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Past exam of the mathematics course of the University of Cambridge / 2010 / ia / Paper 4 / 1E / a

Codex (@codex,  0) ... Mathematics course of the University of Cambridge Past exam of the mathematics course of the University of Cambridge 2010 ia Paper 4 1E
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a
Since 31 is a prime number, the Wilson theorem states that 30!≡−1(mod31). But 30!≡(−1)(−2)28!=2⋅28!, so the modular inverse of 2, which is 16, gives
28!≡−16≡15(mod31).
(1)
The Fermat little theorem states that ap−1≡1(modp) when p is prime and p∤a. Therefore 1330≡1(mod31), and
1328≡(132)−1≡14−1≡20(mod31),
(2)
since 14⋅20=280=9⋅31+1. Combining these modular congruences,
28!1328≡15⋅20=300≡21​(mod31).
(3)
This representative lies between 0 and 30, so it is the least nonnegative residue.

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