What a large number you have

Note that $709!\equiv (-10)(-11)\cdots(-718) \pmod {719}$ Since there is an odd number of terms, this is also equivalent to $709!\equiv -\dfrac{718!}{9!}\pmod {719}$ By Wilson's Theorem (iff $p$ is prime, $(p-1)!\equiv -1\pmod p$ ), $709!\equiv \dfrac{1}{9!}\pmod {719}$ Notice that $6!\equiv 720\equiv 1\pmod {719}$ , and so we calculate as follows: $\dfrac{1}{9!}\equiv\dfrac{1}{6!}\cdot\dfrac{1}{7\cdot 8\cdot 9}\equiv\dfrac{1}{7\cdot 8\cdot 9}\pmod{719}$ Now, notice that $8\cdot 9\cdot 10\equiv 1\pmod {719}$ , and so $\dfrac{1}{7\cdot 8\cdot 9}\equiv \dfrac{10}{7}\pmod {719}$ Finally, $7\cdot 103\equiv 2\pmod {719}$ , and we find our answer: $\dfrac{10}{7}\equiv\dfrac{1030}{2}\equiv\boxed{515}\pmod {719}$

Since $719$ is prime, we can apply Wilson's Theorem to get that $718! \equiv -1$ $mod$ $719$ . This can be written as

$709! \cdot 710 \cdot 711 \cdots 718 \equiv 709! \cdot -9 \cdot -8 \cdots -1 \equiv 709! \cdot -9! \equiv -1$ $mod$ $719$ .

This give us $709! \cdot 72 \cdot 7 \cdot 720 \equiv 1$ $mod$ $719$ .

We now reduce the expression $mod$ $719$ :

$709! \cdot 72 \cdot 7 \cdot 720 \equiv 1+719 \equiv 720$

$709! \cdot 72 \cdot 7 \equiv 1+719 \equiv 720$

$709! \cdot 7 \equiv 10+ 5 \cdot 719 \equiv 3605$

$709! \equiv 515$

Thus the remainder when $709!$ is divided by $709$ is $515$

We are given that $719$ is prime, hence by Wilson's theorem $718! \equiv -1 \pmod{719}$ Therefore $\begin{aligned} 709! \cdot 710 \cdot 711 \cdots 718 &\equiv 709! \cdot (-9)(-8)(-7) \cdots (-1) \pmod{719} \\ &\equiv -9! \cdot 709! \pmod{719} \\ &\equiv 215 \cdot 709! \pmod{719} \end{aligned}$ Hence $215 \cdot 709! \equiv -1 \pmod{719}$ This is linear congruence generally solvable by Euclidean algorithm. Thus using Euclidean algorithm we get $709! \equiv 515 \pmod{719}$

Let p=719. By wilsons theorem, 718! == 1 mod p == 709! * (-9)(-8)(-7)...(-1) == 709! (504) where == denotes equivalence. We can solve for the residue of 709! trivially from here.

(p-1)!=-1 mod p by Wilson's theorem. Hence 709!x710x711...718=718 mod 719. To make the arithmetic simpler equivalently we have: 709!x(-9)(-8)...(-3)(-2)=1 mod 719 or 709!x(-1)^8x9x8x7x6!=1 mod 719 but note that 719 = 6!-1 hence 709!x9x8x7x(719+1)=1 mod 719 i.e. 709!x9x8x7=1 mod 719. Hence we just need to find the multiplicative inverse of 9x8x7 = 504 in the group mod 719. This can be done by Euclid's Algorithm since hcf(504,719)=1 means there exist a,b in Z such that 504xa+719xb=1, in which case a = 504^(-1), hence 709! = a mod 719. This yields the answer given.

The question can be translated into: "What is 709! mod 719?"

As 719 is prime, use Wilson's Theorem, and we get 718! mod 719 = 718. We then work backwards.

717! mod 719 = a, where a is an integer such that 718a mod 719 = 718. We solve the equation and get a = 1. Work backwards similarly, until we get the desired answer, that is: 709! mod 719 = 515.

We calculate the remainder by using a simple loop in javascript:

var n=1;
for(var i=2 ; i <= 709 ; i++)
    n = n * i % 719;

which will multiply all 709 number and get the remainder for each turn. Because 709 is rather small, we can afford this.

By Wilson's Theorem, since 719 is prime we have 718! = -1 (mod 719). Now, 718! = 709! * 710 * 711 * ... * 718. Since we're working mod 719, we can rewrite 710 as -9, 711 as -8, and so on up to 718 as -1. Therefore, 709! * (-9!) = -1 (mod 719). Hence 709! * (9!) = 1 (mod 719). Now observe that 6! = 720 = 1 (mod 719), so we can cancel a 6! out of the 9 factorial. We get 709! * (7 * 8 * 9) = 1 (mod 719), so 709! = (504)^(-1) (mod 719). To find the inverse of 504 we simply use the Bezout algorithm, and get 515.