Follow the clues to find the mystery number.
You can trace over all of the diagonals of a pentagon without
lifting your pencil and without going over any more than once. Can
the same thing be done with a hexagon or with a heptagon?
How many different sets of numbers with at least four members can
you find in the numbers in this box?
Correct solutions were received from Holly (Nottingham High
School for Girls), Andrei (School 205 Bucharest) and Michael (home
I liked Holly's explanation. It is hard to write the solution to
such a visual problem and it showed a lot of insight so I have
included it here.
The answer to the first part is 1, because there are, logically,
only three places that the first red counter could go:
If a red counter was put in the middle then two more would be
needed to go in corners (to give the diagonals an even number), and
two to go on the middle column and the middle row. This is five red
counters which are not available.
If a counter was placed in the middle of an edge then one would be
needed on a corner, and then another to go on a diagonal and
another to go along the side. The one along the side would then
need another in its column/row, again 5 red counters.
However, if one were placed in a corner then another would be
placed in that row and another in that column. the only way to do
it, to suit the diagonals, is to have a red counter in each
If the diagonals are not included, then two additional ways are
possible. To create a solution the board must have symmetry, this
meaning that the only other ways possible are having all four in
one corner or having two on each side(opposite each other). If
there was not symmetry then the counters would not work because
each counter needs two more on its row and column, therefore they
must also be able to form a square or a rectangle.