lotterypuzzle atlas

Bitcoin Puzzle record

Bitcoin Puzzle #140

open · public key on chain

14 BTC nominal2139 candidate keys

Where it sits on the line

Puzzles #128–#152: search work One column per puzzle; height is the base-2 logarithm of the search steps. Puzzle #140 is highlighted. The line sits at 2^69. #128 · 2^127 #129 · 2^128 #130 · 2^65.5 #131 · 2^130 #132 · 2^131 #133 · 2^132 #134 · 2^133 #135 · 2^68 #136 · 2^135 #137 · 2^136 #138 · 2^137 #139 · 2^138 #140 · 2^70.5 #141 · 2^140 #142 · 2^141 #143 · 2^142 #144 · 2^143 #145 · 2^73 #146 · 2^145 #147 · 2^146 #148 · 2^147 #149 · 2^148 #150 · 2^75.5 #151 · 2^150 #152 · 2^151
  • found
  • open · address only
  • open · public key on chain
  • the line · 269

This puzzle needs about 270.5 search steps — above the line (269).

No puzzle above the line has been found yet.

Model: kangaroo. The public key has been on chain since 2019, so the square-root shortcut applies: about 2 × 2(N−1)/2 group operations.

How the work is estimated

Record

P2PKH address
1QKBaU6WAeycb3DbKbLBkX7vJiaS8r42Xo
mempool.space Blockstream
Known public key
031f6a332d3c5c4f2de2378c012f429cd109ba07d69690c6c701b6bb87860d6640
HASH160
ffbb35a7bb9bbe16c1aa2534f7ff11d59c8e3d1a
Search class
public-key
A public key is on record, so the square-root shortcut (kangaroo) works inside the interval. A key on record is not a spend.
Candidate interval
2139 → 2140 − 1
start0x80000000000000000000000000000000000
end0xfffffffffffffffffffffffffffffffffff
Interval size696898287454081973172991196020261297061888 scalars. Exact BigInt bounds, not a floating estimate.

Snapshot 2 Oct 2026. The reward is the catalog's nominal allocation (N × 0.1 convention from 2023-04-16; earlier spends predate it), not a live balance. A key that matches the address proves the math, not that the coins are still there: check an explorer. Missing fields are omitted, not invented.

Mentioned in the atlas

Where #140 comes up in the room pages. rooms 12 · mentions 35

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Method

Kangaroo searches only the window you choose. Being inside the window (containment) and being found by the solver are two separate odds, and finishing a bounded run stays probabilistic.