Number Play

Given: A row of numbers. A cell is a supercell if its number is greater than all its adjacent (left and right) neighbours.

Step 1 – List the numbers with their positions:
Position 1: 6828, Position 2: 670, Position 3: 9435, Position 4: 3780, Position 5: 3708, Position 6: 7308, Position 7: 8000, Position 8: 5583, Position 9: 52

Step 2 – Check each cell:
- 6828 (pos 1): only right neighbour is 670. Since $6828 > 670$, it IS a supercell. ✓
- 670 (pos 2): neighbours 6828 and 9435. Since $670 < 6828$, it is NOT a supercell.
- 9435 (pos 3): neighbours 670 and 3780. Since $9435 > 670$ and $9435 > 3780$, it IS a supercell. ✓
- 3780 (pos 4): neighbours 9435 and 3708. Since $3780 < 9435$, it is NOT a supercell.
- 3708 (pos 5): neighbours 3780 and 7308. Since $3708 < 7308$, it is NOT a supercell.
- 7308 (pos 6): neighbours 3708 and 8000. Since $7308 < 8000$, it is NOT a supercell.
- 8000 (pos 7): neighbours 7308 and 5583. Since $8000 > 7308$ and $8000 > 5583$, it IS a supercell. ✓
- 5583 (pos 8): neighbours 8000 and 52. Since $5583 < 8000$, it is NOT a supercell.
- 52 (pos 9): only left neighbour is 5583. Since $52 < 5583$, it is NOT a supercell.

Final Answer: The supercells are 6828, 9435, and 8000.

Given: The coloured (supercell) positions are positions 1 (5346), 4 (1258), and 8 (9635). We must fill the blanks with 4-digit numbers so that ONLY these three cells are supercells.

Concept: A cell is a supercell only if it is greater than both its neighbours. The non-coloured cells must NOT be supercells.

Strategy:
- Position 1 (5346) must be greater than position 2. So position 2 < 5346.
- Position 4 (1258) must be greater than positions 3 and 5. So positions 3 and 5 must be less than 1258, i.e., between 1000 and 1257.
- Position 8 (9635) must be greater than positions 7 and 9. So positions 7 and 9 < 9635.
- Non-coloured cells (2, 3, 5, 6, 7, 9) must not be supercells.

Sample filling:
- Position 2: 4000 (less than 5346 ✓; must not be supercell, so position 3 ≥ 4000)
- Position 3: 1100 (less than 1258 ✓; less than 4000, so position 2 = 4000 > 1100, position 2 is not a supercell since 4000 < 5346 ✓)
- Position 5: 1200 (less than 1258 ✓)
- Position 6: 3000 (greater than 1200, so position 5 is not a supercell ✓; position 6 must not be supercell, so position 7 ≥ 3000)
- Position 7: 8000 (greater than 3000, so position 6 is not a supercell ✓; position 7 must not be supercell, so 8000 < 9635 ✓ but position 6 = 3000 < 8000, so position 7 would be a supercell — adjust)
- Adjust position 7: 9000 (less than 9635 ✓). Position 6 = 3000 < 9000, so position 7 is not a supercell only if position 6 > 9000 — contradiction.
- Better: Position 6: 9100, Position 7: 9200 — but then 9200 > 9635 is false, and 9200 > 9100 makes position 7 a potential supercell.

Simpler valid filling:
| 5346 | 2000 | 1100 | 1258 | 1050 | 2500 | 9100 | 9635 | 1000 |

Verification:
- 5346: right neighbour 2000 < 5346 ✓ (supercell)
- 2000: neighbours 5346 > 2000, not supercell ✓
- 1100: neighbours 2000 > 1100, not supercell ✓
- 1258: neighbours 1100 < 1258 and 1050 < 1258 ✓ (supercell)
- 1050: neighbours 1258 > 1050, not supercell ✓
- 2500: neighbours 1050 < 2500 and 9100 > 2500, not supercell ✓
- 9100: neighbours 2500 < 9100 and 9635 > 9100, not supercell ✓
- 9635: neighbours 9100 < 9635 and 1000 < 9635 ✓ (supercell)
- 1000: neighbour 9635 > 1000, not supercell ✓

Answer: One valid filling is: | 5346 | 2000 | 1100 | 1258 | 1050 | 2500 | 9100 | 9635 | 1000 |

Given: A single row of 9 cells. Numbers must be between 100 and 1000, no repetitions. Maximise the number of supercells.

