C639 — Solve the two-calendar compatibility equation

Question

What count family permits an Enochian corresponding span and a Priestly minimum fourteen years shorter?

Inputs

{
  "equation": "182m−14=168n",
  "positive_integer_counts": true
}

Sources

[
  "Derived from File70 §6 source geometry"
]

Opened utc

2026-09-28T00:58:22.881303+00:00

Results

{
  "reduced_equation": "13m−12n=1",
  "general_solution": "m=1+12q; n=1+13q",
  "examples_m_n": [
    [
      1,
      1
    ],
    [
      13,
      14
    ],
    [
      25,
      27
    ],
    [
      37,
      40
    ]
  ],
  "source_choice_q": 1
}

Finding

The pair of calendar requirements has an infinite integer family. Its source member is m=13,n=14; the smaller m=n=1 also fits the two-calendar equation. The exposure condition must supply any stronger selection.

Reassessment

Add the fixed thirteen-year exposure flanks and solve the complete three-calendar system.

Checks

{
  "examples": true,
  "source_member": true,
  "not_unique_pair": true
}

Closed utc

2026-09-28T00:58:22.881481+00:00

Record provenance

Exact record hash: 2d75bc9bfb77ad832fbb24b55007d44225e7432b192eeef77c2166cdfd940f33

Source journal: Research_Cycles/C0632_C0731/journal.json