Dan Lewis

The Twelve Coins

1) The Heavier Coin

You have a balance and twelve coins, one of which is heavier than the rest, but you don’t know which one it is. Using just 3 weighings on the balance, can you identify which coin is the odd one out?

SOLUTION

This one is pretty easy. There are a few ways to solve this but they all follow the same principle and pattern. I’m going to pick one approach — probably the most straight-forward one.

Step 1: Put four coins on either side of the balance, reserving four others. If the balance stays even, you know that the heavier coin is among the four reserved. If the balance tips, you know that the heavier coin is on the side that went down. Either way, you now have isolated the heavier coin within a subset of four.

Step 2: Put two coins of the four identified in Step 1 on either side of the balance, reserving none. Whichever side goes down has the heavier coin.

Step 3: You have two coins left and can use the scale once more. That’s a great situation to be in! Put one coin on either side and whichever sides drops contains theheavier coin. You win.

Pretty simple, right? The trick is to realize that if the balance stays level, you’ve still gained key information.

2) The Different Coin

You have a balance and twelve coins, one of which weighs a different amount than the rest, but you don’t know which one it is — and you don’t know if the different coin is heavier or lighter than the rest. Using just 3 weighings on the balance, can you identify which coin is the odd one out — and whether it’s heavier and lighter than the others?

SOLUTION

Here’s a framework for thinking through the solution. (The full solution is pages and pages long.)

First, let’s arbitrarily label all the coins. 1 through 9 works for most, and let’s use T, J, and Q for ten, eleven, and twelve. (That’s Ten, Jack, Queen, if you want to know why I chose those letters.)

Second, let’s put coins 1-2-3-4 on one side of the balance and coins 5-6-7-8 on the other, keeping 9-T-J-Q in reserve. One of three things will happen: The 1-2-3-4 side will go down; the 1-2-3-4 side will go up; or the balance will stay level. For sake of example, let’s say that it stays level. Good news: you know that the odd coin out is one of 9-T-J-Q. You’ve eliminated two-thirds of the possible answers. Had the balance tipped, you’d have only eliminated a third.

The bad news is that if you follow your instinct and balance 9-T versus J-Q, you’re not able to eliminate any more coins — it’s going to tip. Balancing, say, 9 versus T has a similar problem — if it doesn’t tip, you’re left with J and Q, and putting both on the balance tells you very little.

There are three things you have to realize in order to get this one done.

1) If a coin goes up in one weighing and down in another, it’s not the coin you’re looking for.

2) You’re going to need to test an odd numbers of coins, so you’ll need to add in a coin of known weight (that is, one that’s already been identified as one of the 11 normals) at some point. (As it turns out, you’ll always need to test an odd number of coins at one point or another, making this critical.)

3) The first weighing isn’t about eliminating coins. It’s about collecting information. The second is similar, but more importantly, about getting down to no more than three choices left, and knowing something about each of those choices.

Let’s say that the 6 coin is the odd coin out and you’ll see what I mean. Again, et’s put coins 1-2-3-4 on one side of the balance and coins 5-6-7-8 on the other, keeping 9-T-J-Q in reserve for our first weighing. The 1-2-3-4 side goes up and the 5-6-7-8 side goes down. So we know that, if the odd coin is one of the 1-2-3-4, it’s lighter than the rest; and if the odd coin is in the 5-6-7-8 group, it’s heavier than the rest. The 9-T-J-Q? All normals.

Second weighing: Put 1-2-5 on the left. On the right, put 3-6-Q. Note that the 4, 7, and 8 aren’t being tested again and the Q is being used as a placeholder, to make sure the scale can balance in case the 4, 7, or 8 are the odd duck. The right side goes down again, meaning that the if the coin is the 1, 2, or 5, it’s lighter than the rest; or if it’s the 3, 6, or Q, it’s heavier. We can remove the Q from the solution set because it’s already been proven to be a normal coin from the first weighing. We can also remove the 3 and 5 because they can’t be heavy in one weighing and light in the other. They’re both normals.

So we’re left with 1, 2, and 6. We know if the answer is 1 or 2, that’s because it’s too light. We know if the answer is 6, it’s because it’s too heavy. Our third weighing: 1 versus 2. If it tips, whichever side goes up holds the mystery coin. If it stays balanced, the mystery coin is the too-heavy 6. That’s what happens and we’ve identified our coin.

Let’s do this again, starting with step 2, but this time, let’s make the 4 coin lighter than the other eleven. To recap, the 4 was on the left side of the balance for the first weighing but not on the balance at all for the second. So this time, the balance stays level on the second weighing. We now know that our mystery coin is one of the 4, 7, and 8. If the answer is 7 or 8, it’s because the coin is too heavy, so we weight 7 versus 8. Had it tipped, whichever side goes down contains the maladjusted coin. Otherwise, the balance will stay level and the 4 is our answer.

You can try other permutations if you’d like. The biggest trick is to make sure you use a dummy coin in the second weighing to make sure that the third weighing has only three options, max, to test.