{
  "schema_version": "ocme-reviewed-question-reader-v1",
  "question_id": "q-20261013-advance-20261018-r03-gcd-coprime-0092",
  "pilot_number": "R0016-08992",
  "number": 8992,
  "family_id": "qf-ext18-gcd-associative-triple",
  "title": {
    "zh-Hant": "第 8992 題：最大公因數與互質",
    "en": "Question 8992: Both Parenthesizations of a Three-Input GCD"
  },
  "original_question": {
    "schema_version": "ocme-question-v0.1",
    "id": "q-20261013-advance-20261018-r03-gcd-coprime-0092",
    "batch_id": "batch-20261013-advance-20261018-r03-gcd-coprime-foundations",
    "language": "zh-Hant",
    "source_mko_ids": [
      "mko-gcd-coprime"
    ],
    "title_zh": "最大公因數與互質",
    "stem_zh": "自然數gcd。操作=gcd-associative-triple；資料={\"a\":28,\"b\":50,\"c\":78}。請依 [a,b,c,left_gcd,right_gcd,equal] 的欄序回答。",
    "answer": {
      "kind": "expression",
      "value": "(28, 50, 78, 2, 2, 是)",
      "explanation_zh": "擴展基礎核對=(28, 50, 78, 2, 2, 是)；依完整明示資料逐欄精確重算；有限／合成資料核對不是一般形式證明；gcd(0,0)=0；互質不表示兩數皆質數。"
    },
    "difficulty": {
      "band": "upper_secondary",
      "level": 2,
      "skills": [
        "最大公因數與互質"
      ]
    },
    "variant": {
      "family_id": "qf-ext18-gcd-associative-triple",
      "seed": 202610183092,
      "parameters": {
        "a": 28,
        "b": 50,
        "c": 78,
        "domain": "Nat_gcd",
        "operation": "gcd-associative-triple",
        "semantic_template": "qf-ext18-gcd-associative-triple/{\"a\":28,\"b\":50,\"c\":78}"
      }
    },
    "verification": {
      "status": "candidate",
      "checks": [
        "producer_exact_finite_arithmetic",
        "producer_source_snapshot_check",
        "producer_substantive_template_deduplication",
        "producer_json_schema"
      ],
      "warning_zh": "僅為產生端候選及自檢；主端須另做驗收，不構成人工審定、普遍證明、Lean執行或形式Evidence。"
    },
    "provenance": {
      "generator_role": "question_producer",
      "generator_model": "model_unknown",
      "generated_at": "2026-10-13",
      "contract_version": "ocme-question-factory-v0.1"
    }
  },
  "original_english_edition": {
    "schema_version": "ocme-question-localization-v0.1",
    "question_id": "q-20261013-advance-20261018-r03-gcd-coprime-0092",
    "batch_id": "batch-20261013-advance-20261018-r03-gcd-coprime-foundations",
    "locale": "en",
    "translation_state": "ai_candidate",
    "source": {
      "question_sha256": "ad0a508984c7c0dfd70f2757b5cf8e9df5d47f954ab110469bda022a5a3d0fe4",
      "family_id": "qf-ext18-gcd-associative-triple",
      "seed": 202610183092,
      "semantic_template": "qf-ext18-gcd-associative-triple/{\"a\":28,\"b\":50,\"c\":78}"
    },
    "fields": {
      "title": "Both Parenthesizations of a Three-Input GCD",
      "stem": "Natural inputs a=28, b=50, c=78, with gcd(0,0)=0. Give (a, b, c, gcd(gcd(a,b),c), gcd(a,gcd(b,c)), are the results equal?).",
      "answer": {
        "kind": "expression",
        "value": "(28, 50, 78, 2, 2, Yes)",
        "explanation": "Compute both complete gcd routes rather than merely repeating an asserted associativity law. Exact finite/synthetic checks are not new general formal Evidence. The exact result, retaining the requested order, is (28, 50, 78, 2, 2, Yes)."
      },
      "skills": [
        "Both Parenthesizations of a Three-Input GCD"
      ],
      "verification_warning": "This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."
    }
  },
  "givens": [
    {
      "id": "G001",
      "text": {
        "zh-Hant": "自然數gcd。操作=gcd-associative-triple；資料={\"a\":28,\"b\":50,\"c\":78}。請依 [a,b,c,left_gcd,right_gcd,equal] 的欄序回答。",
        "en": "Natural inputs a=28, b=50, c=78, with gcd(0,0)=0. Give (a, b, c, gcd(gcd(a,b),c), gcd(a,gcd(b,c)), are the results equal?)."
