \(\gcd \left( {f(3),{\text{ }}f(4)} \right)\);

(ii)     \(\gcd \left( {f(4),{\text{ }}f(5)} \right)\).

Explain why \(f(n)\) is always exactly divisible by \(5\).

By factorizing \(f(n)\) explain why it is always exactly divisible by \(6\).

Determine the values of \(n\) for which \(f(n)\) is exactly divisible by \(60\).

\(240,{\rm{ }}1020,{\rm{ }}3120\)     A2

Note:     Award A2 for three correct answers, A1 for two correct answers.

[2 marks]

(i)     \(1020 = 240 \times 4 + 60\)     (M1)

\(240 = 60 \times 4\)

\(\gcd (1020,{\text{ }}240) = 60\)     A1

(ii)     \(3120 = 1020 \times 3 + 60\)     (M1)

\(1020 = 60 \times 17\)

\(\gcd (1020,{\text{ }}3120) = 60\)     A1

Note:     Must be done by Euclid’s algorithm.

[4 marks]

by Fermat’s little theorem with \(p = 5\)     R1

\({n^5} \equiv n(\bmod 5)\)

so 5 divides \(f(n)\)

[1 mark]

\(f(n) = n({n^2} - 1)({n^2} + 1) = n(n - 1)(n + 1)({n^2} + 1)\)     (A1)A1

\(n - 1,{\text{ }}n,{\text{ }}n + 1\) are consecutive integers and so contain a multiple of \(2\) and \(3\)     R1R1

Note:     Award R1 for justification of \(2\) and R1 for justification of \(3\).

And therefore \(f(n)\) is a multiple of \(6\).     AG

[4 marks]

from (c) and (d) \(f(n)\) is always divisible by \(30\)     R1

considering the factorization, it is divisible by \(60\) when \(n\) is an odd number     A1

or when \(n\) is a multiple of \(4\)     A1

Note:     Accept answers such as when \(n\) is not congruent to \(2(\bmod 4)\).

[3 marks]

Total [14 marks]

Well answered.

Also well answered. A few candidates did not use the Euclidean algorithm to find the gcd as instructed.

Many candidates essential said it was true because it was! There is only one mark which means one minute, so it must be a short answer which it is by using Fermat’s Little Theorem.

Some good answers but too many did not factorize as instructed, so that they could then spot the consecutive numbers.

Only the better candidates realised that they had to find another factor of \(2\).