Treechat ·
!quoted by C2C_NFTmarket
Bsvcrypto ·
🤔
🔓 SOLUTION: 4 and 13 (Sum 17 · Product 52)
Last round I noted the speech bubbles in #01 needed swapping for the chain to resolve. This time Donkey speaks first, the product-holder, so the conversation resolves exactly as written. Same logic, higher level. 🫏
Start: every pair of distinct integers from 2 to 99 → 2–3, 2–4, 2–5 … 98–99. That's 4,753 possible pairs.
🧩 STEP ONE
Which products could Donkey immediately identify? Remove them.
Any product that splits into two valid factors in only ONE way. Donkey would know instantly, so he could never say "I don't know."
Examples: 6 = 2×3 only · 8 = 2×4 only · 22 = 2×11 only · 77 = 7×11 only
Every product of two different primes dies here, as does any product containing a prime over 49 (like 53×4 = 212), since doubling that prime is already over 99.
🧩 STEP TWO
Which SUM could allow Bull to say "I knew you didn't know"?
Bull was CERTAIN from his sum alone, before Donkey spoke.
Certain means NO exceptions: if even ONE split of his sum had a uniquely-factoring product, Donkey might have known, and Bull would only be "probably" certain.
So EVERY split of Bull's sum must be ambiguous. That kills:
• Every even sum up to 54 → it splits into two different primes (8 = 3+5, 28 = 5+23, 54 = 7+47…). Sum 6 dies too: 2+4 → 8.
• Every odd sum where (sum − 2) is prime → 2×prime splits one way only (13 = 2+11 → 22).
• 51 → 17+34, and 578 = 17×34 only (2×289 is out of range).
• Every sum from 55 to 196 except 194 → pair it with a prime from 53 to 97 (57 = 53+4 → 212 = 53×4 only; for 106 use 59+47).
• 194 → 95+99 → 9405 = 95×99 only.
• 197 → 98+99 → 9702 = 98×99 only.
Only ten sums survive: 11, 17, 23, 27, 29, 35, 37, 41, 47, 53
Check for 17, every split ambiguous:
30 = 2×15 = 3×10 · 42 = 3×14 = 2×21 · 52 = 4×13 = 2×26 · 60 = 5×12 = 3×20
66 = 6×11 = 2×33 · 70 = 7×10 = 2×35 · 72 = 8×9 = 3×24
("Certain" is the only reading that works: if Bull merely meant "you just told me," his statement adds nothing, and Donkey could never go on to say "N…