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 "Now I know.")
🧩 STEP THREE
After Bull's statement, what possibilities remain available to Donkey?
Only pairs whose sum is on the list of ten: 145 pairs out of 4,753.
Donkey takes his product and examines every factorization, keeping only those whose sum is on the list.
For Donkey's product, 52:
52 = 2×26 → sum 28 (even, not on the list) ✗
52 = 4×13 → sum 17 ✓
🧩 STEP FOUR
Donkey says "Now I know." Why?
Exactly one factorization of his product survives Bull's statement.
For 52, only 4×13 is left. Donkey knows.
(Across all 145 pairs, only 86 have a product where exactly one factorization survives. Those are the only pairs where Donkey could say "Now I know.")
🧩 STEP FIVE
Bull says "Now I know them too." Why can Bull identify the exact pair from his sum?
Bull checks each split of his sum and asks, "would Donkey now be down to ONE?"
He can only name the pair if exactly ONE split passes.
For sum 17:
2+15 → 30: 2×15 (17), 5×6 (11) → two left ✗
3+14 → 42: 3×14 (17), 2×21 (23) → two left ✗
4+13 → 52: 4×13 (17) only ✅
5+12 → 60: 5×12 (17), 3×20 (23) → two left ✗
6+11 → 66: 6×11 (17), 2×33 (35) → two left ✗
7+10 → 70: 7×10 (17), 2×35 (37) → two left ✗
8+9 → 72: 8×9 (17), 3×24 (27) → two left ✗
Exactly one. Bull knows: 4 and 13.
Every other sum fails, because each has at least two splits where Donkey could have known:
11 → 2+9, 3+8, 4+7 · 23 → 4+19, 7+16 · 27 → 2+25, 4+23 · 29 → 2+27, 4+25
35 → 3+32, 4+31 · 37 → 5+32, 6+31 · 41 → 3+38, 4+37 · 47 → 4+43, 6+41 · 53 → 5+48, 6+47
(e.g. 76 = 2×38, sum 40, dead, or 4×19, sum 23, alive → Donkey would have known)
Only 17 leaves Bull with exactly one. That is the final proof.
⚠️ THE GOLDEN RULE — THE COMPLETE CHAIN
🫏 What Donkey knows: the product, 52, which splits two ways (2×26, 4×13), so he can't know at first.
🐂 What Bull knows: the sum, 17, which splits seven ways, every one with an ambiguous product.
🫏 What Donkey knows about Bull: Bull's certainty means his sum is one of the ten, which kills 2×26 (sum 28) and leaves 4×13.
🐂 What Bull knows about Donkey: of the seven splits of 17, only 4+13 would let Donkey become certain, so Donkey's "Now I know" points to exactly one pair.
🚫 Why there is nowhere else to go: every other surviving sum leaves Bull with two or more candidates, so 17 is the only sum where the whole conversation can happen, and 4+13 is the only pair within it.
🔓 THE NUMBERS ARE 4 AND 13.
Donkey never said them. He didn't have to. 🫏🤪