      }
    }
  ],
  "steps": [
    {
      "id": "S001",
      "uses": [
        "G001"
      ],
      "before": "natural triple=[28, 50, 78];gcd(0,0)=0",
      "after": "compute both complete parenthesizations",
      "relation": "implies",
      "rule_id": "gcd_two_routes_scope",
      "conditions": [],
      "explanation": {
        "zh-Hant": "此題要求兩條實際Euclid路線，不只重複結合律結論；輸入可含0。",
        "en": "Compute both actual Euclidean routes rather than repeating the associativity conclusion.Inputs may include0."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S002",
      "uses": [
        "G001",
        "S001"
      ],
      "before": "left inner gcd(a,b) Euclid row0:divide natural 28 by positive 50",
      "after": "28=0·50+28",
      "relation": "implies",
      "rule_id": "natural_euclidean_division_witness",
      "conditions": [],
      "explanation": {
        "zh-Hant": "正除數給唯一商餘，餘數滿足0≤r<divisor；這個完整等式同時核商與餘數，不作除以0。",
        "en": "A positive divisor gives a unique quotient/remainder with0≤r<divisor.The full equality audits both values without division by zero."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": {
        "operation": "euclidean_division",
        "inputs": [
          "28",
          "50",
          "0"
        ],
        "output": "28",
        "label": "left inner gcd(a,b) Euclid row0"
      }
    },
    {
      "id": "S003",
      "uses": [
        "G001",
        "S002"
      ],
      "before": "phase=left inner gcd(a,b);pair=[28, 50];quotient=0;remainder=28",
      "after": "next pair=[50, 28];0<=remainder<divisor",
      "relation": "implies",
      "rule_id": "gcd_common_divisor_invariance",
      "conditions": [],
      "explanation": {
        "zh-Hant": "x=q*y+r表示共同因數同時整除x,y當且僅當同時整除y,r；正除數下餘數嚴格變小，因此迴圈會到第二項0。",
        "en": "From x=q*y+r,a number divides both x,y exactly when it divides both y,r.The remainder is strictly smaller than the positive divisor,so the second argument eventually becomes0."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S004",
      "uses": [
        "G001",
        "S003"
      ],
      "before": "left inner gcd(a,b) Euclid row1:divide natural 50 by positive 28",
      "after": "50=1·28+22",
      "relation": "implies",
      "rule_id": "natural_euclidean_division_witness",
      "conditions": [],
      "explanation": {
        "zh-Hant": "正除數給唯一商餘，餘數滿足0≤r<divisor；這個完整等式同時核商與餘數，不作除以0。",
        "en": "A positive divisor gives a unique quotient/remainder with0≤r<divisor.The full equality audits both values without division by zero."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": {
        "operation": "euclidean_division",
        "inputs": [
          "50",
          "28",
          "1"
        ],
        "output": "22",
        "label": "left inner gcd(a,b) Euclid row1"
      }
    },
    {
      "id": "S005",
      "uses": [
        "G001",
        "S004"
      ],
      "before": "phase=left inner gcd(a,b);pair=[50, 28];quotient=1;remainder=22",
      "after": "next pair=[28, 22];0<=remainder<divisor",
      "relation": "implies",
      "rule_id": "gcd_common_divisor_invariance",
      "conditions": [],
      "explanation": {
        "zh-Hant": "x=q*y+r表示共同因數同時整除x,y當且僅當同時整除y,r；正除數下餘數嚴格變小，因此迴圈會到第二項0。",
        "en": "From x=q*y+r,a number divides both x,y exactly when it divides both y,r.The remainder is strictly smaller than the positive divisor,so the second argument eventually becomes0."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S006",
      "uses": [
        "G001",
        "S005"
      ],
      "before": "left inner gcd(a,b) Euclid row2:divide natural 28 by positive 22",
      "after": "28=1·22+6",
      "relation": "implies",
      "rule_id": "natural_euclidean_division_witness",
      "conditions": [],
      "explanation": {
        "zh-Hant": "正除數給唯一商餘，餘數滿足0≤r<divisor；這個完整等式同時核商與餘數，不作除以0。",
        "en": "A positive divisor gives a unique quotient/remainder with0≤r<divisor.The full equality audits both values without division by zero."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": {
        "operation": "euclidean_division",
        "inputs": [
          "28",
          "22",
          "1"
        ],
        "output": "6",
        "label": "left inner gcd(a,b) Euclid row2"
      }
    },
    {
      "id": "S007",
      "uses": [
        "G001",
        "S006"
      ],
      "before": "phase=left inner gcd(a,b);pair=[28, 22];quotient=1;remainder=6",
      "after": "next pair=[22, 6];0<=remainder<divisor",
      "relation": "implies",
      "rule_id": "gcd_common_divisor_invariance",
      "conditions": [],
      "explanation": {
