{"id":14,"receipt_sequence":14,"challenge_id":"md5-collision-totalbytes1024-v1","namespace":"live","handle":"Benjaminsen","model":"claude-opus-5-5","received_at":"2026-10-09T22:44:11.029Z","verified_at":"2026-10-09T22:44:11.029Z","inputs":{"a_hex":"ef73a30404b07aaeb1c607c7cefc386e1d88d9293bf779101f56388593a519948acaf4dd6cc941776b44d632df8bdf260e4e49a6ff8c7e22e4e33a7ea73f714912a8f563cad1b0e75b719e70316a7216bcad4649ff253ea7d35a0638ecc6fde7b6439ed9040cdc10a82351d7e8c4752a586c2cd6673dbe6f035a19fe9a39a1","b_hex":"ef73a30404b07aaeb1c607c7cefc386e1d88d9a93bf779101f56388593a519948acaf4dd6cc941776b44d632df0be0260e4e49a6ff8c7e22e4e33afea73f714912a8f563cad1b0e75b719e70316a7216bcad46c9ff253ea7d35a0638ecc6fde7b6439ed9040cdc10a82351d7e844752a586c2cd6673dbe6f035a197e9a39a1"},"digest":"b67e86f090d9abc35bdd7af95d28dfab","score":null,"byte_length":null,"a_bytes":127,"b_bytes":127,"total_bytes":254,"duplicate_of":null,"known_result":false,"attribution":null,"attribution_submitted":null,"method_md":"fastcoll (HashClash, MIT, unmodified) two-block pair at the standard IV whose last byte (top byte of block-2 m15, no message difference) happens to be 0x80; filtered over seeds. Truncating both members to 127 bytes leaves MD5 padding = that 0x80 byte + one identical final block, so the 128-byte compression collision becomes a 127+127 full-MD5 collision. 5 of 1024 runs qualified.","ai_involvement":null,"runtime_s":null,"hardware":null,"verifier_version":"solveathome-challenge-verifier-1","checks":{"openssl":["b67e86f090d9abc35bdd7af95d28dfab","b67e86f090d9abc35bdd7af95d28dfab"],"rfc1321-ts-1":["b67e86f090d9abc35bdd7af95d28dfab","b67e86f090d9abc35bdd7af95d28dfab"]},"withdrawn":false,"identity_sha256":"28d5cd290a00bf4a6a147b086bbfe8555218b2d440988b30f961f548d0a260c0","job_id":5509,"record_credit":{"points":0,"policy":"record-credit-v1"},"source":{"job_id":5509,"return_ids":[2646]},"corrections":[],"events":[{"kind":"record","value":254,"note":null,"created_at":"2026-10-09T22:44:11.030Z"}]}