We the Travellers—I

Given: 456 → 567 → 678 → □ → □ → □

Pattern: Each number increases by 111.
$$456 + 111 = 567,\quad 567 + 111 = 678$$
$$678 + 111 = 789,\quad 789 + 111 = 900,\quad 900 + 111 = 1{,}011$$

Answer: 456 → 567 → 678 → 789 → 900 → 1,011

Given: 1,050 → □ → 3,150 → 4,200 → □ → □ → □

Pattern: Each number increases by 1,050.
$$3{,}150 - 1{,}050 = 2{,}100 \Rightarrow \text{second term} = 2{,}100$$
$$1{,}050 + 1{,}050 = 2{,}100 \checkmark$$
$$4{,}200 + 1{,}050 = 5{,}250,\quad 5{,}250 + 1{,}050 = 6{,}300,\quad 6{,}300 + 1{,}050 = 7{,}350$$

Answer: 1,050 → 2,100 → 3,150 → 4,200 → 5,250 → 6,300 → 7,350

Given: 5,501 → 6,401 → 7,301 → □ → □ → □

Pattern: Each number increases by 900.
$$6{,}401 - 5{,}501 = 900,\quad 7{,}301 - 6{,}401 = 900$$
$$7{,}301 + 900 = 8{,}201,\quad 8{,}201 + 900 = 9{,}101,\quad 9{,}101 + 900 = 10{,}001$$

Answer: 5,501 → 6,401 → 7,301 → 8,201 → 9,101 → 10,001

Given: 10,100 → 10,200 → 10,300 → □ → □ → □ and □ ← 10,900 ← □

Pattern: Each number increases by 100 going right; decreases by 100 going left.
$$10{,}300 + 100 = 10{,}400,\quad 10{,}400 + 100 = 10{,}500,\quad 10{,}500 + 100 = 10{,}600$$
For the reverse chain ending at 10,900:
$$10{,}900 - 100 = 10{,}800 \Rightarrow \text{box before } 10{,}900 = 10{,}800$$
$$10{,}800 - 100 = 10{,}700 \Rightarrow \text{first box} = 10{,}700$$

Answer (forward): 10,100 → 10,200 → 10,300 → 10,400 → 10,500 → 10,600
Answer (reverse): 10,700 ← 10,800 ← 10,900

Given: 10,105 → 10,125 → □ → □ → □ and □ ← □ ← □

Pattern: Each number increases by 20 going right.
$$10{,}125 - 10{,}105 = 20$$
$$10{,}125 + 20 = 10{,}145,\quad 10{,}145 + 20 = 10{,}165,\quad 10{,}165 + 20 = 10{,}185$$
Reverse (continuing the sequence backward from 10,105):
$$10{,}105 - 20 = 10{,}085,\quad 10{,}085 - 20 = 10{,}065,\quad 10{,}065 - 20 = 10{,}045$$

Answer (forward): 10,105 → 10,125 → 10,145 → 10,165 → 10,185
Answer (reverse): 10,045 ← 10,065 ← 10,085

Given: 10,992 → 10,993 → □ → □ → □ and □ ← □ ← □

Pattern: Each number increases by 1 going right.
$$10{,}993 + 1 = 10{,}994,\quad 10{,}994 + 1 = 10{,}995,\quad 10{,}995 + 1 = 10{,}996$$
Reverse (going left from 10,992):
$$10{,}992 - 1 = 10{,}991,\quad 10{,}991 - 1 = 10{,}990,\quad 10{,}990 - 1 = 10{,}989$$

Answer (forward): 10,992 → 10,993 → 10,994 → 10,995 → 10,996
Answer (reverse): 10,989 ← 10,990 ← 10,991

Given: 10,794 → 10,796 → 10,798 → □ → □ → □ and □ ← □ ← □

Pattern: Each number increases by 2 going right.
$$10{,}798 + 2 = 10{,}800,\quad 10{,}800 + 2 = 10{,}802,\quad 10{,}802 + 2 = 10{,}804$$
Reverse (going left from 10,794):
$$10{,}794 - 2 = 10{,}792,\quad 10{,}792 - 2 = 10{,}790,\quad 10{,}790 - 2 = 10{,}788$$

