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SOME NUMERICAL puzzles involve implicit constraints imposed by the world rather than by the maths itself. Take last week's bouquet puzzle, for example. In general terms this goes:

Q: The flowers in a bouquet are all red bar X, all blue bar Y and all white bar Z. How many are there of each?

Flowers are physical objects and are countable, so each of X, Y and Z must be a positive whole number. But there are also constraints on the group of numbers as a whole, as well as on each individual number. X, Y & Z can't be just any old positive whole numbers. The total number of flowers is half of (X+Y+Z), which means that (X+Y+Z) must be even, which was where the following question fell down:

Q: My sheep are all mottled bar 223, all black bar 232 and all white bar 322. How many of each are there?

Back with the flowers, it is easy to show that the number of red flowers is half (X+Y+Z)-X. This must be a positive number if this is really a puzzle about flowers. This means that any two of the numbers must together be greater than the third, which is the same condition for three sticks to make a triangle: the lengths of any two must together exceed the third.

The duck question also involved hidden constraints:

Q: I have ducks and pigs only. The number of heads multiplied by the number of beaks gives the total number of legs. How many ducks? How many pigs?

Trial and error shows three ducks and three pigs is a solution. To show that it is the only solution requires algebra. There are (d+p) heads and (2d+4p) legs, as ducks have two legs and pigs four. Also (d+p)Xd = (2d+4p), we are told. This equation has two unknowns, but a little directed manipulation leads to d/p = (4-p)/(p-2). Now d and p are both positive. So (4-p)/(p-2) must be positive too. This means that both (4-p) and (p-2) must be positive. We are looking for a whole number between four and two. This is three. So there are three pigs and we see that there must be three ducks too. To double-check this: three pigs and three ducks gives six heads. Six heads times three beaks gives 18, which is, as required, the total number of legs. This nut is simple to crack and in isolation is unworthy of such a sledgehammer. But then it is as well to practise sledgehammer technique on manageable nuts.

Last week's puzzles

The shape is half a half-hexagon. It is divisible into four similar shapes, as shown. Note that two of them are flipped over (see diagram).

Points to ponder

1 I have one stick a metre long and another two metres long. I break the longer stick in two at a point chosen at random along its length. What's the probability that the three sticks will make a triangle?

2 A pig and a pog costs pounds 1, and a pog and a pug costs 52p. A pug and a pig costs even less. How much is a pig? A pog? A pug?

The puzzles from programmes 1-6 of `Puzzle Panel' are now available at the website at www.puzzlemaster.co.uk

comments to: indy@puzzlemaster.co.uk