Concept: In a single row, a supercell must be greater than both its neighbours (or its only neighbour for end cells). To maximise supercells, we use an alternating high-low-high-low pattern.

Strategy: Place large numbers at odd positions and small numbers at even positions so that every odd-position number is greater than its neighbours.

Sample filling (alternating peaks):
| 900 | 101 | 800 | 102 | 700 | 103 | 600 | 104 | 500 |

Verification:
- 900 (pos 1): neighbour 101 < 900 ✓ supercell
- 101 (pos 2): neighbours 900 > 101, not supercell ✓
- 800 (pos 3): neighbours 101 < 800 and 102 < 800 ✓ supercell
- 102 (pos 4): neighbours 800 > 102, not supercell ✓
- 700 (pos 5): neighbours 102 < 700 and 103 < 700 ✓ supercell
- 103 (pos 6): neighbours 700 > 103, not supercell ✓
- 600 (pos 7): neighbours 103 < 600 and 104 < 600 ✓ supercell
- 104 (pos 8): neighbours 600 > 104, not supercell ✓
- 500 (pos 9): neighbour 104 < 500 ✓ supercell

Total supercells = 5 (positions 1, 3, 5, 7, 9)

Answer: | 900 | 101 | 800 | 102 | 700 | 103 | 600 | 104 | 500 |

From Question 3 above, using the alternating pattern, the supercells are at positions 1, 3, 5, 7, and 9.

Answer: 5 supercells.

Let us find the maximum number of supercells for different row sizes:

| Number of cells (n) | Maximum supercells |
|---|---|
| 1 | 1 |
| 2 | 1 |
| 3 | 2 |
| 4 | 2 |
| 5 | 3 |
| 6 | 3 |
| 7 | 4 |
| 8 | 4 |
| 9 | 5 |

Pattern observed: The maximum number of supercells for $n$ cells is $\left\lceil \dfrac{n}{2} \right\rceil$ (ceiling of $n/2$), i.e., $\dfrac{n+1}{2}$ when $n$ is odd, and $\dfrac{n}{2}$ when $n$ is even.

Strategy: Arrange numbers in an alternating high–low–high–low pattern. Place the largest numbers at positions 1, 3, 5, 7, 9 (odd positions) and the smallest numbers at positions 2, 4, 6, 8 (even positions). Every odd-position number will be greater than its even-position neighbours, making it a supercell. This gives the maximum possible supercells.

Answer: No, it is not possible to fill a table with distinct numbers such that there are no supercells.

Reason: Consider any table (row or grid) filled with distinct numbers. The cell containing the largest number in the entire table will always be greater than all its neighbours (since all numbers are distinct and it is the maximum). Therefore, it will always be a supercell. Hence, there will always be at least one supercell. It is impossible to have zero supercells when all numbers are distinct.

Largest number: Yes, the cell with the largest number will always be a supercell. Since all numbers are distinct, the largest number is greater than every other number in the table, and in particular greater than all its neighbours. So it is always a supercell.

Smallest number: No, the cell with the smallest number can never be a supercell (when all numbers are distinct). A supercell must be greater than all its neighbours. But the smallest number is less than every other number, including all its neighbours. So it cannot be a supercell.

*Exception note:* If a table has only one cell, that single number is both the largest and smallest, and it is trivially a supercell (no neighbours to compare with).

