Determine $(n− 1)! \pmod {n}$ for each of the following $n$:
(a) $n = 15$
(b) $n = 21$
(c) $n = 30$
What do you notice about your results?
(a) Since $n=15$ is not prime, we can't immediately use Wilson's Theorem.
We note that
$ 14! \equiv 0 \pmod 3$ because 3 is a factor of 14!
$ 14! \equiv 0 \pmod 5$ because 5 is a factor of 14!
Using the result from Exercise (3.1).24(d) we have
$$ 14! \equiv 0 \pmod {15} $$
(b) Again, since $n=21$ is not prime, we can't use Wilson's Theorem.
We note that
$ 20! \equiv 0 \pmod 3$ because 3 is a factor of 20!
$ 20! \equiv 0 \pmod 7$ because 7 is a factor of 20!
Using the result from Exercise (3.1).24(d) we have
$$ 20! \equiv 0 \pmod {21} $$
(b) Again, since $n=30$ is not prime, we can't use Wilson's Theorem.
We note that
$ 29! \equiv 0 \pmod 2$ because 2 is a factor of 29!
$ 29! \equiv 0 \pmod 3$ because 3 is a factor of 29!
$ 29! \equiv 0 \pmod 7$ because 7 is a factor of 29!
Using the result from Exercise (3.1).24(d) we have
$$ 29! \equiv 0 \pmod {30}$$
We notice that the expression $(n− 1)! \pmod {n}$ is congruent to $0$. This is because the selected $n$ are not prime, and the prime factors of $n$ are factors of $(n-1)!$. This will always be the case because those factors are always less than or equal to $(n-1)$.