Math Detours · Grade 7 · Around the corner · 11

The party puzzle

At any party, the number of people who shook an odd number of hands is even. Always. Count the hands two ways to see why.

Name ____________________Date ____________

Tonight's idea

At a party some pairs of people shake hands and some do not. No pair shakes twice. Afterwards each person says how many hands they shook.

The number of people who shook an odd number of hands is always even.

A: 3 B: 2 C: 4 D: 1 E: 2 F: 2
Each line is one handshake. Orange dots shook an odd number of hands.

Why? Count the hands that were shaken, in two ways. First way: add up everyone's number. Second way: every handshake uses exactly two hands, so the total is 2 × (number of handshakes).

Both ways count the same hands, so they agree. The sum of everyone's numbers is twice something, so it is even.

Now look at that sum. Even numbers cannot make it odd. Odd numbers can, but odd + odd is even, so they cancel in pairs. If one odd number were left without a partner, the sum would be odd. It is not. So the odd numbers pair off exactly: the number of odd people is even.

Watch one: the party above

The numbers 3, 2, 4, 1, 2, 2 add to 14. There are 7 lines, and \(2 \times 7 = 14\). The odd people are A and D.

Now B and D shake hands. B goes from 2 to 3 and D from 1 to 2. The odd people are A and B. Still two.

Each new handshake changes two people's numbers by 1. So the count of odd people goes up by 2, down by 2, or stays put. It starts at 0 and never becomes odd.

Your turn

  1. Five people shook 1, 2, 2, 3 and 4 hands. (a) How many handshakes were there? (b) How many people have an odd number? (c) Draw a party that fits: five dots, and a line for each handshake.
  2. Seven people each say, "I shook exactly 3 hands." Show that at least one of them has miscounted.
  3. Nine people are at a party. (a) Can every one of them shake hands with exactly 5 of the others? (b) With exactly 4 of the others? If it can be done, say how. If not, say why.
  4. A solid shape has flat faces that are all triangles. Each edge is where exactly two faces meet, like a handshake between two faces. (a) Count the sides of all the triangles in two ways, and show that no solid has exactly 7 triangle faces. (b) A solid has 8 triangle faces. How many edges does it have?

Stretch: passing the test is not enough

Four people say they shook 3, 3, 1 and 1 hands. The sum is even, and the number of odd people is even. Show that this party still cannot have happened. What does that tell you about the rule?

Answers

  1. (a) \(1 + 2 + 2 + 3 + 4 = 12\) hands, so \(12 \div 2 = 6\) handshakes. (b) Two: the people with 1 and 3. (c) One way: call them V, W, X, Y, Z. V shakes with all four others. W also shakes with X and with Y. Then V has 4, W has 3, X and Y have 2, Z has 1.
  2. The numbers would add to \(7 \times 3 = 21\). That is odd, but the sum must be twice the number of handshakes. Or: seven people with an odd number is an odd number of odd people.
  3. (a) No. \(9 \times 5 = 45\) is odd. (b) Yes. \(9 \times 4 = 36\) is even, which allows it, and here is a way: stand in a circle and shake hands with the two nearest people on each side. That is 18 handshakes.
  4. (a) First way: 7 triangles have \(7 \times 3 = 21\) sides. Second way: every edge is a side of exactly two triangles, so the number of sides is 2 × edges, which is even. 21 is odd, so it cannot be done. (b) \(8 \times 3 = 24\) sides, so \(24 \div 2 = 12\) edges. Two square-based pyramids glued base to base make such a solid.

Stretch. With four people, anyone who shook 3 hands shook hands with everyone else. Two people did that. So each of the other two shook at least 2 hands, but they say 1. Impossible. The rule can prove a list of numbers is wrong. Passing it does not prove a list is right.

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