        "zh-Hant": "x=q*y+r表示共同因數同時整除x,y當且僅當同時整除y,r；正除數下餘數嚴格變小，因此迴圈會到第二項0。",
        "en": "From x=q*y+r,a number divides both x,y exactly when it divides both y,r.The remainder is strictly smaller than the positive divisor,so the second argument eventually becomes0."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S008",
      "uses": [
        "G001",
        "S007"
      ],
      "before": "left inner gcd(a,b) Euclid row3:divide natural 22 by positive 6",
      "after": "22=3·6+4",
      "relation": "implies",
      "rule_id": "natural_euclidean_division_witness",
      "conditions": [],
      "explanation": {
        "zh-Hant": "正除數給唯一商餘，餘數滿足0≤r<divisor；這個完整等式同時核商與餘數，不作除以0。",
        "en": "A positive divisor gives a unique quotient/remainder with0≤r<divisor.The full equality audits both values without division by zero."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": {
        "operation": "euclidean_division",
        "inputs": [
          "22",
          "6",
          "3"
        ],
        "output": "4",
        "label": "left inner gcd(a,b) Euclid row3"
      }
    },
    {
      "id": "S009",
      "uses": [
        "G001",
        "S008"
      ],
      "before": "phase=left inner gcd(a,b);pair=[22, 6];quotient=3;remainder=4",
      "after": "next pair=[6, 4];0<=remainder<divisor",
      "relation": "implies",
      "rule_id": "gcd_common_divisor_invariance",
      "conditions": [],
      "explanation": {
        "zh-Hant": "x=q*y+r表示共同因數同時整除x,y當且僅當同時整除y,r；正除數下餘數嚴格變小，因此迴圈會到第二項0。",
        "en": "From x=q*y+r,a number divides both x,y exactly when it divides both y,r.The remainder is strictly smaller than the positive divisor,so the second argument eventually becomes0."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S010",
      "uses": [
        "G001",
        "S009"
      ],
      "before": "left inner gcd(a,b) Euclid row4:divide natural 6 by positive 4",
      "after": "6=1·4+2",
      "relation": "implies",
      "rule_id": "natural_euclidean_division_witness",
      "conditions": [],
      "explanation": {
        "zh-Hant": "正除數給唯一商餘，餘數滿足0≤r<divisor；這個完整等式同時核商與餘數，不作除以0。",
        "en": "A positive divisor gives a unique quotient/remainder with0≤r<divisor.The full equality audits both values without division by zero."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": {
        "operation": "euclidean_division",
        "inputs": [
          "6",
          "4",
          "1"
        ],
        "output": "2",
        "label": "left inner gcd(a,b) Euclid row4"
      }
    },
    {
      "id": "S011",
      "uses": [
        "G001",
        "S010"
      ],
      "before": "phase=left inner gcd(a,b);pair=[6, 4];quotient=1;remainder=2",
      "after": "next pair=[4, 2];0<=remainder<divisor",
      "relation": "implies",
      "rule_id": "gcd_common_divisor_invariance",
      "conditions": [],
      "explanation": {
        "zh-Hant": "x=q*y+r表示共同因數同時整除x,y當且僅當同時整除y,r；正除數下餘數嚴格變小，因此迴圈會到第二項0。",
        "en": "From x=q*y+r,a number divides both x,y exactly when it divides both y,r.The remainder is strictly smaller than the positive divisor,so the second argument eventually becomes0."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S012",
      "uses": [
        "G001",
        "S011"
      ],
      "before": "left inner gcd(a,b) Euclid row5:divide natural 4 by positive 2",
      "after": "4=2·2+0",
      "relation": "implies",
      "rule_id": "natural_euclidean_division_witness",
      "conditions": [],
      "explanation": {
        "zh-Hant": "正除數給唯一商餘，餘數滿足0≤r<divisor；這個完整等式同時核商與餘數，不作除以0。",
        "en": "A positive divisor gives a unique quotient/remainder with0≤r<divisor.The full equality audits both values without division by zero."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": {
        "operation": "euclidean_division",
        "inputs": [
          "4",
          "2",
          "2"
        ],
        "output": "0",
        "label": "left inner gcd(a,b) Euclid row5"
      }
    },
    {
      "id": "S013",
      "uses": [
        "G001",
        "S012"
      ],
      "before": "phase=left inner gcd(a,b);pair=[4, 2];quotient=2;remainder=0",
      "after": "next pair=[2, 0];0<=remainder<divisor",
      "relation": "implies",
      "rule_id": "gcd_common_divisor_invariance",
      "conditions": [],
      "explanation": {
        "zh-Hant": "x=q*y+r表示共同因數同時整除x,y當且僅當同時整除y,r；正除數下餘數嚴格變小，因此迴圈會到第二項0。",
        "en": "From x=q*y+r,a number divides both x,y exactly when it divides both y,r.The remainder is strictly smaller than the positive divisor,so the second argument eventually becomes0."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S014",
