{
  "schema_version": "ocme-reviewed-question-reader-v1",
  "question_id": "q-20261009-advance-20261013-r02-integer-0015",
  "pilot_number": "R0011-06115",
  "number": 6115,
  "family_id": "qf-density13-integer-euclidean-divmod",
  "title": {
    "zh-Hant": "第 6115 題：帶符號整數的歐式商餘",
    "en": "Question 6115: Euclidean Integer Quotient and Remainder"
  },
  "original_question": {
    "schema_version": "ocme-question-v0.1",
    "id": "q-20261009-advance-20261013-r02-integer-0015",
    "batch_id": "batch-20261009-advance-20261013-r02-integer-foundations",
    "language": "zh-Hant",
    "source_mko_ids": [
      "mko-integer"
    ],
    "title_zh": "帶符號整數的歐式商餘",
    "stem_zh": "有限整數協議；a=15、b=6，b>0。求 (q,r) 使a=bq+r且0≤r<b，不用向零截斷的負餘數。",
    "answer": {
      "kind": "expression",
      "value": "(2, 3)",
      "explanation_zh": "歐式商與餘數=(2, 3)。先除，再把負餘調入0至b−1並修正商；有限運算非整數環公理證明。"
    },
    "difficulty": {
      "band": "lower_secondary",
      "level": 2,
      "skills": [
        "帶符號整數的歐式商餘"
      ]
    },
    "variant": {
      "family_id": "qf-density13-integer-euclidean-divmod",
      "seed": 202610132015,
      "parameters": {
        "a": 15,
        "b": 6,
        "operation": "density13_integer_euclidean_divmod",
        "domain": "finite_integer_arithmetic_protocol",
        "semantic_template": "qf-density13-integer-euclidean-divmod/[15,6]"
      }
    },
    "verification": {
      "status": "candidate",
      "checks": [
        "producer_exact_integer_rational_arithmetic",
        "producer_finite_enumeration",
        "producer_source_snapshot_check",
        "producer_substantive_template_deduplication",
        "producer_json_schema"
      ],
      "warning_zh": "僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"
    },
    "provenance": {
      "generator_role": "question_producer",
      "generator_model": "model_unknown",
      "generated_at": "2026-10-09",
      "contract_version": "ocme-question-factory-v0.1"
    }
  },
  "original_english_edition": {
    "schema_version": "ocme-question-localization-v0.1",
    "question_id": "q-20261009-advance-20261013-r02-integer-0015",
    "batch_id": "batch-20261009-advance-20261013-r02-integer-foundations",
    "locale": "en",
    "translation_state": "ai_candidate",
    "source": {
      "question_sha256": "bdf71893fd4175c8b86f3c76e7f7d3c37ae63c00e676070248c1ba44859a7e9c",
      "family_id": "qf-density13-integer-euclidean-divmod",
      "seed": 202610132015,
      "semantic_template": "qf-density13-integer-euclidean-divmod/[15,6]"
    },
    "fields": {
      "title": "Euclidean Integer Quotient and Remainder",
      "stem": "Integer a=15, positive b=6. Give (q,r) with a=bq+r and 0≤r<b, not a negative remainder from truncation toward zero.",
      "answer": {
        "kind": "expression",
        "value": "(2, 3)",
        "explanation": "If a truncated division gives a negative remainder, adjust it into 0,…,b−1 and correct the quotient. Finite arithmetic is not an integer-ring axiom proof. The exact result, retaining the requested order, is (2, 3)."
      },
      "skills": [
        "Euclidean Integer Quotient and Remainder"
      ],
      "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": "有限整數協議；a=15、b=6，b>0。求 (q,r) 使a=bq+r且0≤r<b，不用向零截斷的負餘數。",
        "en": "Integer a=15, positive b=6. Give (q,r) with a=bq+r and 0≤r<b, not a negative remainder from truncation toward zero."
      }
    }
  ],
  "steps": [
    {
      "id": "S001",
      "uses": [
        "G001"
      ],
      "before": "declared exact integer or rational-representation protocol",
      "after": "retain signs,ordered operands and original index order",
      "relation": "implies",
      "rule_id": "integer_arithmetic_scope",
      "conditions": [],
      "explanation": {
        "zh-Hant": "用精確整數運算核對此有限資料，不取浮點近似，也不把有限計算升為全部整數環公理證明。",
        "en": "Check these finite data with exact integer arithmetic,without floating approximations or a claim to prove all integer-ring axioms."