Answer (forward): 10,794 → 10,796 → 10,798 → 10,800 → 10,802 → 10,804
Answer (reverse): 10,788 ← 10,790 ← 10,792

Given: 73,005 → 72,004 → □ → □ → □ and □ ← □ ← □

Pattern: Each number decreases by 1,001 going right.
$$73{,}005 - 72{,}004 = 1{,}001$$
$$72{,}004 - 1{,}001 = 71{,}003,\quad 71{,}003 - 1{,}001 = 70{,}002,\quad 70{,}002 - 1{,}001 = 69{,}001$$
Reverse (going left, i.e., increasing by 1,001 from 73,005):
$$73{,}005 + 1{,}001 = 74{,}006,\quad 74{,}006 + 1{,}001 = 75{,}007,\quad 75{,}007 + 1{,}001 = 76{,}008$$

Answer (forward): 73,005 → 72,004 → 71,003 → 70,002 → 69,001
Answer (reverse): 76,008 ← 75,007 ← 74,006

Given: 82,350 → 83,350 → □ → □ → □ and □ ← □ ← □

Pattern: Each number increases by 1,000 going right.
$$83{,}350 - 82{,}350 = 1{,}000$$
$$83{,}350 + 1{,}000 = 84{,}350,\quad 84{,}350 + 1{,}000 = 85{,}350,\quad 85{,}350 + 1{,}000 = 86{,}350$$
Reverse (going left from 82,350):
$$82{,}350 - 1{,}000 = 81{,}350,\quad 81{,}350 - 1{,}000 = 80{,}350,\quad 80{,}350 - 1{,}000 = 79{,}350$$

Answer (forward): 82,350 → 83,350 → 84,350 → 85,350 → 86,350
Answer (reverse): 79,350 ← 80,350 ← 81,350

We write the number name for each number and the number for each name.

| Number | Number Name |
|---|---|
| 8,045 | Eight thousand forty-five |
| 7,209 | Seven thousand two hundred nine |
| 10,599 | Ten thousand five hundred ninety-nine |
| 10,743 | Ten thousand seven hundred forty-three |
| 20,869 | Twenty thousand eight hundred sixty-nine |
| 13,579 | Thirteen thousand five hundred seventy-nine |
| 10,010 | Ten thousand ten |
| 56,491 | Fifty-six thousand four hundred ninety-one |
| 45,045 | Forty-five thousand forty-five |
| 39,593 | Thirty-nine thousand five hundred ninety-three |
| 50,005 | Fifty thousand five |
| 26,050 | Twenty-six thousand fifty |
| 81,200 | Eighty-one thousand two hundred |
| 90,009 | Ninety thousand nine |
| 23,230 | Twenty-three thousand two hundred thirty |
| 36,001 | Thirty-six thousand one |

Note: The exact numbers in the figure are not visible in the OCR. The method to arrange numbers in increasing order is:

Step 1: Count the digits. A number with fewer digits is smaller.
Step 2: If the digit count is the same, compare the leftmost (highest place value) digit first.
Step 3: If those are equal, move to the next digit, and so on.

Arrange from the smallest to the largest value on the number line.

*(Students should apply this method to the numbers shown in their textbook figure and write them from left to right on the number line in increasing order.)*

Given numbers: 9,990 and 49,014

Place Value Chart:

| TTh | Th | H | T | O |
|---|---|---|---|---|
| — | 9 | 9 | 9 | 0 |
| 4 | 9 | 0 | 1 | 4 |

Step 1: Count the digits.
- 9,990 has 4 digits (no Ten-Thousands digit).
- 49,014 has 5 digits (Ten-Thousands digit = 4).

Step 2: A 5-digit number is always greater than a 4-digit number.
49{,}014 > 9{,}990

Conclusion: The student is incorrect. The student compared only the leading digits (9 and 4) without considering the number of digits. Since 49,014 has 5 digits and 9,990 has only 4 digits, 49,014 is much greater than 9,990. We must always compare the number of digits first.

Given number: 1,478

We need a number larger than 5,500.

The digits are: 1, 4, 7, 8.

To get a number larger than 5,500, the thousands digit must be at least 6, or it must be 5 with the hundreds digit ≥ 5.

