26. Five Solutions and a Puzzle
The last column ended with five puzzles. Here are the simple solutions which were promised, and another slightly more difficult puzzle the solution to which will be given in the next column.
The Two Jars Puzzle: How do we measure four litres of water with just a five-litre jar and a three-litre jar and an unlimited amount of water? First fill the five-litre jar and then fill the three-litre jar from it. Then empty the three-litre jar and pour the remaining two litres from the five-litre jar into it. Next fill the five-litre jar and pour one litre into the three-litre jar. The five-litre jar now contains four litres.
The Rabbit Puzzle: How many days does it take a rabbit to climb out of a thirty-metre hole it has fallen into if it can climb three metres a day only to fall back two metres at the end of the day? Since the rabbit climbs three metres and falls back two metres every day, at the end of the day it is one metre closer to the top. So does it take 30 days to get out of the hole? No, because on the 27th day it is three metres from the top, and on the 28th day the rabbit will climb the remaining three metres to the top and will get out of the hole before it falls back two metres.
The Two Trains Puzzle : How far will a fly travel if it flies continuously at 75 km. per hour between two trains approaching each other on 100 kilometres of track at 50 km. per hour? There is a very elegant mathematical solution which involves the summation of a convergent infinite series. However, there is a much simpler solution if we note that the trains will meet in exactly one hour and during this time the fly will have travelled exactly 75 kilometres.
The Russian Postal System Puzzle: How does Boris who lives in Moscow send Natasha who lives in St. Petersburg a diamond ring through the Russian mails if any letter or small parcel which looks like it may contain something valuable is opened and the contents stolen? (Sending the ring in a locked box and sending the key separately is ruled out because both the box and the key might be stolen together.) Borisís very simple solution, which he communicates to Natasha by telephone, consists in sending the ring in a padlocked box and keeping the key. When Natasha receives the box, she locks it with her padlock and returns the box to Boris. Boris then removes his padlock and returns the box to Natasha who removes her padlock and opens the box.
The Rope Around the Earth Puzzle: We encircle the Earth at the equator with a long piece of rope. Then we cut the rope and insert another piece exactly one metre in length and arrange this slightly longer piece so that it is at the same uniform distance from the Earth all the way around. What is the size of the gap between the rope and the equator? If we let the radius of the Earth be R, then the circumference is 2piR. Now let g represent the length of the gap between the lengthened rope and the Earthís surface. Then the circumference of the lengthened rope is equal to 2pi(R+g) which we know is 1 metre greater than the circumference of the Earth or 2piR+1. Therefore, we have the equation
2pi(R+g) = 2piR+1
which we can solve for the gap size g and obtain
g = 1/2pi .
If we take pi to be 3.16, we have that the gap between the Earthís surface and the lengthened rope is about 16 centimetres. The truly remarkable thing about this solution is that the length of the gap is independent of the size of the sphere we use. It may be a golf ball or a basketball or the Earth itself. If we fit a cord about its circumference and then lengthen the cord by one metre, the size of the gap will always be the same - about 16 centimetres!
The Three Daughters Puzzle: I have a friend who has three daughters, and he has asked me to guess their ages to their nearest birthdays (so that the ages I am looking for will be whole numbers). He says he will give me some clues, one at a time.
The first clue is that the product of the girlsí ages is 36. I think about this briefly and do some simple arithmetic calculations, and then say that this is very helpful but I need further information.
The second clue I am given is that the sum of the ages is the same as the day of the month on which we are having this discussion, but Iím not going to tell you either the day or the month. I think for a minute and say that this clue helps narrow the choice of possible ages but I still need a little more information.
The third clue is that the oldest girl has blue eyes. Given this clue, I immediately know the ages of the three girls.