Past exam of the mathematics course of the University of Cambridge / 2014 / ia / Paper 4 / 6E / iv

Table of contents
- Solution iv

Solution

Fermat's little theorem reduces the exponent modulo

22

because

23

is a prime number and

5

is coprime to it. Compute

5^{2} \equiv 3 (mod 22)

, then

5^{4} \equiv 9 (mod 22)

and

5^{5} \equiv 45 \equiv 1 (mod 22)

. Thus

5^{5} = 22 k + 1

for an integer

k

, and

5^{5^{5}} = 5^{22 k + 1} \equiv (5^{22})^{k} 5 \equiv 5 (mod 23) .

The exponent is

5^{5} = 3125

, rather than an iterated left-associated power.

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