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The Handshake Parity Theorem

Issue 22 Core Published 2 Oct 2026 By The Chalkboard Desk

Tier Core — The Handshake Parity Theorem

The problem

At a meeting, people shake hands with some of the others. Nobody shakes their own hand and the meeting is not necessarily complete. Show that the number of people who shook hands an odd number of times is even, and explain why a meeting of people cannot have exactly people shaking an odd number of hands.
  • Each handshake involves exactly two people.
  • No self-handshakes, and two people may shake hands more than once.
  • The argument must not assume anything about who shook whose hand.

Take a minute

Reveal hint 1
Count the total number of handshake-ends rather than the handshakes.
Reveal hint 2
Each handshake contributes exactly to the sum , whatever the individual degrees are.

Solution

AnswerThe parity statement is forced: the number of odd-degree people is always even, so is impossible.

Derivation

  1. Model the meeting as a graph.

  2. Double-count the endpoints.

  3. Read off the parity.

  4. Apply it to the number .

  5. Handle multiple handshakes.