# C638 — Explain the Priestly minimum by factor structure

## Question

Why does subtracting the14-year rail width from13×182 produce14×168?

## Inputs

```json
{
  "corresponding_count": 13,
  "Enochian_half": 182,
  "rail_width": 14
}
```

## Sources

```json
[
  "File70 §§6.2–6.5"
]
```

## Opened utc

2026-09-28T00:57:56.936051+00:00

## Results

```json
{
  "corresponding_factorization": [
    14,
    169
  ],
  "minimum_factorization": [
    14,
    168
  ],
  "Priestly_half": 168,
  "identity": "13×182−14=14(169−1)=14×168"
}
```

## Finding

The Priestly minimum follows from182=14×13 and168=13²−1. The common14-year width turns the corresponding vector into fourteen Priestly halves. This explains the change from13 to14 calendar units through the rail geometry.

## Reassessment

Test whether the combined half-calendar conditions select a unique count or a congruence family.

## Checks

```json
{
  "factorization": true,
  "minimum": true,
  "difference_of_squares": true
}
```

## Closed utc

2026-09-28T00:57:56.936249+00:00

## Record provenance

Exact record hash: `48e03ab500abe657bfcd30dd81ee12ec3744e5ab24b2d6c021bf1052845dc452`

Source journal: `Research_Cycles/C0632_C0731/journal.json`