You have 24 coins, 23 of which are the same weight, but 1 coin is either heavier or lighter

You have 24 coins. 23 coins are the same weight, but 1 coin is either heavier or lighter. All you are given is a set of BALANCE scales, which can compare the weight of any two sets of coins out of the total set of 24 coins.

What is the minimum number of weighings required to identify the coin, and why?

There are a few different ways to do it in five weighings. I have not found any that are fewer than five except by luck. Then it is 2, and you have to be really lucky with that.

3 Weighings

1. Put the coins in two stacks of 12 coins. Place on stack on each arm of the balance scale. One arm will higher. That is the lightest stack of coins.

2. Remove one coin from each stack. If they are now balanced you know you have the different coin separated, but you do not know which one of the two coins it is.

3. To determine determine which coin is the different one

a. Remove all coins but one from the scale.

b. Place one of your two separated coins on the other arm of the balance.

c. If the coins balance then the coin you are looking for is the other separated coin. If the coins do not balance it is the coin you placed on the scale last.

To take this one step farther the minimum number required guarantee that you find the correct coin is – 8 weighings.

1. Start the same way as before.

2. Remove 6 coins from each stack. If they balance remove those coins and place the top three from each of the other two stacks. If they do not balance remove the top three from the stacks still on the scale. [Note: keep track of which groups of six coins balanced]

4. Repeat the weighing-in with stacks of three coins. This time if not balanced remove the top coin. If they do balance get the other three stack from the unbalanced six.

5. Confirm the weigh-in with this stack of 3 coins. It will be unbalanced.

6. Remove the top coin from each stack. If the new stacks of two balance you have now reduced the number of coins you need to test to two. Go to step 8.

7. Remove the top coins from each stack. If unbalanced they odd coin is still on the scale. If balanced the odd coin has just been removed.

8. Remove all coins from the scale. Then go to Step 3 in the previous process using one of the known standard weight coins.

Niel Leon

Community Developer – Engineering.com

I believe it can be guaranteed in 5 weighings.

First split the coins in piles of 8. Weigh the first against the second and the second against the third. The pair that doesn’t balance will be the contain the coin that is of different weight. For example if pile 1 and 2 balance then pile three contains the coin, if pile 2 and 3 balance pile then pile 1 contains the coin, else pile 2 contains the coin.

Then split the pile into piles of two coins and follow a similar weighing to the first one. This give which two coins contains the different weight coin in two weighings. This is because after 2 weighings you can identify which pile has different weight. For example if the first pile and the second pile are equal and the second pile and the third pile is equal then the one that is different is the fourth pile otherwise follow step ones logic.

After this take one coin from the rest of the eliminated and compare it against one of the ‘Unknown’ coins. If it balances there the other coin is the different weight. If it doesn’t balance then that one is the different weight coin.

This gives 2(the 8 coin piles) + 2(the 2 coin piles) + 1(the one coin comparison) which is 5 weighings.