The robot IMSO–20 moves from the left side of the $5\times7$ board to the right side as shown on the left. It collects the scores on all cells that it passes. At the beginning, it may start in any cell of the first column from the left. Then it must move to the next column in the same row, one row above, or one row below, as shown on the right. Sketch a path for each route that gives (a) the lowest score (2 marks), (b) the highest score (4 marks).
Each box in the diagram contains a multiple of five from 5 to 60. List ALL possible arrangements of the numbers if:
(a) 15, 40 and 55, in that order, occupy successive boxes.
(b) 5 is in an odd-numbered box.
(c) The number in box 6 is larger than that in box 8.
(d) The difference between the numbers in boxes 8 and 7 is 5.
(e) Box 2 contains 20.
(f) The number in box 5 ends in zero.
(g) The number in box 4 is twice the number in box 12.
(h) 25 is in a box numbered two lower than the box containing 35.
A $5\times7$ board is to be partitioned into $1\times1$, $2\times2$ and $3\times3$ sub-boards. What is the smallest number of $1\times1$ sub-boards? Sketch one possible diagram. (2 marks)
Now use an $11\times11$ board under the same conditions. What is the smallest number of $1\times1$ sub-boards? Sketch one possible diagram. (4 marks)
**Correct answers with incorrect drawings will be given 0 marks.**
The model robot has six chambers: head (HD), raised right arm (RA), left arm (LA), body (BD), right leg (RL), left leg (LL). Insert one control block into each chamber, chosen without repetition from 1,2,…,8; the number is its weight. The robot is operational only when:
- The body is the heaviest and the head the lightest of the chosen blocks.
- Balance: right arm plus right leg equals left arm plus left leg.
- The right arm is heavier than the left arm.
- Stability: the right leg is heavier than the right arm; the left leg is heavier than the left arm and heavier than 5.
One example is shown on the right. Find ALL other operational arrangements.
Place each number from 1 to 10 without repetition in the empty circles so that the sum along every straight line is 30. Fill in the diagram for ALL possible arrangements.
A $5\times5$ square grid with a unit-square hole at the center can be cut along grid lines into 4 congruent pieces (the same shape and size). The central hole allows each piece to have area 6 square units.
The following is one way to do this. Sketch ALL other possible ways, excluding the example. Two ways are equivalent if one can be obtained from the other by a rotation or reflection.