      "uses": [
        "G001",
        "S013"
      ],
      "before": "phase=left inner gcd(a,b);terminal pair=[2, 0]",
      "after": "gcd=2",
      "relation": "implies",
      "rule_id": "gcd_terminal_zero_convention",
      "conditions": [],
      "explanation": {
        "zh-Hant": "終點gcd(x,0)=x；x=0時採gcd(0,0)=0的自然數慣例。不取模0，也不由gcd值推出兩數皆為質數。",
        "en": "At termination gcd(x,0)=x,including the natural convention gcd(0,0)=0.Never take a remainder modulo0,and do not infer that both inputs are prime."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S015",
      "uses": [
        "G001",
        "S014"
      ],
      "before": "left outer gcd(ab,c) Euclid row0:divide natural 2 by positive 78",
      "after": "2=0·78+2",
      "relation": "implies",
      "rule_id": "natural_euclidean_division_witness",
      "conditions": [],
      "explanation": {
        "zh-Hant": "正除數給唯一商餘，餘數滿足0≤r<divisor；這個完整等式同時核商與餘數，不作除以0。",
        "en": "A positive divisor gives a unique quotient/remainder with0≤r<divisor.The full equality audits both values without division by zero."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": {
        "operation": "euclidean_division",
        "inputs": [
          "2",
          "78",
          "0"
        ],
        "output": "2",
        "label": "left outer gcd(ab,c) Euclid row0"
      }
    },
    {
      "id": "S016",
      "uses": [
        "G001",
        "S015"
      ],
      "before": "phase=left outer gcd(ab,c);pair=[2, 78];quotient=0;remainder=2",
      "after": "next pair=[78, 2];0<=remainder<divisor",
      "relation": "implies",
      "rule_id": "gcd_common_divisor_invariance",
      "conditions": [],
      "explanation": {
        "zh-Hant": "x=q*y+r表示共同因數同時整除x,y當且僅當同時整除y,r；正除數下餘數嚴格變小，因此迴圈會到第二項0。",
        "en": "From x=q*y+r,a number divides both x,y exactly when it divides both y,r.The remainder is strictly smaller than the positive divisor,so the second argument eventually becomes0."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S017",
      "uses": [
        "G001",
        "S016"
      ],
      "before": "left outer gcd(ab,c) Euclid row1:divide natural 78 by positive 2",
      "after": "78=39·2+0",
      "relation": "implies",
      "rule_id": "natural_euclidean_division_witness",
      "conditions": [],
      "explanation": {
        "zh-Hant": "正除數給唯一商餘，餘數滿足0≤r<divisor；這個完整等式同時核商與餘數，不作除以0。",
        "en": "A positive divisor gives a unique quotient/remainder with0≤r<divisor.The full equality audits both values without division by zero."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": {
        "operation": "euclidean_division",
        "inputs": [
          "78",
          "2",
          "39"
        ],
        "output": "0",
        "label": "left outer gcd(ab,c) Euclid row1"
      }
    },
    {
      "id": "S018",
      "uses": [
        "G001",
        "S017"
      ],
      "before": "phase=left outer gcd(ab,c);pair=[78, 2];quotient=39;remainder=0",
      "after": "next pair=[2, 0];0<=remainder<divisor",
      "relation": "implies",
      "rule_id": "gcd_common_divisor_invariance",
      "conditions": [],
      "explanation": {
        "zh-Hant": "x=q*y+r表示共同因數同時整除x,y當且僅當同時整除y,r；正除數下餘數嚴格變小，因此迴圈會到第二項0。",
        "en": "From x=q*y+r,a number divides both x,y exactly when it divides both y,r.The remainder is strictly smaller than the positive divisor,so the second argument eventually becomes0."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S019",
      "uses": [
        "G001",
        "S018"
      ],
      "before": "phase=left outer gcd(ab,c);terminal pair=[2, 0]",
      "after": "gcd=2",
      "relation": "implies",
      "rule_id": "gcd_terminal_zero_convention",
      "conditions": [],
      "explanation": {
        "zh-Hant": "終點gcd(x,0)=x；x=0時採gcd(0,0)=0的自然數慣例。不取模0，也不由gcd值推出兩數皆為質數。",
        "en": "At termination gcd(x,0)=x,including the natural convention gcd(0,0)=0.Never take a remainder modulo0,and do not infer that both inputs are prime."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S020",
      "uses": [
        "G001",
        "S019"
      ],
      "before": "right inner gcd(b,c) Euclid row0:divide natural 50 by positive 78",
      "after": "50=0·78+50",
      "relation": "implies",
      "rule_id": "natural_euclidean_division_witness",
      "conditions": [],
      "explanation": {
        "zh-Hant": "正除數給唯一商餘，餘數滿足0≤r<divisor；這個完整等式同時核商與餘數，不作除以0。",
        "en": "A positive divisor gives a unique quotient/remainder with0≤r<divisor.The full equality audits both values without division by zero."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": {
