Use something to find the remainder

By Prime factorizing $301$ we get $301=7\times 43$

Therefore, we can apply Chinese remainder theorem since, $\gcd(7,43)=1$ and Fermat's little theorem since, $7$ and $43$ are PRIME numbers.

So, by these theorems we get:

$123^{42}\equiv 1\pmod{43}$ and $123^{6}\equiv 1\pmod{7}$

Now, $123^{6}\equiv 1\pmod{7}$

$\implies (123^6)^7=123^{42}\equiv 1^7\equiv 1\pmod {7}$

Let $123^{42} =x$ then $x\equiv 1\pmod{43}$ and $x\equiv 1\pmod {7}$

$\therefore x=43m+1=7n+1$

Thus, by “Chinese remainder theorem” we have $7n+1\equiv 1\pmod {43}$

$\implies n\equiv 0\pmod {43}$

$\therefore n=43k$ [where $k$ is any integer]

Thus, we get $x=7(43k)+1=301k+1$

$\therefore 123^{42}\equiv 1\pmod {301}$

By multiplying both sides by $123^3$ we get:

$123^{42}\times 123^3 = 123^{45}\equiv 123^3\pmod {301}$

$\implies 123^{45}\equiv 1860867\equiv 85\pmod {301}$

Hence, $123^{45}\equiv \boxed{85}\pmod{301}$