1. The rectangle has a total area of 1,056 sq. cm.
2. The rectangle is 33cm x 32cm.
Extension
We do not know of any other set of squares which create a rectangle. Nor can we prove that there is no other. You might be able to find another, or, you might be able to prove that there cannot be another set.

Action	Moves	Total So Far
Blocks 1 & 2 to Position C.	3
Blocks 3 to Position B.	1
Blocks 1 & 2 back onto Block 3.	3	7 moves
Block 4 to Position C.	1
Blocks 1 - 3 onto Block 4.	7	15 moves
Block 5 to Position B.	1
Blocks 1 - 4 onto Block 5.	15	31 moves
Block 6 to Position C.	1
Blocks 1 - 4 to Position A.	15	47 moves
Block 5 onto Block 6.	1
Blocks 1- 3 to Position C.	7	55 moves
Block 4 to Position B.	1
Blocks 1 & 2 to Position A.	3
Blocks 3 onto Block 4.	1	60 moves

Extensions
1. 5 moves
2. 57 moves

Seven Queens ...
Six Queens ...
Five Queens ...

Difference	Escape rolls
0	6 ways to escape
1	10 ways to escape
2	8 ways to escape
3	6 ways to escape
4	4 ways to escape
5	2 ways to escape
	36 possible rolls

Extension
So far it is not certain which is the best placement strategy. Some possibilities are:

1 2 2 1 0 0

On average this takes about 13.7 rolls

2 2 2 0 0 0

On average this takes about 15.5 rolls

1 3 2 0 0 0

On average this takes about 14.4 rolls

A table of results for different numbers of sheep shows an unusual pattern:

Sheep in Front	1	2	3	4	5	6	7	8	9	...	41	50	100	211
Sheep shorn	1	1	1	2	2	2	3	3	3	...	14	17	34	71

• One rule is: Divide by 3 and round up if there is a remainder.
• Another way to find the answer is expressed symbolically as: INT[^{(n - 1)}/₃] + 1
• INT means to take the whole number (or integer) part of the an answer. So these symbols mean first subtract 1 from the number of sheep, then divide by 3, then use only the whole number part of the answer, and finally, add 1.
• Another symbolic expression which gives the same result is: INT[^{(n + 2)}/₃]
• Sneaking past three at a time becomes: INT[^{(n + 3)}/₄]
• Sneaking past x at a time becomes: INT[^{(n + x)}/_{(x + 1)}]
• With 2 shearers, and sneaking past 3 we get: 2 x INT[^{(n - 1)}/_{(2 + 3)}]

• Peter: (5, 6, 9) = 20
• Paul: (7, 8, 10, 15) = 40
• Mary: (11, 14, 16, 17, 40) = 98

To find all the solutions one could systematically test all combinations for Peter and Paul from the lowest for Peter (5, 6, 7) = 18, to the highest for Paul (15, 16, 17, 40) = 88. Remember Paul's total must be twice that of Peter.

A systematic list would look something like this:

Peter	Paul
(5, 6, 7) = 18	( _, _, _, _ ) = 36	Lowest for Peter.
(5, 6, 8) = 19	( _, _, _, _ ) = 38
...	...
( _, _, _ ) = 44	(15, 16, 17, 40) = 88	Highest for Paul.

Some results of this systematic search are:

• Lowest:
  Peter (5, 6, 9) = 20, Paul (7, 8, 10, 15) = 40, Mary (11, 14, 16, 17, 40) = 98
• Next Lowest:
  Peter (5, 6, 10) = 21, Paul (7, 8, 11, 16) = 42 or (7, 9, 11, 15) = 42
• Highest:
  Peter (10, 14, 17) = 41, Paul (11, 15, 16, 40) = 82

Total	2	3	4	5	6	7	8	9	10	11	12
No. of Ways	1	2	3	4	5	6	5	4	3	2	1

seems to have been built into the symmetry of the board. We could reason that every climber has an equal chance of reaching the second last row, then, since Climber 7 has more chance of being rolled, it has more chance of reaching the winning line first.

This intuitive reasoning seems reasonable, but it doesn't hold up when compared to the empirical data gained by playing the game many times. Climbers 2 and 12 win most often. How can this be explained?

In fact, the explanation is not easy, but an indication of the reason comes by considering a race between the 2 and the 7, ignoring all other rolls, where the 2 has to go one step and the 7 has to go six steps.

⁶/₇ x ⁶/₇ x ⁶/₇ x ⁶/₇ x ⁶/₇ x ⁶/₇ = 39.6%.

Examples

This arrangement for one colour scores 28 points.

This arrangement for one colour scores 40 points.

This is one arrangement for getting zero points.

1. 8       2. 10

3. A table of results shows:
   

   Length of Bed (L)	1	2	3	4	5
   No. of Tiles (T)	8	10	12	14	16

4. For any size bed the number of tiles is twice the length of the bed, plus another six. That is:

   T = 2 x L + 2

   It is twice because you need to surround both sides of the bed, then six extras allows for three at each end. If the bed was 100 units long, you would need 100 on one side, 100 on the other side, and 3 at each end. Therefore if N = 100, T = 206.
   There are other ways to 'see' the bed which will help you work out the number of tiles. Can you 'see' things a different way?

Extensions

Length of Bed (L)	23	38	100	247	1,396	40	140	982	5.5
No. of Tiles (T)	52	82	226	500	2,798	86	286	1,970	17

Place the central dip of the heart into this loop.   Bring the large hoop(on the end of the rod) through the dip, from position A to position B.   Lift the heart out of the loop to detach the pieces.

