Monks with Green Noses: Answer

First we attempt to derive a possible solution.

Suppose one nose.
First day, green-nose monk sees no green noses. Knows he has one, and leaves. Others saw a green nose, but did not know that there were no others, until they infer this on the second day, by the monk leaving. Solution: one day <=> one nose.

Suppose two noses.
Second day, green nosed monks see other green nosed monk has not left. Realise there is an additional green nose but cannot see one, they know they have a green nose and leave. Solution: two days <=> two noses

So we potulate that n noses = n days.

If true for a particular n, then suppose n+1 noses. Green nosed monk realises on day n+1 that monks have not left, so has green nose. He leaves. (n+1 days = n+1 noses.) Thus, if true for n, then true for n+1. Since one can repeatedly substitute n+1>n, giving the statement "true for n+1 then true for n+2", etc., then this means that:

If true for some given n, then true for all n. We know postulate true for n=1, so true for all n.

QED (proof by induction)

The answer to the problem is, therefore, if the monks realise on the seventh day there are SEVEN GREEN NOSES.

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