        "operation": "euclidean_division",
        "inputs": [
          "50",
          "78",
          "0"
        ],
        "output": "50",
        "label": "right inner gcd(b,c) Euclid row0"
      }
    },
    {
      "id": "S021",
      "uses": [
        "G001",
        "S020"
      ],
      "before": "phase=right inner gcd(b,c);pair=[50, 78];quotient=0;remainder=50",
      "after": "next pair=[78, 50];0<=remainder<divisor",
      "relation": "implies",
      "rule_id": "gcd_common_divisor_invariance",
      "conditions": [],
      "explanation": {
        "zh-Hant": "x=q*y+r表示共同因數同時整除x,y當且僅當同時整除y,r；正除數下餘數嚴格變小，因此迴圈會到第二項0。",
        "en": "From x=q*y+r,a number divides both x,y exactly when it divides both y,r.The remainder is strictly smaller than the positive divisor,so the second argument eventually becomes0."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S022",
      "uses": [
        "G001",
        "S021"
      ],
      "before": "right inner gcd(b,c) Euclid row1:divide natural 78 by positive 50",
      "after": "78=1·50+28",
      "relation": "implies",
      "rule_id": "natural_euclidean_division_witness",
      "conditions": [],
      "explanation": {
        "zh-Hant": "正除數給唯一商餘，餘數滿足0≤r<divisor；這個完整等式同時核商與餘數，不作除以0。",
        "en": "A positive divisor gives a unique quotient/remainder with0≤r<divisor.The full equality audits both values without division by zero."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": {
        "operation": "euclidean_division",
        "inputs": [
          "78",
          "50",
          "1"
        ],
        "output": "28",
        "label": "right inner gcd(b,c) Euclid row1"
      }
    },
    {
      "id": "S023",
      "uses": [
        "G001",
        "S022"
      ],
      "before": "phase=right inner gcd(b,c);pair=[78, 50];quotient=1;remainder=28",
      "after": "next pair=[50, 28];0<=remainder<divisor",
      "relation": "implies",
      "rule_id": "gcd_common_divisor_invariance",
      "conditions": [],
      "explanation": {
        "zh-Hant": "x=q*y+r表示共同因數同時整除x,y當且僅當同時整除y,r；正除數下餘數嚴格變小，因此迴圈會到第二項0。",
        "en": "From x=q*y+r,a number divides both x,y exactly when it divides both y,r.The remainder is strictly smaller than the positive divisor,so the second argument eventually becomes0."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S024",
      "uses": [
        "G001",
        "S023"
      ],
      "before": "right inner gcd(b,c) Euclid row2:divide natural 50 by positive 28",
      "after": "50=1·28+22",
      "relation": "implies",
      "rule_id": "natural_euclidean_division_witness",
      "conditions": [],
      "explanation": {
        "zh-Hant": "正除數給唯一商餘，餘數滿足0≤r<divisor；這個完整等式同時核商與餘數，不作除以0。",
        "en": "A positive divisor gives a unique quotient/remainder with0≤r<divisor.The full equality audits both values without division by zero."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": {
        "operation": "euclidean_division",
        "inputs": [
          "50",
          "28",
          "1"
        ],
        "output": "22",
        "label": "right inner gcd(b,c) Euclid row2"
      }
    },
    {
      "id": "S025",
      "uses": [
        "G001",
        "S024"
      ],
      "before": "phase=right inner gcd(b,c);pair=[50, 28];quotient=1;remainder=22",
      "after": "next pair=[28, 22];0<=remainder<divisor",
      "relation": "implies",
      "rule_id": "gcd_common_divisor_invariance",
      "conditions": [],
      "explanation": {
        "zh-Hant": "x=q*y+r表示共同因數同時整除x,y當且僅當同時整除y,r；正除數下餘數嚴格變小，因此迴圈會到第二項0。",
        "en": "From x=q*y+r,a number divides both x,y exactly when it divides both y,r.The remainder is strictly smaller than the positive divisor,so the second argument eventually becomes0."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S026",
      "uses": [
        "G001",
        "S025"
      ],
      "before": "right inner gcd(b,c) Euclid row3:divide natural 28 by positive 22",
      "after": "28=1·22+6",
      "relation": "implies",
      "rule_id": "natural_euclidean_division_witness",
      "conditions": [],
      "explanation": {
        "zh-Hant": "正除數給唯一商餘，餘數滿足0≤r<divisor；這個完整等式同時核商與餘數，不作除以0。",
        "en": "A positive divisor gives a unique quotient/remainder with0≤r<divisor.The full equality audits both values without division by zero."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": {
        "operation": "euclidean_division",
        "inputs": [
          "28",
          "22",
          "1"
        ],
        "output": "6",
