To divide \(10^{30}\) by 97 and find the quotient and remainder, we can use long division or modular arithmetic. Here, I will use modular arithmetic to find the remainder and calculate the quotient from that:

We want to find the remainder of \(10^{30} \mod 97\). 

### Step 1: Use Fermat's Little Theorem

Fermat's Little Theorem states that if \(p\) is a prime number and \(a\) is an integer not divisible by \(p\), then:

\[ a^{p-1} \equiv 1 \pmod{p} \]

In this case, let \(a = 10\) and \(p = 97\). Applying the theorem:

\[ 10^{96} \equiv 1 \pmod{97} \]

### Step 2: Simplify using Fermat's Theorem

Since \(10^{96} \equiv 1 \pmod{97}\), consider \(10^{30}\):

We want to find \(10^{30} \mod 97\). To simplify \(10^{30}\), let's explore powers of 10 mod 97.

1. \( 10^1 \equiv 10 \pmod{97} \)
2. \( 10^2 \equiv 100 \equiv 3 \pmod{97} \)
3. \( 10^3 \equiv 30 \pmod{97} \)
4. And so forth.

Keep reducing:

Continue until you find a pattern or can simplify \(10^{30}\).

Using modular properties and calculations, we simplify powers iteratively:

### Step 3: Powers of 10 Reduction

By iterating further, find powers close to having repeated patterns or less than moderately:
\[ 10^{96} \equiv 1 \Rightarrow (10^{30})^3 = 10^{90} \]
Hence, \(10^{30}\) is part of a repeating cycle or uses symmetry:
To simplify:

- \( 10^{30} = (10^{10})^3 \equiv (1)^3 = 1 \pmod{97}\) because of Fermat’s theorem.

### Conclusion:

Remainder: Since \(10^{30} \equiv 1 \pmod{97}\), the remainder when dividing \(10^{30}\) by 97 is 1.

Finally, calculate the quotient:
\[ \text{Quotient} = \frac{10^{30} - 1}{97} \]

Note - Calculation of actual huge power results can be complex, approximation suggests using reduced manageable exponents or efficient computational tools for practical contextual use of direct result verification.

Result: Quotient is a large integer, remainder is 1.
