There's a constant struggle between ideals and getting things done practically within time limits. Just because I don't fully understand space and time doesn't mean I haven't a plane to catch. Travel may not be impossible, but it certainly has its attendant difficulties. Arranging for all those miraculous coincidences, culminating in your arrival at Heathrow with a valid passport and tickets in one hand and the appropriate luggage in the other, is an exercise in logical problem-solving that would be a mathematical nightmare if treated on a logical basis. What size of suitcase do you need? It depends on how much you want to take. How much do you want to take? As much as you can fit into a suitcase. Then there's the weight limit. Do you take lots of light things you don't need very much, or a few heavy things that are really indispensable? Ideally I'd take everything; so it's more a question of what I leave out. Even for a small number of omissions from a small set of objects, this can produce a large and unwieldy number of combinations.
Points to ponder:
1. You have 10 pairs of socks, each of a subtly different shade, and you love then all, but you have room for only seven pairs. How many trips would you have to make to exhaust all the possible combinations of pairs?
2. Professor Shiftless won't travel anywhere without his breakfast cereal All Brain. He has a set of reinforced Heavy Weighton suitcases all made from the same thickness of metal and all the same shape, but different sizes. The 10cm Weighton weighs 21kg and will carry .25kg of All Brain. The weight limit on Icarus Airlines is 325kg. What size suitcase:
a) would oblige him to travel without any cereal? b) would maximise the amount of cereal he took?
Solutions to last week's puzzles:
1. The number 1.999..., in which the 9s carry on for ever, and we are to imagine that there is no final digit, is indeed equal to 2.0. To see this, let 1.999... (as here described) equal X, where X is the value you wish to determine:
X + 18 = 1.999... + 18
which equals 10x, whence 9X = 18; and so X = 2.
2. Given the conditions in the problem, the light in the lighthouse goes on and off an infinite number of times before two seconds have elapsed, each change of state lasting half as long as the previous one.
There is no last end-state. It is impossible to say whether the light is on or off after exactly two seconds.Reuse content