### Calendar Capers

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!

### Rotating Triangle

What happens to the perimeter of triangle ABC as the two smaller circles change size and roll around inside the bigger circle?

# Master Minding

##### Stage: 3 Challenge Level:

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 possible arrangements.

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?

MMMmmmm!

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 which?

Case 2

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 which one?.

Swap one of the beads for another of the same colour as the remaining bead (you now have two the same colour).

Case b

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.

Case c

Either

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.

Or

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.

Case 3

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 wrong positions.

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) colour.

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 them..

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.