Strategy: Place the second largest number next to the largest number. Then the second largest number has the largest number as a neighbour, and since the largest > second largest, the second largest is not a supercell.

Example (row of 5 cells):
| 300 | 500 | 490 | 200 | 400 |

Here:
- Largest = 500 (pos 2): neighbours 300 and 490, both less than 500 → supercell ✓
- Second largest = 490 (pos 3): neighbours 500 and 200. Since $490 < 500$, it is not a supercell ✓
- 400 (pos 5): only neighbour 200 < 400 → supercell
- 300 (pos 1): only neighbour 500 > 300 → not a supercell

Answer: | 300 | 500 | 490 | 200 | 400 | — the second largest number (490) is not a supercell.

Yes, it is possible.

Strategy:
- Place the second largest next to the largest so it is not a supercell.
- Place the second smallest between two numbers that are both smaller than it (i.e., only the smallest number is its neighbour, or arrange so its neighbours are all smaller).

Example (row of 5 cells):
| 101 | 100 | 500 | 490 | 200 |

Numbers in order: 100 (smallest), 101 (second smallest), 200, 490 (second largest), 500 (largest).

Check:
- 101 (pos 1, second smallest): only neighbour is 100. Since $101 > 100$ → supercell ✓
- 100 (pos 2, smallest): neighbours 101 and 500, both > 100 → not a supercell ✓
- 500 (pos 3, largest): neighbours 100 and 490, both < 500 → supercell ✓
- 490 (pos 4, second largest): neighbours 500 and 200. Since $490 < 500$ → not a supercell ✓
- 200 (pos 5): only neighbour 490 > 200 → not a supercell ✓

Answer: Yes, it is possible. Example: | 101 | 100 | 500 | 490 | 200 |

This is an open-ended creative question. Here are some example variations students can create:

Variation 1: Fill a 2-row × 5-column grid with numbers 1–10 (no repetition) such that exactly 3 cells are supercells (neighbours include top, bottom, left, right).

Variation 2: Fill a row of 7 cells with numbers between 50 and 150 (no repetition) such that the supercells are exactly at positions 2, 4, and 6.

Variation 3: Fill a row of 6 cells such that the number of supercells equals the number of non-supercells.

Students should create their own puzzle, verify the solution, and then present only the empty grid (with coloured cells marked) to classmates as a challenge.

Given: All numbers use digits 1, 0, 6, 3, 9 in some order (each digit used exactly once). Coloured cells (supercells) must be greater than all neighbours; non-coloured cells must not be supercells.

From Table 1, the coloured positions correspond to the same pattern. The coloured cells in Table 2 are at positions that mirror Table 1's supercells: (1,2)=96301, (2,3)=60319, (3,2) and (3,3)=60193 area, etc. Based on the given numbers and the constraint:

The numbers already given are: 96,301 | 36,109 | 13,609 | 60,319 | 19,306 | 60,193 | 10,963.

Other possible 5-digit numbers using digits {0,1,3,6,9}: 90,631; 30,916; 10,369; 63,190; 39,601; 16,390; 93,610; 31,096; 19,630; 61,390; 90,163; 13,069; etc.

For the blanks, we fill non-coloured cells with numbers smaller than their coloured neighbours:

Sample complete table:
| 30,916 | 96,301 | 36,109 | 13,069 |
| 19,630 | 13,609 | 60,319 | 19,306 |
| 31,096 | 90,163 | 60,193 | 16,390 |
| 39,601 | 10,963 | 13,906 | 10,369 |

(Coloured/supercells shown in bold: 96,301; 60,319; 90,163 — these are greater than all their neighbours.)

(i) The biggest number in the table: $96,301$ (or $90,163$ depending on filling — the largest among all entries). With the filling above, the largest is 96,301.