      },
      "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": "divide a=15 by positive b=6",
      "after": "15=2·6+3",
      "relation": "implies",
      "rule_id": "euclidean_division_witness",
      "conditions": [],
      "explanation": {
        "zh-Hant": "商與餘數滿足整數等式，且0≤r<b；負被除數也不能使用向零截斷後的負餘數。",
        "en": "The quotient and remainder satisfy the integer identity and0≤r<b;negative dividends still do not use a negative truncated remainder."
      },
      "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": [
          "15",
          "6",
          "2"
        ],
        "output": "3",
        "label": "Euclidean remainder"
      }
    },
    {
      "id": "S003",
      "uses": [
        "G001",
        "S002"
      ],
      "before": "0≤3<6;a=bq+r",
      "after": "unique pair(q,r)=(2,3)",
      "relation": "implies",
      "rule_id": "bounded_euclidean_remainder_uniqueness",
      "conditions": [],
      "explanation": {
        "zh-Hant": "兩個合法餘數之差絕對值小於b，又等於b的整數倍，只能是0，故商餘數唯一。",
        "en": "The difference of two legal remainders has absolute value below b but is an integer multiple of b,so it is0 and the quotient-remainder pair is unique."
      },
      "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": [
      2,
      3
    ]
  },
  "conclusion": {
    "canonical_value": "(2, 3)",
    "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-20261009-advance-20261013-r02-integer-foundations",
      "batch_path": "public/data/questions/batches/batch-20261009-advance-20261013-r02-integer-foundations/questions.jsonl",
      "batch_sha256": "fe7ab53bd896704506979e7b4bde27359c6a58a2382c94c9e728ab4500a3982d",
      "original_line": 15,
      "original_line_sha256": "ed5840ba90e87cee47d3bf3f0789a82c39fce81fa7417787558219170ad8b013",
      "question_compact_sha256": "bdf71893fd4175c8b86f3c76e7f7d3c37ae63c00e676070248c1ba44859a7e9c",
      "english_row_compact_sha256": "01254d57afc0b79447e88c6e4e57484dc06d84dcf04015825fe9f80a79c69348",
      "english_manifest_path": "public/data/i18n/editions/batch-20261009-advance-20261013-r02-integer-foundations.en.json",
      "english_rows_path": "public/data/i18n/questions/batch-20261009-advance-20261013-r02-integer-foundations.en.jsonl"
    },
    "mko_bindings": {
      "mko-integer": {
        "sha256": "e1e3776eaf56e003f2eb07a66a39c7e221d26759df8fbc4db2652a2733318d66",
        "english_sha256": "24d4c8918b430c9dea2f942bb60a89a295d2386bf3bd9c14731ad56e0fb49034",
        "original_path": "public/data/mko/integer.json",
        "english_path": "public/data/i18n/editions/mko-integer.en.json"
      }
    },
    "selected_manifest_sha256": "5233edba9654a6cffd8b6b2d83c932f82ded15f9c0b7c733bd049943e672acff",
    "selected_part_sha256": "5b2ce631281eb24b279908f47c6de66fa60d326520852879d324c4f3ac653970",
    "original_candidate_sha256": "3db657d964551777c3f47adb4990fb6b90f0944bdd0d8a65bbe1db8893ac8cf7",
    "original_candidate_line_sha256": "3db657d964551777c3f47adb4990fb6b90f0944bdd0d8a65bbe1db8893ac8cf7"
  },
  "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-p12",
    "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": "b9bbc1c947f477a9c4bb5a579368ac5265f2da4572ce612c30f3633ecaf4ae80",
      "question_first_witness_sha256": "93f720df3ab3af13a792a3e6b443728e3c2a56276df839911f36569333236c55"
    },
    "status": "second_context_reviewed",
    "mathematical_semantics": true,
    "bilingual_semantics": true,
    "steps_reviewed": 3,
    "method": "family_template_semantic_review_plus_independent_per_instance_reconstruction",
    "exact_question_first_check": {
      "model_sha256": "252873a394dd8a211f91a12d359ea288d4cab66e1f283f00a9499ebff34a9b6a",
      "arithmetic_steps": 1
    },
    "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."
  }
}