        "label": "right inner gcd(b,c) Euclid row3"
      }
    },
    {
      "id": "S027",
      "uses": [
        "G001",
        "S026"
      ],
      "before": "phase=right inner gcd(b,c);pair=[28, 22];quotient=1;remainder=6",
      "after": "next pair=[22, 6];0<=remainder<divisor",
      "relation": "implies",
      "rule_id": "gcd_common_divisor_invariance",
      "conditions": [],
      "explanation": {
        "zh-Hant": "x=q*y+r表示共同因數同時整除x,y當且僅當同時整除y,r；正除數下餘數嚴格變小，因此迴圈會到第二項0。",
        "en": "From x=q*y+r,a number divides both x,y exactly when it divides both y,r.The remainder is strictly smaller than the positive divisor,so the second argument eventually becomes0."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S028",
      "uses": [
        "G001",
        "S027"
      ],
      "before": "right inner gcd(b,c) Euclid row4:divide natural 22 by positive 6",
      "after": "22=3·6+4",
      "relation": "implies",
      "rule_id": "natural_euclidean_division_witness",
      "conditions": [],
      "explanation": {
        "zh-Hant": "正除數給唯一商餘，餘數滿足0≤r<divisor；這個完整等式同時核商與餘數，不作除以0。",
        "en": "A positive divisor gives a unique quotient/remainder with0≤r<divisor.The full equality audits both values without division by zero."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": {
        "operation": "euclidean_division",
        "inputs": [
          "22",
          "6",
          "3"
        ],
        "output": "4",
        "label": "right inner gcd(b,c) Euclid row4"
      }
    },
    {
      "id": "S029",
      "uses": [
        "G001",
        "S028"
      ],
      "before": "phase=right inner gcd(b,c);pair=[22, 6];quotient=3;remainder=4",
      "after": "next pair=[6, 4];0<=remainder<divisor",
      "relation": "implies",
      "rule_id": "gcd_common_divisor_invariance",
      "conditions": [],
      "explanation": {
        "zh-Hant": "x=q*y+r表示共同因數同時整除x,y當且僅當同時整除y,r；正除數下餘數嚴格變小，因此迴圈會到第二項0。",
        "en": "From x=q*y+r,a number divides both x,y exactly when it divides both y,r.The remainder is strictly smaller than the positive divisor,so the second argument eventually becomes0."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S030",
      "uses": [
        "G001",
        "S029"
      ],
      "before": "right inner gcd(b,c) Euclid row5:divide natural 6 by positive 4",
      "after": "6=1·4+2",
      "relation": "implies",
      "rule_id": "natural_euclidean_division_witness",
      "conditions": [],
      "explanation": {
        "zh-Hant": "正除數給唯一商餘，餘數滿足0≤r<divisor；這個完整等式同時核商與餘數，不作除以0。",
        "en": "A positive divisor gives a unique quotient/remainder with0≤r<divisor.The full equality audits both values without division by zero."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": {
        "operation": "euclidean_division",
        "inputs": [
          "6",
          "4",
          "1"
        ],
        "output": "2",
        "label": "right inner gcd(b,c) Euclid row5"
      }
    },
    {
      "id": "S031",
      "uses": [
        "G001",
        "S030"
      ],
      "before": "phase=right inner gcd(b,c);pair=[6, 4];quotient=1;remainder=2",
      "after": "next pair=[4, 2];0<=remainder<divisor",
      "relation": "implies",
      "rule_id": "gcd_common_divisor_invariance",
      "conditions": [],
      "explanation": {
        "zh-Hant": "x=q*y+r表示共同因數同時整除x,y當且僅當同時整除y,r；正除數下餘數嚴格變小，因此迴圈會到第二項0。",
        "en": "From x=q*y+r,a number divides both x,y exactly when it divides both y,r.The remainder is strictly smaller than the positive divisor,so the second argument eventually becomes0."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S032",
      "uses": [
        "G001",
        "S031"
      ],
      "before": "right inner gcd(b,c) Euclid row6:divide natural 4 by positive 2",
      "after": "4=2·2+0",
      "relation": "implies",
      "rule_id": "natural_euclidean_division_witness",
      "conditions": [],
      "explanation": {
        "zh-Hant": "正除數給唯一商餘，餘數滿足0≤r<divisor；這個完整等式同時核商與餘數，不作除以0。",
        "en": "A positive divisor gives a unique quotient/remainder with0≤r<divisor.The full equality audits both values without division by zero."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": {
        "operation": "euclidean_division",
        "inputs": [
          "4",
          "2",
          "2"
        ],
        "output": "0",
        "label": "right inner gcd(b,c) Euclid row6"
      }
    },
    {
      "id": "S033",
      "uses": [
        "G001",
        "S032"
      ],
      "before": "phase=right inner gcd(b,c);pair=[4, 2];quotient=2;remainder=0",
      "after": "next pair=[2, 0];0<=remainder<divisor",
