How can you find one bad coin among 12 using only three weighings in a pan balance? We published the classic solution in November 2013, but a clever alternative appeared in Eureka in October 1950. Here the aspiring counterfeiter is named Felix Fiddlesticks, and he’s trying to find the coin for this mother:
F set the coins out in a row
And chalked on each a letter, so,
To form the words: “F AM NOT LICKED”
(An idea in his brain had clicked).
And now his mother he’ll enjoin:
This plan will reveal the bad coin no matter which it is. For instance, if the coin marked O is heavy, then in the first two weighings the left pan will drop, and in the third weighing it will rise; no other bad coin will produce this result. The plan will also reveal reliably whether the bad coin is heavy or light.
Such cases number twenty-five,
And by F’s scheme we so contrive
No two agree in their effect,
As is with pen and patience checked:
And so the dud is found. Be as it may
It only goes to show CRIME DOES NOT PAY.
(Cedric Smith, “The Twelve Coin Problem,” The Mathematical Gazette, Vol. 89, No. 515 [July 2005], pp. 280-281. A generalized algorithm for solving such a problem for any number of coins is given by Michael Weatherfield in the same issue, pp. 275-279.)
In 1904, a 12-year-old J.R.R. Tolkien sent this rebus to a family friend, Father Francis Morgan. What does it say?
Three coins are lying on a table: a quarter, a half dollar, and a silver dollar. You claim one coin, I’ll claim the other two, and then we’ll toss all three. A coin that lands tails counts zero, and a coin that lands heads wins its value (in cents, 25, 50, or 100) for its owner. Whichever of us has the larger score wins all three coins. If all three coins land tails then we toss again.
Which coin should you claim to make the game fair — that is, so that each of us has an expected win of zero?
A puzzle by A. Savin: Using each of the digits 1, 2, 3, and 4 twice, write out an eight-digit number in which there is one digit between the 1s, two digits between the 2s, three digits between the 3s, and four digits between the 4s.
Prove that every convex polyhedron has at least two faces with the same number of sides.
You and I have to travel from Startville to Endville, but we have only one bicycle between us. So we decide to leapfrog: We’ll leave Startville at the same time, you walking and I riding. I’ll ride for 1 mile, and then I’ll leave the bicycle at the side of the road and continue on foot. When you reach the bike you’ll ride it for 1 mile, passing me at some point, then leave the bike and continue walking. And so on — we’ll continue in this way until we’ve both reached the destination.
Will this save any time? You say yes: Each of us is riding for part of the distance, and riding is faster than walking, so using the bike must increase our average speed.
I say no: One or the other of us is always walking; ultimately every inch of the distance between Startville and Endville is traversed by someone on foot. So the total time is unchanged — leapfrogging with the bike is no better than walking the whole distance on foot.
A puzzle from J.A.H. Hunter’s Fun With Figures (1956):
A man paddling a canoe upstream sees a glove in the water as he passes under a bridge. Fifteen minutes later, he turns around and paddles downstream. He passes under the bridge and travels another mile before reaching the rock from which he started, which the glove is just passing. If he paddled at the same speed the whole time and lost no time in turning around, what is the speed of the current?