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 &lt; 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 &gt; 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 &lt; 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 &gt; 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 &lt; 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 &gt; 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 &lt; 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

a. A few thousand in number (examples):
- Pages in all the books in a school library
- Bricks in a classroom wall
- Words in a chapter of a textbook
- Leaves on a small tree
- Tiles on the school floor

b. More than ten thousand in number (examples):
- Grains of rice in a 1 kg bag (approximately 25,000–30,000 grains)
- Hairs on a human head (approximately 1,00,000)
- Words in a full textbook
- Blades of grass in a school ground
- Grains of sand in a small sandbox

Method of estimation:
- A typical page has about 200–300 words.
- A Class 6 Maths textbook has about 200–250 pages.
- Estimated words: $250 \times 250 = 62,500$

Answer: a. More than 5000 ✓

The number of words in a maths textbook is far more than 5000 — it is likely around 50,000–70,000 words.

Method: This depends on the school size and location.
- A school with 500–1000 students in an urban area: many students use buses.
- Roughly 30–50% may travel by bus.
- For a school of 600 students: $600 \times 0.4 = 240$ students.

Answer: For a medium to large school, likely a. More than 200.

However, for a small school (under 400 students) or a school where most students walk, the answer could be b. Less than 200. Students should estimate based on their own school.

Estimation:
- Milk (500 ml): approximately ₹25–30
- 3 types of fruit (e.g., banana, apple, mango): approximately ₹20 + ₹40 + ₹30 = ₹90
- Custard powder: approximately ₹20–30
- Sugar: approximately ₹5

Total estimated cost: ₹25 + ₹90 + ₹25 + ₹5 = ₹145–₹170

Answer: No, ₹100 is likely not enough. The actual cost would be closer to ₹150–₹200 depending on the quantity and type of fruits chosen. Roshan's estimate of ₹100 is an underestimate.

Method: Using the map of India.
- Gandhinagar is in western India (Gujarat), approximately at 23°N, 72°E.
- Kohima is in northeastern India (Nagaland), approximately at 25°N, 94°E.
- Horizontal distance (longitude difference): $94° - 72° = 22°$. At this latitude, 1° longitude ≈ 100 km. So horizontal ≈ 2200 km.
- Vertical distance (latitude difference): $25° - 23° = 2°$. 1° latitude ≈ 111 km. So vertical ≈ 222 km.
- Straight-line distance ≈ $\sqrt{2200^2 + 222^2} \approx \sqrt{4840000 + 49284} \approx \sqrt{4889284} \approx 2211$ km.

Answer: The straight-line distance is approximately 2200 km. By road, it would be considerably more (around 3000–3500 km).

Estimation:
- Sheetal is in Grade 6, so she has completed about 6 years of schooling (Grades 1–5, now in Grade 6 ≈ half year done). Let's say approximately 5.5 years.
- School days per year: approximately 200 days (accounting for holidays, weekends).
- School hours per day: approximately 6 hours.

Total hours:
$$5.5 \times 200 \times 6 = 5.5 \times 1200 = 6600 \text{ hours}$$

Answer: No, 13,000 hours is an overestimate. A more realistic estimate is about 6,000–7,000 hours. Sheetal's estimate of 13,000 hours would require about 10–11 years of schooling at 6 hours/day, 200 days/year, which does not match Grade 6.

Average walking speed: approximately 4–5 km/h.

a. To a favourite nearby place:
- Assume the place is about 1–2 km away.
- Time = $\dfrac{1.5 \text{ km}}{4.5 \text{ km/h}} \approx 0.33$ hours ≈ 20 minutes.

b. To a neighbouring state's capital:
- Example: Delhi to Jaipur (Rajasthan) ≈ 280 km by road.
- Time = $\dfrac{280}{4.5} \approx 62$ hours ≈ about 2.5 days of continuous walking (or about 8–10 days walking 8 hours/day).

c. Southernmost to northernmost point of India:
- Kanyakumari (southernmost) to Ladakh/Kashmir (northernmost) ≈ 3,200 km.
- Time = $\dfrac{3200}{4.5} \approx 711$ hours ≈ about 30 days of continuous walking, or roughly 3–4 months walking 8 hours/day.

This is an open-ended creative question. Here are some sample estimation questions students can create:

1. How many steps would it take to walk around the boundary of your school?
2. How many glasses of water do you drink in a week?
3. How many books are there in your school library?
4. How many heartbeats does a person have in one year?
5. How many rotis/chapatis does your family eat in a month?
6. How many seconds have you been alive?
7. How many tiles are there on the floor of your classroom?

Students should estimate the answers first, then try to find the actual values and compare.