      "relation": "implies",
      "rule_id": "gcd_common_divisor_invariance",
      "conditions": [],
      "explanation": {
        "zh-Hant": "x=q*y+r表示共同因數同時整除x,y當且僅當同時整除y,r；正除數下餘數嚴格變小，因此迴圈會到第二項0。",
        "en": "From x=q*y+r,a number divides both x,y exactly when it divides both y,r.The remainder is strictly smaller than the positive divisor,so the second argument eventually becomes0."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S034",
      "uses": [
        "G001",
        "S033"
      ],
      "before": "phase=right inner gcd(b,c);terminal pair=[2, 0]",
      "after": "gcd=2",
      "relation": "implies",
      "rule_id": "gcd_terminal_zero_convention",
      "conditions": [],
      "explanation": {
        "zh-Hant": "終點gcd(x,0)=x；x=0時採gcd(0,0)=0的自然數慣例。不取模0，也不由gcd值推出兩數皆為質數。",
        "en": "At termination gcd(x,0)=x,including the natural convention gcd(0,0)=0.Never take a remainder modulo0,and do not infer that both inputs are prime."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S035",
      "uses": [
        "G001",
        "S034"
      ],
      "before": "right outer gcd(a,bc) Euclid row0:divide natural 28 by positive 2",
      "after": "28=14·2+0",
      "relation": "implies",
      "rule_id": "natural_euclidean_division_witness",
      "conditions": [],
      "explanation": {
        "zh-Hant": "正除數給唯一商餘，餘數滿足0≤r<divisor；這個完整等式同時核商與餘數，不作除以0。",
        "en": "A positive divisor gives a unique quotient/remainder with0≤r<divisor.The full equality audits both values without division by zero."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": {
        "operation": "euclidean_division",
        "inputs": [
          "28",
          "2",
          "14"
        ],
        "output": "0",
        "label": "right outer gcd(a,bc) Euclid row0"
      }
    },
    {
      "id": "S036",
      "uses": [
        "G001",
        "S035"
      ],
      "before": "phase=right outer gcd(a,bc);pair=[28, 2];quotient=14;remainder=0",
      "after": "next pair=[2, 0];0<=remainder<divisor",
      "relation": "implies",
      "rule_id": "gcd_common_divisor_invariance",
      "conditions": [],
      "explanation": {
        "zh-Hant": "x=q*y+r表示共同因數同時整除x,y當且僅當同時整除y,r；正除數下餘數嚴格變小，因此迴圈會到第二項0。",
        "en": "From x=q*y+r,a number divides both x,y exactly when it divides both y,r.The remainder is strictly smaller than the positive divisor,so the second argument eventually becomes0."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S037",
      "uses": [
        "G001",
        "S036"
      ],
      "before": "phase=right outer gcd(a,bc);terminal pair=[2, 0]",
      "after": "gcd=2",
      "relation": "implies",
      "rule_id": "gcd_terminal_zero_convention",
      "conditions": [],
      "explanation": {
        "zh-Hant": "終點gcd(x,0)=x；x=0時採gcd(0,0)=0的自然數慣例。不取模0，也不由gcd值推出兩數皆為質數。",
        "en": "At termination gcd(x,0)=x,including the natural convention gcd(0,0)=0.Never take a remainder modulo0,and do not infer that both inputs are prime."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S038",
      "uses": [
        "G001",
        "S037"
      ],
      "before": "gcd(a,b)=2;gcd(b,c)=2",
      "after": "left=2;right=2;equal=是",
      "relation": "implies",
      "rule_id": "gcd_parenthesizations_actual_comparison",
      "conditions": [],
      "explanation": {
        "zh-Hant": "分別拿兩個內層結果進外層，最後比較；一般理由是兩路都有同一組三數共同因數，但本題仍保留實際商餘證據。",
        "en": "Feed each inner result into its own outer route and compare.The general reason is equality of the three-number common-divisor sets,but this instance retains the actual quotient/remainder witnesses."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。",
          "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence."
        }
      },
      "arithmetic": null
    }
  ],
  "math_model": {
    "result": [
      28,
      50,
      78,
      2,
      2,
      true
    ]
  },
  "conclusion": {
    "canonical_value": "(28, 50, 78, 2, 2, 是)",
    "explanation": {
      "zh-Hant": "依上列完整代值及條件核對得到原題答案；原題編號、格式與狀態保持不變。",
      "en": "The full substitutions and premise checks give the original answer; source identity,format and status are unchanged."
    }
  },
  "definition_note": {
    "zh-Hant": "沿用原題明示的論域、假設與來源 MKO 定義。本題已完成第二對話數理與中英語義覆核，並由原始參數逐題獨立重算模型、答案與每個算術步驟；原題仍為 candidate。",
    "en": "Use the original question domain, assumptions and cited MKO definitions. This item has second-context mathematical and bilingual review, with the model, answer and every arithmetic step independently recomputed from the original parameters; the source question remains a candidate."
