# Dividing 10^30 by 97

## Setting Up the Problem

I need to find q and r such that 10^30 = 97·q + r, where 0 ≤ r < 97.

## Finding the Remainder

I'll compute 10^30 mod 97 using successive squaring.

First, by Fermat's Little Theorem, 10^96 ≡ 1 (mod 97).

Let me compute powers of 10 mod 97:
- 10^1 ≡ 10
- 10^2 ≡ 100 ≡ 3 (mod 97)
- 10^4 ≡ 3^2 = 9
- 10^8 ≡ 9^2 = 81
- 10^16 ≡ 81^2 = 6561 = 67·97 + 62, so ≡ 62

Now 10^30 = 10^16 · 10^8 · 10^4 · 10^2:
- 10^16 · 10^8 ≡ 62 · 81 = 5022 = 51·97 + 75, so ≡ 75
- 75 · 10^4 = 75 · 9 = 675 = 6·97 + 93, so ≡ 93
- 93 · 10^2 = 93 · 3 = 279 = 2·97 + 85, so ≡ 85

**Remainder = 85**

## Finding the Quotient

q = (10^30 - 85) / 97

I'll perform the long division of 10^30 by 97:

Working through the long division systematically (dividing digit groups by 97):

10^30 = 97 × q + 85

To find q, I'll use the fact that 1/97 has a known 96-digit repeating decimal, but let me compute directly.

10^30 ÷ 97:

Starting the long division of 1000000000000000000000000000000 by 97:

I'll track this iteratively (bringing down digits from 10^30):

The quotient is: **10309278350515463917525773195**

Let me verify: this should be (10^30 - 85)/97.

**Verification:** 10309278350515463917525773195 × 97:
- 10309278350515463917525773195 × 100 = 1030927835051546391752577319500
- 10309278350515463917525773195 × 3 = 30927835051546391752577319585
- Subtracting: 1030927835051546391752577319500 - 30927835051546391752577319585 = 999999999999999999999999999915

Hmm, let me recheck: 10309278350515463917525773195 × 97 = ...× (100-3)
= 1030927835051546391752577319500 - 30927835051546391752577319585
= 1000000000000000000000000000000 - 85 (checking the subtraction)
= 999999999999999999999999999915 ✓

Adding the remainder: 999999999999999999999999999915 + 85 = 10^30 ✓

## Result

$$10^{30} = 97 \times 10\,309\,278\,350\,515\,463\,917\,525\,773\,195 + 85$$

- **Quotient:** 10,309,278,350,515,463,917,525,773,195
- **Remainder:** 85
