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The Counterfeit Penny
Submitted by Janine Jones, 10/07/1996.
This article by Valerio De Angelis and Allen Stenger.
You have eight pennies that look the same, but one is counterfeit.
The counterfeit penny is known to be heavier than the others.
How can you tell which one is counterfeit by using a balance scale only twice?
(The balance scale has two pans but no weights; you can place any
combination of pennies in either pan and then observe whether the
pennies in each pan weigh the same, or if one collection is
heavier than the other.)
The original question did not include the information that the
counterfeit penny is heavier, but without that information not
even the MathNerds could do it. Can you?
Hint 1
Start small. Try to solve this problem instead: You have 3 pennies,
and one is counterfeit and heavier than the others. Find it in one weighing.
(If this is too hard, start even smaller: suppose you only have 2 pennies; find the
counterfeit in one weighing.)
Hint 2
With three pennies, the only reasonable thing you can do is pick any two of
them and put one in each pan. There are three possibilities:
- left pan is heavy
- the pans balance
- right pan is heavy
Explain how you can deduce the heavy coin in each case.
Hint 3
The first and third cases are easy: we know the counterfeit coin is heavy,
so the pan that is heavy has the counterfeit coin. The second case is only
a little harder: The counterfeit coin is heavier than the others, so if the
two coins in the balance weigh the same, neither can be counterfeit.
Therefore by a process of elimination the
third coin (the one not in the pans) is counterfeit.
We started small and solved the small problem, now let's try to leverage
that solution to solve a bigger problem. Explain how to perform two
weighings on eight coins, so that the last weighing matches the step
we used for three coins.
Click here for rest of the solution.
The Rest of the Solution
To apply the solution for three coins, we need to isolate the counterfeit
coin to a group of three coins, using only one weighing, and in fact we
can use the same idea. Divide the eight coins into groups of three, three,
and two, then balance the two groups of three against
each other. Just as in the three-coin problem, if one side is heavier,
it contains the counterfeit coin, and if the sides balance, the other
group of two coins contains the counterfeit coin. If one of the groups
of three contains the counterfeit, we can find it in one more weighing
using the three-coin solution. If the group of two contains the
counterfeit, we can find it in one more weighing by balancing the
two coins against each other.
Notice that we could have isolated a counterfeit coin in two weighings
even if we were given nine coins instead of eight, but if there
were ten coins we could not have found the counterfeit by this method.
Can you guess a generalization of this method? How large a collection
of coins could you handle in N weighings?
What If The Counterfeit Could Be Heavy Or Light? The Twelve Coins Problem
What if we only knew that the counterfeit coin was a
different
weight that the true coins, so it might be heavier or
lighter? Could we still find it?
If we knew it was lighter, we could use the same
method (except that we would pick the lighter side as
counterfeit instead of the heavier side), but if we only knew
the counterfeit was a different weight, we couldn't use the
same method. In the three-coin problem, if the pans balanced
we would know the third coin was counterfeit, but if the
pans did not balance we would only know that one of the two
was counterfeit, but not which one.
It turns out that there are methods for this problem too, but they are
harder to discover. This problem is often stated as the Twelve Coins or
Twelve Bags of Gold, where you have twelve items and must discover the false item in three
weighings. See if you can discover a method for this. Starting small
is still a good idea, so start by discovering how to isolate the
counterfeit out of three coins using two weighings.
In general you won't be able to narrow down
the false item until the end, so you should think of the weighings as
methods for collecting information about collections of coins, where
at the end you will put together all the information and draw a conclusion.
This problem is written up in another Best of MathNerds article:
The Twelve Coins (or Twelve Bags of Gold).
Another Popular Balance Problem
Show that using four known weights of
your choice, you can weigh any number of pounds, from one through forty pounds.
You put the item to be weighed in one pan, and you are allowed to place known
weights in
either
pan; they don't have to all be in the other pan.
Hint: Start small. Show that with two known weights you can measure one
through four pounds.
Another hint: Many balance problems (including the ones we have looked at here)
involve ternary (base 3) arithmetic, either implictly or explicitly, because
there are three possible outcomes to each weighing.
References
-
Mario Martelli and Gerald Gannon,
"Weighing Coins: Divide and Conquer to Detect a Counterfeit",
College Mathematics Journal, Vol. 28, No. 5 (Nov., 1997), pp. 365-367.
This article gives simples methods to solve the problems,
both if the relative weight of the false coin is known and if it is not.
-
Ronald L. Graham, Donald E. Knuth, Oren Patashnik,
Concrete Mathematics.
Addison-Wesley, 2nd edition, 1994.
On page 2 we find "we'll see repeatedly in this book that it's advantageous to
LOOK AT SMALL CASES first". Many mathematical puzzles (and even serious problems)
are overwhelming because of the number of items involved; see if you can state
a similar problem using smaller numbers and solve that, as we did here with the number
of coins. This is helpful for nearly any puzzle that involves a large number of
things, such as the Towers of Hanoi (64 disks), the Locker Problem (1000 lockers),
finding the sum of the first 100 integers, and many more. Do not stubbornly
say "My job is to solve 8 coins, do not distract me with 3-coin problems."
-
Wikipedia has an article on
balance puzzles.
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