Competition problem — Number theory

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Number theory Difficulty 5.4 AIME, harder Find the answer

27. Find the least positive residues of
a) 6! modulo 7
b) 10 ! modulo 11
c) 12! modulo 13
d) 16! modulo 17 .
e) Can you propose a theorem from the above congruences?

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solutions — 2

Solution 1

27. a) -1
b) -1
c) -1
d) -1
e) $(p-1)!\equiv-1(\bmod p)$ when $p$ is prime

Solution 2

27. a) -1
b) -1
c) -1
d) -1
e) $(p-1)!\equiv-1(\mod p)$ when $p$ is prime

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.