Choose any three by three square of dates on a calendar page...
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!
What happens to the perimeter of triangle ABC as the two smaller
circles change size and roll around inside the bigger circle?
The first part of the solution was sent in anonymously.
As there are three colours and two pegs it is possible to have any one
of the three colours on each of the pegs, making a total of 3x3 = 9
The following solution owes much to the excellent work of Edith from The
Mount School who thinks the best strategy is to use two beads of hte
same colour to start with and then use beads of different colours
according to the feedback. She thinks the minimum number of goes to be
sure of winning is four.
However, is the suggested variation on Edith's solution the most
efficient? How difficult would it be to come up with a strategy for 4
colours or 3 pegs?
Start with two beads of the same colour.
There are three possibilities:
1) The hidden beads are the same as the selected beads - 4 points - you
have a match.
2) Neither of the hidden beads are the same as the two selected beads -
0 points - you now know that there are no beads of the selected colour
behind the screen.
3) One of the hidden beads is the same colour as the selected beads - 2
points - as one of the selected beads must be in the same position - but
Choose two beads, one of each of the remaining two colours...
Then there are only three possibilities (the first two happen if the two
hidden beads are different colours and the third only if the two hidden
beads are the same colour).
a) They are the right colour beads in the right positions - 4 points -
you have a match.
b) They are the right colour beads in the wrong positions - 2 points
c) One bead is the right colour in the right place - 2 points - but
Swap one of the beads for another of the same colour as the remaining
bead (you now have two the same colour).
You will have substituted one of the right colour beads in the wrong
place with a right colour bead in the right place - 2 points
Next place the two beads in the opposite arrangement to their positions
at b) above.
you remove the right colour bead in the right place with a bead of the
wrong colour - 0 points
Next replace both beads with beads of the same (and last) colour.
you replace the wrong colour bead two beads of the same colour as the
one you had just removed - 4 points - you have a match.
Replace one of the beads with a different colour
a) The bead that was in the right place is left and the other bead is
replaced by the correct colour - 4 points - you have made a match.
b) The bead that was in the right place was left and the other bead is
the wrong colour - 2 points
c) The bead that was the same (colour and position) is removed and
replaced by another bead that has a matching colour to the other hidden
bead but it is in the wrong place - 2 points - right colour beads in the
d) The bead that was the same (colour and position) is removed and
replaced by a bead that does not match the second hidden bead - 1 point
- one bead that is the right colour in the wrong place.
In case d) Put the bead of the colour that has not been changed on the
other peg and replace the "new" bead with the third (as yet unused)
We now have a problem distinguishing b from c as they are both awarded
two points. We need a strategy in the next move that will distinguish
In the next go swap the bead that had just been substituted for one of
the third (unused) colour.
b) This will now give the correct arrangement - 4 points.
c) You are left with one bead of the right colour in the wrong place - 1
point. Next swap the beads around and replace the last changed colour
with a bead of the third colour. This means you have reversed the
positions they were in c above.
Minimum to be certain of success is four moves.