Future Factorials

We generalize it to find the remainder for $(p-1)! - (p-2)!$ when it's divided by prime $p$

By Wilson's Theorem, $(p-1)! \pmod p \equiv -1$

$\Rightarrow (p-1) (p-2)! \pmod p \equiv -1$

$\Rightarrow (p-2)! \pmod p \equiv - \left ( \frac {1}{p-1} \right ) \equiv 1$ , because $1(p) - 1(p-1) = 1$

Thus, $(p-1)! - (p-2)! \pmod p \equiv -2 \equiv p-2$ .

In this case, $p = 2017 \Rightarrow p - 2 = \boxed{2015}$

First note that for a prime $p$ , Wilson's theorem states that: $(p-1)!\equiv p-1\equiv -1\pmod p$ . As a direct consequence (by dividing by $p-1$ ), we can see that $(p-2)!\equiv 1\pmod p$ . As $2017$ is a prime, we have: $2016!-2015!=(2017-1)!-(2017-2)!\equiv -1-1=-2\equiv \fbox{2015}\pmod{2017}$

Since $2016 \equiv -1 \mod 2017$ . So, $2016(-1) \equiv 1 \mod 2017$ . It implies that $-1$ is the inverse of 2016 modulo 2017. So, $2015! \equiv 2016! \times 2016^{-1} \equiv 2016! \times (-1) \equiv -2016! \equiv 1 \mod 2017$ . So, $2016! - 2015! \equiv -1-1 \equiv -2 \equiv 2015 \mod 2017$ .

$2016! - 2015! = 2015*2015!$

By Wilson's Theorem, $2016! \equiv 2016 \pmod {2017}$

So, $2015! \equiv 1 \pmod {2017}$

$\implies (2015!)2015 \equiv \boxed {2015} \pmod {2017}$

So this solution relies more on seeing a pattern:

The first few factorials are as follows: 3! - 2! = 4 4! - 3! = 18 5! - 4! = 96 6! - 5! = 600 7! - 6! = 4320

The next thing to do is to divide the equations by the next factor and see what the remainders are, for example:

(3! - 2!)/4 = 1 with no remainder. However (4! - 3!)/5 has a remainder of 3, which is the b in (a! - b!)/c. Notice that 5 is a prime. (5! - 4!)/6 = 16 with no remainder. (6! - 5!)/7 has a remainder of 5, again the b and 7 being a prime.

Thus, I figured that if the equation is (a! - b!)/c and c is a prime, then the remainder should be b.

(2016! - 2015!)/2017 has a remainder of 2015.

since 2017 is prime, it has no divisor less than 2017. 2016! -2015! = (2016 - 1) * 2015! = 2015 * 2015! since 2015! is the product of all positive integers less than or equal to 2015, the remainder is the sum of all divisors times 2015 = -2 (mod 2017) the sum is $\frac{(2015+1)(2015)}{2}$ , and when multiplied by -2, the result is $\frac{2016}{2}*(-2)^2=2016*2=-1*2=-2=2015$

2015 or -2. From wilson 2016!=-1mod2017. Hence 2016 2015!=-1mod2017 => (2017-1) 2015!=-1mod2017 => 2015!=1mod2017. Overall the remainder is -2