ORIGINAL FIELD NOTE / 001

Two different inputs. The same check digit.

Using the 7–3–1 method, A returns 0 and K also returns 0. The strings differ. The digit agrees. This tiny example shows why a match cannot establish document authenticity.

Method and result

For a one-character input, the first weight is 7. A has value 10: 10 × 7 = 70, giving remainder 0. K has value 20: 20 × 7 = 140, also giving remainder 0. The calculation follows the check-digit method specified by ICAO. ICAO Doc 9303, Part 3: check-digit method

A directly reproducible collision
InputValueWeighted totalDigit
A10700
K201400
0000
<000

Why the collision occurs

If a single-position value increases by 10, the weighted total increases by a multiple of 10. Its remainder modulo 10 stays the same. This is a property of the arithmetic, not a finding about a particular passport or vendor.

Reproduce the demonstration

  1. Open the check-digit calculator and enter A.
  2. Record the displayed total and digit.
  3. Replace the input with K.
  4. Observe a different total and the same digit.
  5. Use the dataset below to check the other listed inputs.

Scope and limitations

This dataset contains ten deliberately chosen synthetic data elements. It is not sampled from identity documents. It supports the narrow claim that different input strings can share a check digit. It does not estimate error-detection rates, fraud prevalence or the performance of any verification service.

The automated tests in this project's source check the listed expected totals and digits and confirm the illustrated collision. No document-upload service, commercial vendor or government-document authentication process was tested.