Number Play — NCERT Solutions

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

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).