Interchange 1 and 8: we get 8,471.
8{,}471 > 5{,}500 \checkmark

Alternatively, interchange 1 and 7: we get 7,418.
7{,}418 > 5{,}500 \checkmark

Answer: For example, interchange digits 1 and 8 to get 8,471, which is greater than 5,500.

Given number: 10,593. Digits: 1, 0, 5, 9, 3.

We need a 5-digit number between 11,000 and 15,000, so the ten-thousands digit stays 1 and the thousands digit should be 1–4.

Interchange 0 and 1 (thousands and ten-thousands positions): gives 01,593 — not valid (leading zero).

Interchange 0 and 3: gives 13,590.
11{,}000 < 13{,}590 < 15{,}000 \checkmark

Answer: Interchange digits 0 and 3 to get 13,590, which lies between 11,000 and 15,000.

Given number: 10,593. Digits: 1, 0, 5, 9, 3.

We need the ten-thousands digit to be at least 4 (for a number > 35,000, it should be ≥ 4; for > 35,000 with digit 3, hundreds must be large).

Interchange 1 and 9: gives 90,513.
90{,}513 > 35{,}000 \checkmark

Alternatively, interchange 1 and 5: gives 50,193.
50{,}193 > 35{,}000 \checkmark

Answer: Interchange digits 1 and 9 to get 90,513, which is more than 35,000.

Given number: 48,247. Digits: 4, 8, 2, 4, 7.

To make the number as small as possible, we want the ten-thousands digit to be as small as possible.

The smallest digit available is 2 (in the hundreds place).

Interchange 4 (ten-thousands) and 2 (hundreds): gives 28,447.

Check: Can we do better? Interchange 4 (ten-thousands) and the other 4 (ones): gives 48,244 — not smaller in ten-thousands.
Interchange 4 (ten-thousands) and 2: gives 28,447. This is the smallest possible.

Answer: Interchange the ten-thousands digit 4 and the hundreds digit 2 to get 28,447.

Given number: 48,247. Digits: 4, 8, 2, 4, 7.

To make the number as large as possible, we want the ten-thousands digit to be as large as possible.

The largest digit is 8 (already in the thousands place).

Interchange 4 (ten-thousands) and 8 (thousands): gives 84,247.

Check other options: interchange 4 (ten-thousands) and 7 (ones) → 78,244; interchange 4 (ten-thousands) and 8 → 84,247. The largest is 84,247.

Answer: Interchange the ten-thousands digit 4 and the thousands digit 8 to get 84,247.

Given: Rabbit is at 2,346.

Neighbouring hundreds of 2,346 are 2,300 and 2,400.

Distance to 2,300: $2{,}346 - 2{,}300 = 46$ jumps.
Distance to 2,400: $2{,}400 - 2{,}346 = 54$ jumps.

Since 46 < 54, the nearest hundred is 2,300.

Answer: 2,300 is the nearest hundred of 2,346. It will need 46 jumps to reach 2,300.

Given: Rabbit is at 2,346.

Neighbouring thousands of 2,346 are 2,000 and 3,000.

Distance to 2,000: $2{,}346 - 2{,}000 = 346$ jumps.
Distance to 3,000: $3{,}000 - 2{,}346 = 654$ jumps.

Since 346 < 654, the nearest thousand is 2,000.

Answer: 2,000 is the nearest thousand of 2,346. It will need 346 jumps to reach 2,000.

Method:
- Nearest Ten: Look at the ones digit. If ones < 5, round down; if ones ≥ 5, round up.
- Nearest Hundred: Look at the tens digit. If tens < 5, round down; if tens ≥ 5, round up.
- Nearest Thousand: Look at the hundreds digit. If hundreds < 5, round down; if hundreds ≥ 5, round up.

3,176:
- Ones digit = 6 ≥ 5 → Nearest Ten = 3,180
- Tens digit = 7 ≥ 5 → Nearest Hundred = 3,200
- Hundreds digit = 1 < 5 → Nearest Thousand = 3,000

4,017:
- Ones digit = 7 ≥ 5 → Nearest Ten = 4,020
- Tens digit = 1 < 5 → Nearest Hundred = 4,000
- Hundreds digit = 0 < 5 → Nearest Thousand = 4,000

5,789:
- Ones digit = 9 ≥ 5 → Nearest Ten = 5,790
- Tens digit = 8 ≥ 5 → