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1000
Skibidi Sequences — Editorial
combinatoricsprobabilitysequences
There are 33 choices for a1a_1 and 22 choices for each subsequent term, so
N=3⋅27=384.N = 3 \cdot 2^7 = 384.
The die has N/48=8N/48 = 8 faces, labeled 11 through 8.8. Rolling twice gives 82=648^2 = 64 equally likely outcomes. We want the product divisible by 6,6, meaning the two results must contribute at least one factor of 22 and one factor of 3.3. By inclusion-exclusion: let AA = both rolls odd, BB = neither roll divisible by 3.3. Then
∣A∣=42=16,∣B∣=62=36,∣A∩B∣=32=9.|A| = 4^2 = 16, \quad |B| = 6^2 = 36, \quad |A \cap B| = 3^2 = 9.
Favorable outcomes: 64−16−36+9=21.64 - 16 - 36 + 9 = 21. The probability is 2164,\tfrac{21}{64}, so p+q=21+64=85.p + q = 21 + 64 = 85.