Identify the Odd Weight Coin in 1,092 coins using 7 Weighings by Jack Wert

The Technique:

Divide the coins into 3 main groups of 4 coins, each group into
sub-groups of 1 and 3 coins. Place two of the main groups on the pans
of a pan balance, the third on the table, and observe the condition of
the balance. Then rotate the sub-groups of 3 and again observe the
condition of the balance. If the condition of the balance changes, the
odd coin is in one of the 3 coin groups, and you will know the relative
weight. If the condition of the balance does not chanage, you will know
the odd coin is one of the single coins. This provides enough
information to allow you to use the third weighing to identify the odd
coin and its relative weight - whether its in one of the sub-groups of 3
or one of the single coins. No labeling is necessary. Note that this
can easily be done mentally, picturing each move.

In Depth:

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: ((3^n)-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): ((3^n)-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.