Let $N = 10^{30}$. We want to divide $N$ by 97 and find the quotient and remainder.

We can write $10^2 \equiv 100 \equiv 3 \pmod{97}$.
Then $10^3 \equiv 10 \cdot 10^2 \equiv 10 \cdot 3 \equiv 30 \pmod{97}$.
$10^4 \equiv (10^2)^2 \equiv 3^2 \equiv 9 \pmod{97}$.
$10^5 \equiv 10^4 \cdot 10 \equiv 9 \cdot 10 \equiv 90 \equiv -7 \pmod{97}$.
$10^6 \equiv (10^3)^2 \equiv 30^2 \equiv 900 \equiv 900 - 9 \cdot 97 = 900 - 873 = 27 \pmod{97}$.
Also, we have $\phi(97) = 96$ since 97 is prime. By Euler's theorem, we have $10^{96} \equiv 1 \pmod{97}$.

We have $10^{30} = (10^3)^{10} = (10^2)^{15}$.
We know $10^2 \equiv 3 \pmod{97}$, so $10^{30} = (10^2)^{15} \equiv 3^{15} \pmod{97}$.
We also know that $3^4 = 81 \equiv -16 \pmod{97}$.
$3^8 \equiv (-16)^2 \equiv 256 \equiv 256 - 2 \cdot 97 = 256 - 194 = 62 \equiv -35 \pmod{97}$.
$3^{12} \equiv (3^4)^3 \equiv (-16)^3 \equiv -16 \cdot 256 \equiv -16 \cdot 62 \equiv -992 \equiv -992 + 10 \cdot 97 \equiv -992 + 970 = -22 \pmod{97}$.
Then $3^{15} = 3^{12} \cdot 3^3 \equiv (-22) \cdot 27 \equiv -594 \pmod{97}$.
We have $-594 + 6 \cdot 97 = -594 + 582 = -12 \pmod{97}$.
Since $-12 + 97 = 85$, we have $10^{30} \equiv 3^{15} \equiv 85 \pmod{97}$.
Therefore, $10^{30} = 97q + 85$ for some integer $q$.
The remainder is 85.
We can write $10^{30} = 10^2 \cdot 10^2 \cdot 10^2 \dots 10^2 \cdot 10^2 \cdot 10^2$, where there are 15 factors of $10^2$.
So we have $10^{30} = (10^2)^{15} \equiv 3^{15} \pmod{97}$.
$3^1 = 3$, $3^2 = 9$, $3^3 = 27$, $3^4 = 81 \equiv -16 \pmod{97}$.
$3^5 = 243 = 2 \cdot 97 + 49$, so $3^5 \equiv 49 \pmod{97}$.
$3^6 = 729 = 7 \cdot 97 + 50$, so $3^6 \equiv 50 \pmod{97}$.
$3^{10} = (3^5)^2 \equiv 49^2 \equiv 2401 = 24 \cdot 97 + 73 \equiv 73 \pmod{97}$.
$3^{15} = (3^5)^3 \equiv 49^3 \pmod{97}$.
We have $49^2 = 2401 \equiv 73 \pmod{97}$.
So $49^3 \equiv 49 \cdot 73 = 3577 = 36 \cdot 97 + 85$.
Hence $49^3 \equiv 85 \pmod{97}$.
The remainder is 85.

Final Answer: The final answer is $\boxed{85}$
