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Solution

Each digit 6 → 9, increase = 3. Count how many times digit 6 appears from 1 to 100. Two-digit numbers (10–99): Tens place = 6 → numbers 60–69 → 10 times Units place = 6 → appears once in each decade (16, 26, ..., 96) → 9 times Total digit 6 occurrences = $10 + 9 = 19$ Additionally, number 6 itself contributes 1 more occurrence. So total 6s = $19 + 1 = 20$ Increase in sum = $20 \times 3 = 60$ But this does not match options → Check again carefully. Correct counting: Tens place occurrences: 60–69 → 10 Units place occurrences: 6, 16, 26, 36, 46, 56, 66, 76, 86, 96 → 10 Total = 20 occurrences. Increase = $20 \times 3 = 60$ But total sum change involves positional value: A “6” in tens place increases number by 30 A “6” in ones place increases by 3 Count tens-place 6s = 10 → total = $10 \times 30 = 300$ Count ones-place 6s = 10 → total = $10 \times 3 = 30$ Total change = $300 + 30 = 330$