Write 57128 as a product of primes.

Prove that \(\left. {22} \right|{5^{11}} + {17^{11}}\).

\(457\,128 = 2 \times 228\,564\)

\(228\,564 = 2 \times 114\,282\)

\(114\,282 = 2 \times 57\,141\)

\(57\,141 = 3 \times 19\,047\)

\(19\,047 = 3 \times 6349\)

\(6349 = 7 \times 907\)     M1A1

trial division by 11, 13, 17, 19, 23 and 29 shows that 907 is prime     R1

therefore \(457\,128 = {2^3} \times {3^2} \times 7 \times 907\)     A1

[4 marks]

by a corollary to Fermat’s Last Theorem

\({5^{11}} \equiv 5(\bmod 11){\text{ and }}{17^{11}} \equiv 17(\bmod 11)\)     M1A1

\({5^{11}} + {17^{11}} \equiv 5 + 17 \equiv 0(\bmod 11)\)     A1

this combined with the evenness of LHS implies \(\left. {25} \right|{5^{11}} + {17^{11}}\)     R1AG

[4 marks]

Some candidates were obviously not sure what was meant by ‘product of primes’ which surprised the examiner.

There were some reasonable attempts at part (c) using powers rather than Fermat’s little theorem.