Given grid:
$$\begin{array}{|c|c|c|}\hline 16200 &amp; 39344 &amp; 29765 \\\hline 23609 &amp; 62871 &amp; 45306 \\\hline 19381 &amp; 50319 &amp; 38408 \\\hline\end{array}$$

Step 1 – Find the current supercell:
Check 62,871 (centre): neighbours are 39344, 45306, 23609, 50319. All are less than 62871. So 62,871 is a supercell. ✓

Check others: 39344 has neighbour 62871 > 39344, not supercell. 50319 has neighbour 62871 > 50319, not supercell. 29765 has neighbour 39344 > 29765, not supercell. 45306 has neighbour 62871 > 45306, not supercell. 16200 has neighbour 23609 > 16200, not supercell. 23609 has neighbour 62871 > 23609, not supercell. 19381 has neighbour 23609 > 19381, not supercell. 38408 has neighbour 50319 > 38408, not supercell.

So currently only 62,871 is a supercell. ✓

Step 2 – Find which digit swap creates 4 supercells:
For 4 supercells in a 3×3 grid, we want the corner cells and/or edge cells to become supercells. The four corners (16200, 29765, 19381, 38408) each have only 2 neighbours, making them easier to be supercells.

If we can make all 4 corners supercells:
- Corner 16200 needs to be > 39344 and > 23609 → needs to be > 39344. Currently 16200 < 39344.
- If we swap digits in 39344 to make it smaller, say swap 3 and 9: 93344 — that makes it larger. Swap 9 and 3 in position: 39344 → 33944? No, we swap two digits of one number.

Try swapping digits in 62871:
Swap 6 and 2: 26871. Now check if this creates more supercells.
- 26871 (centre): neighbours 39344 > 26871, so 26871 is NOT a supercell.
- 39344: neighbours 16200, 29765, 26871, 23609 (wait, 39344 is top-middle). Neighbours: left=16200, right=29765, bottom=26871. Since 39344 > all three → supercell ✓
- 29765 (top-right): neighbours 39344 > 29765, not supercell.

This doesn't give 4 supercells easily.

Try swapping digits in 50319:
Swap 5 and 0: 00319 = 319 (not 5-digit). Not valid.
Swap 5 and 3: 30519. Now 30519 < 62871 ✓, and check if bottom-middle becomes non-supercell.

Systematic approach — swap digits in 62,871 to make it 26,871:
With 26871 in centre:
- 39344 (top-mid): neighbours 16200, 29765, 26871. 39344 > all → supercell ✓
- 23609 (mid-left): neighbours 16200, 26871, 19381. 26871 > 23609, not supercell.
- 50319 (bot-mid): neighbours 19381, 38408, 26871. 50319 > all → supercell ✓
- 29765 (top-right): neighbours 39344 > 29765, not supercell.
- 45306 (mid-right): neighbours 29765, 26871, 38408. 45306 > all → supercell ✓
- 16200 (top-left): neighbours 39344 > 16200, not supercell.
- 19381 (bot-left): neighbours 23609 > 19381, not supercell.
- 38408 (bot-right): neighbours 45306 > 38408, not supercell.
- 26871 (centre): neighbours 39344 > 26871, not supercell.

Supercells: 39344, 50319, 45306 = 3 supercells. Not 4.

Try swapping 6 and 1 in 62,871 → 12,876... wait, swap positions of digits 6 and 1: 62871 has digits 6,2,8,7,1. Swap 6 and 1: 12876.

With 12876 in centre:
- 39344: neighbours 16200, 29765, 12876. 39344 > all → supercell ✓
- 23609: neighbours 16200, 12876, 19381. 23609 > all → supercell ✓
- 50319: neighbours 19381, 38408, 12876. 50319 > all → supercell ✓
- 45306: neighbours 29765, 12876, 38408. 45306 > all → supercell ✓
- 16200: neighbours 39344 > 16200, not supercell.
- 29765: neighbours 39344 > 29765, not supercell.
- 19381: neighbours 23609 > 19381, not supercell.
- 38408: neighbours 50319 > 38408, not supercell.
- 12876: neighbours 39344 > 12876, not supercell.

Supercells: 39344, 23609, 50319, 45306 = 4 supercells ✓

Answer: Swap the digits 6 and 1 in the number 62,871 to get 12,876. This results in exactly 4 supercells: 39,344; 23,609; 50,319; and 45,306.

Note: The answer depends on the student's year of birth. We demonstrate the method using the year 2012 as an example.

Kaprekar Constant = 6174 (for 4-digit numbers).

Starting with 2012:

Round 1: Digits 2,0,1,2 → Largest: 2210, Smallest: 0122=0122
$2210 - 0122 = 2088$

Round 2: Digits 2,0,8,8 → Largest: 8820, Smallest: 0288=0288
$8820 - 0288 = 8532$

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

Answer for 2012: 3 rounds.

Students should apply the same Kaprekar process to their own year of birth and count the rounds needed to reach 6174.

Condition: 5-digit numbers between 35,000 and 75,000 with ALL digits odd.

