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Thank you for all your solutions. I think it is quite difficult to explain the solution to this problem but many of you noticed the pattern of the nodes in the diagram and the importance of the fact that there are four nodes on the diagram, each of which has an odd number of routes from it.
The model of the town, where each island and each side of the river has an odd number of bridges leading to/from it, was included so that you could relate it to the diagrams in the hints. To be able to move from each of the four regions you must have two areas with an even number of bridges so that you can arrive and leave. Solutions received from Andrei of School 205 Bucharest, Alex and Joanna of Woodfall Junior School, Katherine, Phoebe and Katharine of The Mount School and Gowri, Hannah and Sophie of Caistor Grammar School used this idea.
The hints were there to help you to identify the importance of the number of odd nodes. It is worth going back to these and try to generalise your findings from this problem to other similar problems.
Andrei observed that the city is symmetrical in respect to a
longitudinal line.
The journey is impossible.
Suppose the starting point of the walk is on a side of the river
(it could be north or south, as specified the city is symmetrical);
there are three bridges. The people must first go on the first
bridge, return on the second, and then go away by the third, so
they cannot come back! There must be an even number of bridges on
each part so, that they can return home.
After solving the problem this way, I discovered that this is a
famous problem of the history of mathematics, being at the basis of
topology of networks, first developed by Euler in 1735. He
constructed a diagram (essentially the same as yours), and
associated the land with the vertices, and the bridges with the
possible ways of connecting the vertices - arcs.
Using the reciprocal of Euler's theorem: "if a network has two or
less odd vertices, it has at least an Euler path", that says "if a
network has two or less odd vertices, it has at least an Euler
path", it is easy to see that the proposed problem does not admit
an Euler path (i.e. a continuous path that passes through every arc
once and only once).
Make a set of numbers that use all the digits from 1 to 9, once and once only. Add them up. The result is divisible by 9. Add each of the digits in the new number. What is their sum? Now try some other possibilities for yourself!
15 = 7 + 8 and 10 = 1 + 2 + 3 + 4. Can you say which numbers can be expressed as the sum of two or more consecutive integers?