# Solve Recreational Math: Identify Odd Weight Coin in 5 Weighs

• jvwert
In summary, there are 120 visually identical coins, with one coin having an unknown weight compared to the others. By using a two pan balance just five times, it is possible to identify the odd weight coin. This solution reduces it to 2 or 3 possible coins, but does not identify the exact one. A solution must assume the worst possible scenario and it is possible to do this for 121 coins. Removing any non-identical coins will lead to the answer. This method can also be applied to a similar 12 coin problem.
jvwert
You have 120 coins, all of which are visually identical. E
Except for one coin, they all weigh exactly the same.
It is unknown whether the odd weight coin is lighter or heavier then the others.
Using a two pan balance just five times, identify the odd weight coin.

Last edited by a moderator:
Welcome to the forum jvwert! I have deleted your e-mail address because that is one really good way to get spammed.

120 / 3 = 40 ---> 40a, 40b, 40c

Balance 1 40a ---- 40b
Balance 2 40b ---- 40c = you know the 40s different and if they are lighter or heavier

40 / 3 = 15a, 15b, 10

Balance 3 15a ---- 15b = if equal, then Balance 4 is 7----7 and Balance 5 is 3---3

10 / 3 = 5a, 5b

Balance 4 5a ---- 5b
Balance 5 2a ---- 2b

LAF said:

120 / 3 = 40 ---> 40a, 40b, 40c

Balance 1 40a ---- 40b
Balance 2 40b ---- 40c = you know the 40s different and if they are lighter or heavier

40 / 3 = 15a, 15b, 10

Balance 3 15a ---- 15b = if equal, then Balance 4 is 7----7 and Balance 5 is 3---3

10 / 3 = 5a, 5b

Balance 4 5a ---- 5b
Balance 5 2a ---- 2b

This solution reduces it to 2 or 3 coins. Neither identifies the one coin.

LAF said:

120 / 3 = 40 ---> 40a, 40b, 40c

Balance 1 40a ---- 40b
Balance 2 40b ---- 40c = you know the 40s different and if they are lighter or heavier

40 / 3 = 15a, 15b, 10

Balance 3 15a ---- 15b = if equal, then Balance 4 is 7----7 and Balance 5 is 3---3

10 / 3 = 5a, 5b

Balance 4 5a ---- 5b
Balance 5 2a ---- 2b

Your second balance has the chance of being redundant. If group A and B weigh the same, the coin is in group C.

flatmaster said:
Your second balance has the chance of being redundant. If group A and B weigh the same, the coin is in group C.

Well a solution has to assume the worst possible scenario

jvwert said:
You have 120 coins, all of which are visually identical. E
Except for one coin, they all weigh exactly the same.
It is unknown whether the odd weight coin is lighter or heavier then the others.
Using a two pan balance just five times, identify the odd weight coin.

In fact, it can be done for 121 coins.

Please remove non-identifies coin which is 2 or 3. Then, you get answered.

Using your solution to the similar 12 coin problem as a template jvwert:

First split the 120 coins into main groups of 40, and split these into subgroups of 9, 27, 3, and 1 coins.

First weighing: Place one main group on each pan of the balance and leave the other on the table.

Second weighing: Rotate the 27 coin sub groups (from pan A to the table, pan B to pan A, and from the table to pan B) and observe whether the balance changes.

If it does it will identify a 27 coin subgroup that contains the odd coin and whether it is heavy or light. Clear the balance and split this subgroup into three groups of nine, and for the third weighing place one of these groups of nine on each pan of the balance. This will identify the group of nine that contains the odd coin. Split this group into three groups of three. The solution is then trivial.

If the balance does not change then for the third weighing rotate the 9 coin subgroups and observe the balance. If it changes this will identify a 9 coin subgroup that contains the odd coin and whether it is light or heavy. Clear the balance and split this group into three groups of three. The solution is then trivial.

If it does not change then for the fourth weighing rotate the 3 coin subgroups and observe the balance. If it changes this will identify a 3 coin subgroup that contains the odd coin and reveal its relative weight. Clear the balance and split this group into ones. The solution is then trivial.

If it does not change then for the fifth weighing rotate the 1 coin subgroups which will identify the odd coin and its relative weight.

## 1. What is the objective of solving the "Identify Odd Weight Coin in 5 Weighs" problem?

The objective is to determine which coin out of a set of identical-looking coins is the odd one out in terms of weight, using only a balance scale and a maximum of 5 weighings.

## 2. How many coins are typically used in this problem?

This problem can be solved with any number of coins, but it is most commonly seen with 12 coins.

## 3. What is the minimum number of weighings needed to solve the problem?

The minimum number of weighings needed to solve the problem is 3. However, with 12 coins, it is possible to solve it in 4 weighings.

## 4. What is the most efficient strategy for solving this problem?

The most efficient strategy is to divide the coins into 3 groups of 4 and weigh 2 groups against each other. This will narrow down the potential odd coin to one group. Then, weigh 2 coins from that group against each other to identify the odd one.

## 5. Is it possible to solve this problem with a different number of weighings?

Yes, it is possible to solve this problem with a different number of weighings. However, it is not possible to solve it in less than 3 weighings or more than 5 weighings with 12 coins.

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