(ii) The smallest even number: Even numbers end in 0. From the table: 30,916 ends in 6 (even ✓), 19,630 ends in 0 (even ✓), 31,096 ends in 6 (even ✓), 39,601 ends in 1 (odd), 13,069 ends in 9 (odd), 16,390 ends in 0 (even ✓), 10,369 ends in 9 (odd), 13,906 ends in 6 (even ✓), 10,963 ends in 3 (odd), 60,319 ends in 9 (odd), 60,193 ends in 3 (odd), 36,109 ends in 9 (odd), 19,306 ends in 6 (even ✓), 13,609 ends in 9 (odd), 96,301 ends in 1 (odd), 90,163 ends in 3 (odd).

Even numbers: 30,916; 19,630; 31,096; 16,390; 13,906; 19,306. The smallest is 13,906.

(iii) The smallest number greater than 50,000: From all entries, numbers > 50,000 are: 96,301; 60,319; 90,163; 60,193. The smallest among these is 60,193.

After filling, place commas after the thousands digit:
96,301 | 36,109 | 13,609 | 60,319 | 19,306 | 60,193 | 10,963 | 30,916 | 19,630 | 31,096 | 90,163 | 16,390 | 39,601 | 13,906 | 10,369 | 13,069.

Since the number line images are not available in the OCR text, the general method is described:

Method to identify numbers on a number line:
1. Note the two labelled endpoints or marked values.
2. Count the number of equal divisions between them.
3. Calculate the value of each division = $\dfrac{\text{difference between endpoints}}{\text{number of divisions}}$.
4. Add/subtract this value step by step to find each marked position.

General approach for labelling:
- Find the scale (each small division's value).
- Label each tick mark by adding the scale value to the previous mark.
- Circle the leftmost (smallest) value and box the rightmost (largest) value in each number line.

Example: If a number line goes from 0 to 10,000 with 10 equal divisions, each division = 1,000. Marks at 0, 1000, 2000, … 10,000.

Students should apply this method to each of the four number lines (a), (b), (c), (d) shown in their textbook figures.

Concept: Given 4 distinct digits, the largest 4-digit number is formed by arranging digits in descending order, and the smallest by arranging in ascending order (with the smallest non-zero digit first if 0 is included).

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Part a: Difference > 5085

Choose digits: 1, 5, 8, 9
- Largest: 9851
- Smallest: 1589
- Difference: $9851 - 1589 = 8262 > 5085$ ✓

Answer: Digits 1, 5, 8, 9 give difference = 8262 > 5085.

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Part b: Difference < 5085

Choose digits: 5, 6, 7, 8
- Largest: 8765
- Smallest: 5678
- Difference: $8765 - 5678 = 3087 < 5085$ ✓

Answer: Digits 5, 6, 7, 8 give difference = 3087 < 5085.

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Part c: Sum > 9779

Choose digits: 2, 7, 8, 9
- Largest: 9872
- Smallest: 2789
- Sum: $9872 + 2789 = 12661 > 9779$ ✓

Answer: Digits 2, 7, 8, 9 give sum = 12,661 > 9779.

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Part d: Sum < 9779

Choose digits: 1, 2, 3, 4
- Largest: 4321
- Smallest: 1234
- Sum: $4321 + 1234 = 5555 < 9779$ ✓

Answer: Digits 1, 2, 3, 4 give sum = 5555 < 9779.

What is a palindrome? A number that reads the same forwards and backwards.

Smallest 5-digit palindrome:
A 5-digit palindrome has the form $\overline{abcba}$. To minimise, choose $a=1$ (cannot be 0), $b=0$, $c=0$.
Smallest 5-digit palindrome = 10001.

Largest 5-digit palindrome:
To maximise, choose $a=9$, $b=9$, $c=9$.
Largest 5-digit palindrome = 99999.

Sum:
$$10001 + 99999 = 110000$$

Difference:
$$99999 - 10001 = 89998$$

Answer: Sum = 1,10,000 and Difference = 89,998.

Given: Current time = 10:01 (which is itself a palindrome: 1-0-0-1 reads same forwards and backwards).

