Math Detours · Grade 5 · Around the corner · 18

Find the fake coin

Nine coins, one lighter, a balance with no weights. Two weighings are enough if you split into three piles, not two.

Name ____________________Date ____________

Tonight's idea

You have nine coins that look the same. Eight are real. One is a fake, and it is a little lighter. You have a balance with two pans and no weights. How few weighings find the fake?

Most people split the coins in half. But a balance can do three things, not two. The left pan can rise, the right pan can rise, or the beam can stay level.

Left pan rises fake is on the left Level fake was left out Right pan rises fake is on the right
One weighing, three possible results. Each result points at a different pile of three.

So make three piles of three. Put one pile on each pan and leave one pile on the table. The lighter pan rises. If the left pan rises, the fake is on the left. If the right rises, it is on the right. If the beam stays level, the fake is in the pile you left out.

One weighing cuts nine suspects down to three.

A second weighing does the same job with the last three coins.

Watch one: coins numbered 1 to 9

First weighing. Coins 1, 2, 3 against coins 4, 5, 6. The beam stays level. So all six are real, and the fake is 7, 8 or 9.

Second weighing. Coin 7 against coin 8, with coin 9 left out. The pan with coin 8 rises. Coin 8 is the fake.

If that beam had stayed level, the fake would be coin 9, and you would never have weighed it.

Try it with 3, 9 or 27 coins: mathdetours.com/demos/fakecoin.html.

Your turn

  1. You have three coins and one is lighter. Explain how one weighing finds it. Say what you do for each of the three results.
  2. Nine coins are numbered 1 to 9. You weigh 1, 2, 3 against 4, 5, 6 and the left pan rises. Then you weigh coin 1 against coin 2 and the beam stays level. Which coin is the fake?
  3. Now there are 27 coins and one is lighter. (a) What should you put on each pan for the first weighing? (b) How many weighings do you need in all?
  4. There are eight coins and one is lighter. (a) Sam starts by weighing four against four. Which of the three results can never happen? Why can Sam not be sure of finishing with one more weighing? (b) Find piles that always work in two weighings.

Stretch: why not ten?

Two weighings, each with three possible results. How many different pairs of results can you see? Use that to explain why no plan can always find the fake among ten coins in two weighings. How many coins can four weighings handle?

Answers

  1. Put one coin on each pan and leave one out. If the left pan rises, the left coin is the fake. If the right pan rises, the right coin is. If the beam stays level, the fake is the coin you left out.
  2. The left pan rose, so the fake is 1, 2 or 3. Coins 1 and 2 balanced, so both are real. The fake is coin 3.
  3. (a) Nine coins on each pan, nine left out. That leaves nine suspects. (b) Three weighings: 27 suspects, then 9, then 3, then 1.
  4. (a) The beam can never stay level, because the fake is always on one of the pans. So this weighing has only two results, and it leaves four suspects. One weighing can sort out at most three coins, so four is too many. (b) Piles of 3, 3 and 2. Weigh three against three. If a pan rises, you have three suspects and one weighing does it. If the beam is level, weigh the last two against each other.

Stretch. There are \(3 \times 3 = 9\) pairs of results. Ten coins need ten different answers, and nine pairs cannot point to ten coins. Each weighing multiplies the count by 3. So four weighings handle \(3 \times 3 \times 3 \times 3 = 81\) coins.

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