• There are no specific answers to this task.
• One analysis of a strategy, supposing the first two rolls were 5 and then 6, could be:
  5 is rolled first so place it in the centre position because it is in the middle of 1-10.

  5

  
  The next roll is a 6. After that there are 7 numbers left for the third roll. If 6 is placed in the units, four of these (4, 3, 2, 1) would be a poor result, and only three (7, 8, 9) are good. But if the 6 is placed in the hundreds:

  6

  5

  
  The score will 650⁺ whatever the next roll.

  Possible scores the 1st way are:
  156, 256, 356, 456, 756, 856, 956 (Total = 3792)
  Possible scores the 2nd way are:
  651, 652, 653, 654, 657, 658, 659 (Total = 4584)
  Which decision is better in the long run?

Calculating Magic Totals
For a 4 x 4 x 4 cube made from 64 cubes:

• Total from 1 to 64 = 65 x 32 = 2080
• There are 16 distinct rows.
• Magic Total = 2080 ÷ 16 = 130

No. of Sides (N)	3	4	5	6	7	8	9	10
Angle b/n Mirrors (A)	120	90	72	60	51.4	45	40	36

The formula is N x A = 360.

• If A = 100°, then N = 3.6, ie: you can see 3.6 sides.
• If A = 20°, then N = 18, ie: you should be able to see 18 sides.

• The answers to Question 1 are 2A, 1A and 3A
• The finished chart is:

  +	3A	3P	A	P
  3P	0	6P	2P	4P
  2A	5A	P	3A	A
  P	2A	4P	0	2P
  A	4A	2P	2A	0

• Count consistently from the left (or right).
• On move M, the Mth counter is unchanged and all others turn over.
• Continue in order until all counters have been unchanged once and...

Another approach is to draw a diagram like this:

1. 7 hours
2. In the 3rd hour (2 hours, 36 minutes)
3. In the 6th hour (5 hours, 50 minutes, ie: it gets out more quickly.)
4. (Up 3, Slip 2) or (Up 5, Slip 2) for example.

Extensions

1. 12 metres ... 5 hours
2. 15 metres ... 6 hours (5 hours 30 minutes)
3. 100 metres ... 49 hours (48 hours 24 minutes)
4. In the 5th hour (4 hours 30 minutes)
5. The important height (from which it can reach the top in the next hour) is 11 - x.
   It gains (x - y)/2 metres per hour.
   Hence it reaches 11 - x in (11 - x)/[(x - y)/2] hours, rounded up to the next hour.
   Then it reaches the top within the next hour, so the time is:
   (11 - x)/[(x - y)/2] - rounded up - plus 1.

A Size 5 would require 25 pieces.
Size 6 would need 36 pieces - use Size 3 as the unit and build a double Size 3.
A Size N Sphinx would need NxN pieces. This shows the quadratic relationship between length and area.

• It takes 31 moves to relocate the tower made of 5 pieces.
• For various numbers of pieces, the moves are:

  No. of Pieces	2	3	4	5	6	7	8	9	10	...	n
  No. of Moves	3	7	15	31	63	127	511	511	1023	...	2n - 1

• Notice that each answer is 'double the previous answer plus 1'.
  - First move the 'top' 3 pieces to another peg (7 moves).
  - Then move the base piece to the empty peg (1 move).
  - Then move the top 3 pieces back onto the base (7 moves).
• Similarly, to move a tower of 10, first move the top 9 pieces (511) moves, then the base piece (1 move), then the top 9 back again (511 moves), for a total of 1023.
• The formula m = 2ⁿ - 1 works for any number of discs.
  If n = 20, m = 1,048,575.
  Moving one disc per minute would take 104.02 weeks (or two years).
• A tower of 100 pieces would take hundreds of billions of years.
  (2,400,000,000,000,000,000,000,000 years)

• One solution is 3 + 7 - 4 = 6. Another is 6 + 3 - 5 = 4.
• There are 12 solutions altogether.
  (Or 24 if you decide equations like 3 + 7 - 4 = 6 and 7 + 3 - 4 = 6 are different.)
• Tiles 2, 3, 4, 5 and 6 give the same number of solutions.
• One strategy is to recognise pair totals such as 9 = (3 + 6) or (4 + 5).
  Arrangements of these pairs gives 8 solutions, ie: 3 + 6 - 4 = 5
  6 + 3 - 4 = 5
  3 + 6 - 5 = 4
  6 + 3 - 5 = 4
  4 + 5 - 3 = 6
  5 + 4 - 3 = 6
  4 + 5 - 6 = 3
  5 + 4 - 6 = 3
• Similarly, the pair totals of 10 = (3 + 7) or (4 + 6) and 11 = (4 + 7) or (5 + 6) also give 8 solutions each.
• Using 6 numbers gives a total of 28 combinations.
• Using 7 numbers gives 52 combinations.
• Using 8 numbers gives 88 combinations.

1. The most frequent result is 50¢. If you reached the position immediately below the 20¢ prize, the chance of going left or right is ²⁰/₃₆, compared to going ahead (¹⁶/₃₆).
2. The largest prize ($5) is placed where the same rolls are needed one after the other, ie: LLL or RRR. Theoretically, the chance of each of these is:
   ¹⁰/₃₆ x ¹⁰/₃₆ x ¹⁰/₃₆ = 2.14%
   Hence 4.28% of players can expect to win $5.
3. The operator in this setup would LOSE money.
   For 1000 players, the money in is $1000.
   The payout is $1,263.40 for a LOSS of $263.40 or 26.3%.
   If, for example, the $1 prize was changed to NIL, then the operator would just break even.

Extension
There are six pathways to win $4 on either side, hence 12 altogether, eg: RLLL or LULLL.