  },
  "source": {
    "input_commit": "0f7be61e2c22317a4075561dc822c87d881a50db",
    "question_binding": {
      "batch_id": "batch-20261013-advance-20261018-r03-gcd-coprime-foundations",
      "batch_path": "public/data/questions/batches/batch-20261013-advance-20261018-r03-gcd-coprime-foundations/questions.jsonl",
      "batch_sha256": "42ddb0f9d2780c613fbb13b7c105532a263effd3cc83d5abb31fa93c39b1f6e1",
      "original_line": 92,
      "original_line_sha256": "046121b1098fa699fd030fb4146e1a125935622f6b87df82aa80a42262a55916",
      "question_compact_sha256": "ad0a508984c7c0dfd70f2757b5cf8e9df5d47f954ab110469bda022a5a3d0fe4",
      "english_row_compact_sha256": "5956288c3fe2c483c66fdaa0ab158043952c8cd2c17c481fd5cc78abf6657493",
      "english_manifest_path": "public/data/i18n/editions/batch-20261013-advance-20261018-r03-gcd-coprime-foundations.en.json",
      "english_rows_path": "public/data/i18n/questions/batch-20261013-advance-20261018-r03-gcd-coprime-foundations.en.jsonl"
    },
    "mko_bindings": {
      "mko-gcd-coprime": {
        "sha256": "a8f5934d6354ea07a83057c61f045034e2d964850bed4759aedd8721848b3b9b",
        "english_sha256": "9a16e24c872347b1ca2aff7ffd64e3465452e78b46eabfda5a844f785741500e",
        "original_path": "public/data/mko/gcd-coprime.json",
        "english_path": "public/data/i18n/editions/mko-gcd-coprime.en.json"
      }
    },
    "selected_manifest_sha256": "5233edba9654a6cffd8b6b2d83c932f82ded15f9c0b7c733bd049943e672acff",
    "selected_part_sha256": "8e8954cf690c44da1ec1efa08994a485c73ca8706b7865d854e3f2cdfd4ca89f",
    "original_candidate_sha256": "906ff8a5ef527fa8af9a3d0e96b529832a9e24c7cb174f4cc2022189ff751d1f",
    "original_candidate_line_sha256": "906ff8a5ef527fa8af9a3d0e96b529832a9e24c7cb174f4cc2022189ff751d1f"
  },
  "historical_source_state": {
    "review_state": {
      "derivation": "candidate_authored_and_locally_checked",
      "source_question_state": "candidate",
      "independent_AI_or_human_review": "NOT_PERFORMED",
      "formalization": "not_formalized",
      "Lean": "NOT_RUN"
    },
    "scope_note": {
      "zh-Hant": "來源綁定教學推導候選；精確核對不等於獨立AI／人工審查或一般形式證明。原題、答案、MKO、公式、AST與歷史不變。",
      "en": "Source-bound pedagogical derivation candidate.Exact checking is not independent AI/human review or general formal proof.Original questions,answers,MKOs,formulas,ASTs and history remain unchanged."
    },
    "label": "Frozen producer metadata; current per-question review is separate below."
  },
  "revision": {
    "id": "review-20261026-005-p17",
    "review_date": "2026-10-26",
    "owner_authorized_reader_publication": true,
    "scope": "One hundred sixty family prototypes and their mathematical/bilingual reasoning variants, stems, boundary statements and every distinct explanation shape were reviewed in a context independent of the producer; parameter substitutions, source bindings, full math models, both answer strings and every arithmetic step were independently reconstructed per question from the original parameters only, with negative controls. This grouped method does not claim a separate literal AI reading of every repeated sentence or an extra AI voter.",
    "review_receipts": {
      "semantic_review_sha256": "00fcddb10a9fb0f56697308976a52e68e2537472bd8e68ecbe96d3c4553f4ad2",
      "question_first_witness_sha256": "198e04380b2d32aef2375b07ca0df1f8afa454d1184993173fad8a3c9187a3d1"
    },
    "status": "second_context_reviewed",
    "mathematical_semantics": true,
    "bilingual_semantics": true,
    "steps_reviewed": 38,
    "method": "family_template_semantic_review_plus_independent_per_instance_reconstruction",
    "exact_question_first_check": {
      "model_sha256": "c54111eded0f53808f450641b8ad0a79281b5c940f6cb05089bb06eab48210d1",
      "arithmetic_steps": 16
    },
    "uses_answer_as_premise": false,
    "human_review": "NOT_PERFORMED",
    "formalization": "not_formalized",
    "Lean": "NOT_RUN",
    "Canonical_truth_promotion": false,
    "authority": "Owner-authorized reviewed reader publication; original question/candidate states remain historical and unchanged."
  }
}