# Odd Coin Problems

### Jack Wert

I am an retired engineer type, and not capable of the directions of analysis of these problems as are you mathematicians, and therefore must seek simpler ways of solving logic and reasoning problems.  This one was presented to me as the 120 marble problem - five weighings - not knowing if the odd one was heavier or lighter than the others. I pondered it many nights in my slumber inviting mental activity, as I drifted off. Sure enough, one night I woke up in the middle of the night and the solution was there in my brain. After pondering it for a few minutes, I said (to myself) "eureka, I think I have it." And sure enough the next day, after fully digesting it, I was certain it was valid.

For many years - through the fifties, sixties and seventies - I was a loyal follower of Martin Gardner and always looked forward to his column and the periodic puzzle collection. I do not know if he ever presented any of these problems larger than the 12 marble version, and I do not remember who gave me the 120 version, but I think it should be the one to be solved, as the other solutions I have seen for the 12 coin version, get a bit out of hand for anything larger than 39 items - and even the 12 coin size is not easy to do in one's head.

Here is my solution. (A shorter variant is also available.) You might want to skip the early part, as it is for those not "up to date" on the general coin weighing problems.

Different weight coin (marble - ball - etc.) problems using a pan balance to discover the odd item.

This solution is directed at problems wherein the odd coin is either heavier or lighter than the others, but we do not know which. I will call this the "Odd Weight Coin" problem.

First, some fundamentals:

It is assumed that in all problems of this general type, the coins are divided into three equal groups for the weighings - two to be placed on the pans of the balance, and the third placed on the table.

Before going to the subject solution, you must be familiar with the more basic coin weighing problems wherein the relative weight of the odd coin is known (either heavier or lighter). The following chart shows the number of coins for the first five of these - the total number of coins, and the size of the three coin groups. For example, you must know that if you have 81 coins and you know the odd coin is heavier, it will take only 4 weighings to identify it.

I will call this Method A:

Weighings Total coins Coin groups 1 3 1 2 9 3 3 27 9 4 81 27 5 243 81

Now to expand this concept into a system for application to the Odd Weight Coin problems wherein we do not know whether the odd coin is heavier or lighter than the others:

It has been established that these problems are generalized by the equation: (3n - 3)/2 = X, X being the number of coins in the problem in which n weighings will discover the odd coin and its weight relative to the others.

For the first example, I use the very popular 12 coin problem.

Divide the 12 coins into 3 groups of 4 coins, and then divide each of the 4 coin groups into a single coin and a pile of 3 coins. Place the 3 coin piles into paper bags (for easy handling - the weight of the bags will be the same for coins on both pans, so will not affect the weighing results in any way). Place one 4 coin group on each pan of the balance, and the third one on the table.

Observe the condition of the balance. This is the first weighing.

Rotate the bags of 3 coins, moving the one from the right pan to the table, the one from the left pan to the right pan and the one from the table onto the left pan.

Observe the condition of the balance. If it changes, it will identify the bag of 3 coins that has the odd coin and its relative weight. This is the second weighing.

In that case, you have 3 coins one of which is known to be heavier (or lighter). Clear the balance and use Method A, above, to identify the odd coin. That would be the third weighing.

If the condition of the balance does not change, the odd coin is one of the single coins and can be identified and its relative weight determined by rotating them as with the bags of 3. That would be the third weighing.

Problem solved - with no labeling, and not really much of a problem with this method.

Now to apply this system to the larger problems.

Example: Identify the Odd Weight Coin in 1,092 coins using 7 weighing of the balance.

Split the coins into three groups of 364 coins, and each of these groups into separate piles of:

243 / 81 / 27 / 9 / 3 / 1 coins, putting each of the multiple coin piles in a paper bag as with the 3 coin piles in the 12 marble problem. (again, for easy handling)

Put one 364 coin group (five bags and a single coin) on each pan and one group on the table. Observe the condition of the balance. This is the first weighing.

Rotate the bags of 243 coins, and observe the condition of the balance. This is the second weighing.

If it has changed, you will know which bag of 243 coins contains the odd coin, and its relative weight. Taking the 243 coins from that bag, clear the other coins from the balance and use Method A to find the odd coin in the 5 remaining weighings.

If it has not changed, rotate the bags of 81 coins and observe the condition of the balance. This is the third weighing.

If it has changed, you will know which bag of 81 coins contains the odd coin, and its relative weight. Taking the 81 coins from that bag, clear the other coins from the balance and use Method A to find the odd coin in the 4 remaining weighings.

If it has not changed, rotate the bags of 27 coins and observe the condition of the balance. This is the fourth weighing.

If it has changed, you will know which bag of 27 coins contains the odd coin, and its relative weight. Taking the 27 coins from that bag, clear the other coins from the balance and use Method A to find the odd coin in the 3 remaining weighings.

If it has not changed, rotate the bags of 9 coins and observe the condition of the balance. This is the fifth weighing.

If it has changed, you will know which bag of 9 coins contains the odd coin, and its relative weight. Taking the 9 coins from that bag, clear the other coins from the balance and use Method A to find the odd coin in the 2 remaining weighings.

If it has not changed, rotate the bags of 3 coins and observe the condition of the balance. This is the sixth weighing.

If it has changed, you will know which bag of 3 coins contains the odd coin, and its relative weight. Taking the 3 coins from that bag, clear the other coins from the balance and use Method A to find the odd coin in the remaining weighing.

If it has not changed, rotate the remaining single coins to identify the odd coin and its relative weight. This is the seventh weighing.

Problem solved!

You can see that problems employing smaller number of coins (363 / 120 / 39 / 12 / 3) are all sort of sub-sets of this 1,092 coin problem, so no further explanation is needed. You can also see that it is easy to expand it as far in the other direction as wanted - up to 2,391,483 coins in 14 weighings and beyond. All that is needed is a bit of simple arithmetic in order to divide the total into the necessary groups and internal piles.

It has also been stated that given one extra standard weight coin, an extra "puzzle" coin can be added to the problem(s): (3n - 3)/2 + 1 coins using n weighings, employing one extra coin of standard weight.

This is easy. Just place the standard coin on one pan and the extra puzzle coin on the other pan. Proceed as above, and if the pans are balanced at any point during the process, the extra coin is of standard weight. If it is the odd coin, the balance will be unbalanced throughout the process and all of the other coins can be discarded leaving only the extra coin as the odd one. And its weight will be established by the condition of the unbalance.

Note that no labeling of any kind was used in this system, and once you are familiar with it, it is really a breeze. ### Weighing Coins, Balls, What Not ... 