I'm missing 6

$45^{23} \equiv 3^{23} = (3^3)^7 \cdot 3^2 \equiv (-1)^7 \cdot 2 = -2 \equiv \boxed{5} \pmod{7}$

$45 \equiv 3 \pmod{7}$

$45^2 \equiv 2 \pmod{7}$

$45^4 \equiv 4 \pmod{7}$

$45^5 \equiv 5 \pmod{7}$ from 1 and 3

$45^6 \equiv 1 \pmod{7}$ from 1 and 4

$45^{18} \equiv 1 \pmod{7}$

$45^{23} \equiv 5 \pmod{7}$ from 4 and 6

hence the answer is $\boxed{5}$

$3\cdot 45^{23}\equiv 3\cdot 3^{23}\stackrel{\text{Fermat}}\equiv 3^{24\!\pmod{\!6}}\equiv 3^0$

\equiv 1\equiv -6\stackrel{:3}\iff 45^{23}\equiv -2\equiv 5 mod $7$ .

Or written in another way

$45^{23}\equiv 3^{23}\equiv 3^{23\!\pmod{\!6}}\equiv 3^{-1}$

$\equiv \frac{1}{3}\equiv \frac{-6}{3}\equiv -2\equiv 5$ mod $7$ .

$\begin{aligned} 45^{23} & \equiv (42+3)^{23} \pmod{7} \equiv 3^{23} \pmod{7} \equiv 9^{11}\dot{}3 \pmod{7} \\ & \equiv 3\dot{}2^{11} \pmod{7} \equiv 3\dot{}2^2\dot{}8^3 \pmod{7} \equiv 3\dot{}2^2\dot{}1 \pmod{7} \\ & \equiv 12 \pmod{7} \equiv \boxed{5} \pmod{7} \end{aligned}$

45^6=1 (mod 7),¨fermat. 45 = 3 (mod 7) ,
( 45^ 6 ) 3 * 45 ^ 5 = a mod 7. (1 ) ^ 3 * (3) ^ 5 = a mod 7..... 243 = 5 mod 7 a = 5 the remainder