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Question: How many 6-digit positive integers have all digits either 3 or 7, and contain at least one pair of consecutive 3s?
Solution: First, total number of 6-digit numbers with digits only 3 or 7: each digit has 2 choices, so $ 2^6 = 64 $. But 6-digit numbers cannot start with 0, but since digits are only 3 or 7, all 64 are valid.

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