Finding the next palindromic time after 10:01:

Palindromic times on a 12-hour clock have the form where the digits read the same forwards and backwards.

For a time displayed as H:MM or HH:MM:
- 10:01 → digits 1,0,0,1 ✓ palindrome
- Next: 11:11 → digits 1,1,1,1 ✓ palindrome

Time from 10:01 to 11:11:
$$11{:}11 - 10{:}01 = 1 \text{ hour } 10 \text{ minutes} = 70 \text{ minutes}$$

The one after 11:11:
- 12:21 → digits 1,2,2,1 ✓ palindrome

Time from 11:11 to 12:21:
$$12{:}21 - 11{:}11 = 1 \text{ hour } 10 \text{ minutes} = 70 \text{ minutes}$$

Answer:
- Next palindromic time after 10:01 is 11:11, which is 70 minutes later.
- The one after that is 12:21, which is another 70 minutes later.

Kaprekar Constant for 4-digit numbers = 6174.

Rule: Arrange the 4 digits in descending order to form the largest number, and ascending order to form the smallest number. Subtract the smaller from the larger. Repeat.

Starting with 5683:

Round 1:
- Digits: 5, 6, 8, 3
- Largest: 8653, Smallest: 3568 (Note: arrange as 3568)
- $8653 - 3568 = 5085$

Round 2:
- Digits: 5, 0, 8, 5
- Largest: 8550, Smallest: 0558 = 0558
- $8550 - 0558 = 7992$

Round 3:
- Digits: 7, 9, 9, 2
- Largest: 9972, Smallest: 2799
- $9972 - 2799 = 7173$

Round 4:
- Digits: 7, 1, 7, 3
- Largest: 7731, Smallest: 1377
- $7731 - 1377 = 6354$

Round 5:
- Digits: 6, 3, 5, 4
- Largest: 6543, Smallest: 3456
- $6543 - 3456 = 3087$

Round 6:
- Digits: 3, 0, 8, 7
- Largest: 8730, Smallest: 0378 = 0378
- $8730 - 0378 = 8352$

Round 7:
- Digits: 8, 3, 5, 2
- Largest: 8532, Smallest: 2358
- $8532 - 2358 = 6174$ ✓

Answer: The number 5683 takes 7 rounds to reach the Kaprekar constant 6174.

These are estimation questions. Answers will vary based on individual circumstances. Here are reasonable estimates:

a. From seat to classroom door:
A classroom is roughly 8–10 metres long. Average step length ≈ 0.5 m.
Estimated steps: $\dfrac{8}{0.5} = 16$ to $\dfrac{10}{0.5} = 20$ steps.
Estimate: about 15–20 steps.

b. Across the school ground:
A typical school ground is about 50–100 metres across.
Estimated steps: $\dfrac{50}{0.5} = 100$ to $\dfrac{100}{0.5} = 200$ steps.
Estimate: about 100–200 steps.

c. From classroom door to school gate:
Distance might be about 100–200 metres.
Estimate: about 200–400 steps.

d. From school to home:
This varies greatly. If home is 1 km away: $\dfrac{1000}{0.5} = 2000$ steps.
Estimate: about 1000–5000 steps (depending on distance).

Blinks:
- Average blink rate ≈ 15–20 times per minute.

a. In a minute: about 15–20 blinks
b. In an hour: $15 \times 60 = 900$ to $20 \times 60 = 1200$ blinks ≈ about 1000 blinks
c. In a day (16 waking hours): $1000 \times 16 = 16,000$ blinks ≈ about 15,000–16,000 blinks

Breaths:
- Average breathing rate ≈ 15–20 breaths per minute.

a. In a minute: about 15–20 breaths
b. In an hour: $15 \times 60 = 900$ to $20 \times 60 = 1200$ ≈ about 1000 breaths
c. In a day (24 hours): $1000 \times 24 = 24,000$ ≈ about 20,000–25,000 breaths