Odd digits: 1, 3, 5, 7, 9.

The number must be between 35,000 and 75,000, so the ten-thousands digit can be 3, 5, or 7 (but must be odd — all three are odd ✓).

Largest number:
To maximise, use the largest odd digits. Ten-thousands digit can be at most 7 (since number ≤ 75,000, and all digits odd, 79999 > 75000 — check: 79,999 > 75,000, so not valid if strictly less than 75,000).
- If ten-thousands digit = 7: number must be < 75,000. So thousands digit ≤ 4. But digits must be odd, so thousands digit ≤ 3 (odd digits ≤ 4 are 1, 3). Largest with ten-thousands=7: 73,999? Wait, thousands digit must be odd and ≤ 4, so thousands digit = 3. Then remaining digits = 9,9,9. Number = 73,999.
- Check: 73,999 < 75,000 ✓, all digits (7,3,9,9,9) are odd ✓.
- Can we get 75,xxx? 75,000 — digits 7,5,0,0,0 — 0 is not odd. Not valid.
- 75,111 > 75,000 and not in range (must be strictly less than 75,000 or the problem says "between" which typically means exclusive).

Largest = 73,999

Smallest number:
To minimise, use smallest odd digits. Ten-thousands digit = 3 (smallest odd digit that keeps number ≥ 35,000).
- Ten-thousands = 3, thousands digit must make number ≥ 35,000. So thousands digit ≥ 5. Smallest odd digit ≥ 5 is 5. Number = 35,111.
- Check: 35,111 > 35,000 ✓, all digits (3,5,1,1,1) are odd ✓.

Smallest = 35,111

Closest to 50,000:
We need a 5-digit number with all odd digits, between 35,000 and 75,000, closest to 50,000.
- 50,000 has even digits (5,0,0,0,0) — not valid.
- Try 51,111: all digits odd (5,1,1,1,1) ✓. Distance from 50,000 = 1,111.
- Try 49,999: digits 4,9,9,9,9 — 4 is even. Not valid.
- Try 51,111 vs 53,111 vs 55,111: 51,111 is closest to 50,000.
- Can we get closer? 51,111 - 50,000 = 1,111. What about numbers just above 50,000 with all odd digits?
- 50,001: digit 0 is even. Not valid.
- 51,111 seems to be the closest valid number above 50,000.
- What about below 50,000? Largest odd-digit number below 50,000: ten-thousands = 3 or 5 (but 5 gives 5x,xxx ≥ 50,000). Ten-thousands = 3: largest is 39,999. Distance = 50,000 - 39,999 = 10,001. Much farther.
- Ten-thousands = 5, thousands digit must give number < 50,000: thousands digit = 0 (not odd). So no valid number with ten-thousands=5 that is less than 50,000 with all odd digits.

Closest to 50,000 = 51,111

Answers:
- Largest: 73,999
- Smallest: 35,111
- Closest to 50,000: 51,111

Estimation:
- Weekends (Saturdays + Sundays): 52 weeks × 2 = 104 days
- Summer vacation: approximately 45 days
- Winter/Christmas break: approximately 10 days
- Diwali/Puja vacation: approximately 5 days
- National holidays (Republic Day, Independence Day, Gandhi Jayanti): 3 days
- Other gazetted/festival holidays: approximately 10 days

Estimated total: $104 + 45 + 10 + 5 + 3 + 10 = 177$ days

Estimate: approximately 170–180 holidays per year.

Students should check their school calendar for the exact number and compare with their estimate.

Mug:
- A standard drinking mug holds about 200–300 mL.
- Estimate: about 0.25 litres (250 mL)

Bucket:
- A standard household bucket holds about 10–15 litres.
- Estimate: about 10–15 litres

Overhead tank:
- A typical household overhead water tank holds about 500–1000 litres.
- For a family of 4–5 people, a common size is 500 litres or 1000 litres.
- Estimate: about 500–1000 litres

Given: We need one 5-digit number + two 3-digit numbers = 18,670.

Method: Choose two 3-digit numbers, then find the 5-digit number.

Let the two 3-digit numbers be 350 and 420.
$$350 + 420 = 770$$
$$\text{5-digit number} = 18670 - 770 = 17900$$

Verification: $17900 + 350 + 420 = 17900 + 770 = 18670$ ✓

Answer: $17900 + 350 + 420 = 18,670$

*(Other valid answers are possible, e.g., $17500 + 600 + 570 = 18670$)*

Chosen number: 300 (between 210 and 390)

Pattern (like Section 3.9 — consecutive or arithmetic sequences):

Using an arithmetic pattern:
$$50 + 55 + 60 + 65 + 70 = 300$$
Check: $50+55=105$, $105+60=165$, $165+65=230$, $230+70=300$ ✓

Another pattern:
$$1 + 2 + 3 + \ldots + n$$
For $n=24$: $\dfrac{24 \times 25}{2} = 300$ ✓

So: $1+2+3+4+\ldots+24 = 300$

Answer: One valid pattern is $50 + 55 + 60 + 65 + 70 = 300$, or equivalently $1 + 2 + 3 + \cdots + 24 = 300$.

