Don't Use Your Calculator!

Note that $4^3 = 64 = 9\cdot 7 + 1,$ so $4^3$ leaves a remainder of 1 when divided by 9. Using Modular Arithmetic - Exponentiation , $4^18 = (4^3)^6$ also leaves a remainder of 1 when divided by 9.

Note that the remainder upon division by 9 is equivalent to the remainder when the sum of the digits of the number is divided by 9. (Does anyone want to write a quick proof of this?) The sum of the digits is $6+8+7+1+9+4+\color{#D61F06}{X}+6+7+3+6 = 57 +\color{#D61F06}{X}.$

Thus, for the remainder to be 1, we must have $\color{#D61F06}{X}=7$ as $57 + 7= 9\cdot 7+1.$