Powers of 2: $1, 2, 4, 8, 16, 32, 64, 128, \ldots$ i.e., $2^0, 2^1, 2^2, 2^3, \ldots$

Collatz Rule: If even → divide by 2; if odd → multiply by 3 and add 1.

Key observation: Every power of 2 is even (except $2^0 = 1$).

For any power of 2, say $2^n$ (where $n \geq 1$):
- $2^n$ is even → divide by 2 → $2^{n-1}$
- $2^{n-1}$ is even → divide by 2 → $2^{n-2}$
- Continue dividing by 2 each step.
- After $n$ steps: $2^n \to 2^{n-1} \to 2^{n-2} \to \cdots \to 2^1 = 2 \to 2^0 = 1$

The sequence always reaches 1 in exactly $n$ steps (just keep halving).

For $2^0 = 1$: Already at 1.

Conclusion: For all powers of 2, the Collatz sequence simply halves at each step and reaches 1 in a finite number of steps. This is why the Collatz conjecture is definitely true for all starting numbers that are powers of 2.

Starting number: 100

Apply the Collatz rule: even → ÷2; odd → ×3+1.

$$100 \xrightarrow{\div 2} 50 \xrightarrow{\div 2} 25 \xrightarrow{\times 3+1} 76 \xrightarrow{\div 2} 38 \xrightarrow{\div 2} 19$$
$$19 \xrightarrow{\times 3+1} 58 \xrightarrow{\div 2} 29 \xrightarrow{\times 3+1} 88 \xrightarrow{\div 2} 44 \xrightarrow{\div 2} 22$$
$$22 \xrightarrow{\div 2} 11 \xrightarrow{\times 3+1} 34 \xrightarrow{\div 2} 17 \xrightarrow{\times 3+1} 52 \xrightarrow{\div 2} 26$$
$$26 \xrightarrow{\div 2} 13 \xrightarrow{\times 3+1} 40 \xrightarrow{\div 2} 20 \xrightarrow{\div 2} 10 \xrightarrow{\div 2} 5$$
$$5 \xrightarrow{\times 3+1} 16 \xrightarrow{\div 2} 8 \xrightarrow{\div 2} 4 \xrightarrow{\div 2} 2 \xrightarrow{\div 2} 1$$

Full sequence:
100 → 50 → 25 → 76 → 38 → 19 → 58 → 29 → 88 → 44 → 22 → 11 → 34 → 17 → 52 → 26 → 13 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1

Total steps: 25 steps

Conclusion: Yes, the Collatz Conjecture holds for 100. The sequence reaches 1 after 25 steps.

Game: Starting from 0, players alternately add 1, 2, or 3. First to reach exactly 22 wins.

Analysis (working backwards):

A player wins if they can reach 22. A player loses if, no matter what they add (1, 2, or 3), the opponent can reach 22.

Key insight: If it is your turn and the current total is 18, 19, 20, or 21, you can reach 22 by adding 4, 3, 2, or 1 respectively. So reaching 22 is a win.

Losing positions (positions where the player whose turn it is will lose with optimal opponent play):
- If current total = 18: opponent can always reach 22 (add 4 — wait, max add is 3). If total = 18, you can add 1→19, 2→20, 3→21. Opponent then adds 3, 2, or 1 to reach 22. So 18 is a losing position.
- If total = 14: whatever you add (1, 2, or 3), you reach 15, 16, or 17. Opponent can always reach 18. So 14 is a losing position.
- Pattern: losing positions are $18, 14, 10, 6, 2$ (each 4 apart, going backwards from 18).
- Check 2: if total = 2, you add 1→3, 2→4, or 3→5. Opponent reaches 6 (adds 3, 2, or 1). So 2 is a losing position.
- Check 0 (start): if total = 0, first player adds 1→1, 2→2, or 3→3.
- If first player adds 2, total = 2, which is a losing position for the second player!

Winning strategy for Player 1:
- First move: say 2 (reach total = 2, a losing position for Player 2).
- After each move by Player 2, Player 1 should add enough to reach the next multiple in the sequence: 2, 6, 10, 14, 18, 22.
- Rule: After Player 2 adds $k$, Player 1 adds $(4-k)$ to always reach the next target.

Winning positions to hit: 2 → 6 → 10 → 14 → 18 → 22.

Answer: Player 1 always wins with the correct strategy. Player 1 should first say 2, then always add enough to reach 6, 10, 14, 18, and finally 22 (i.e., always add $4 - k$ where $k$ is what Player 2 just added).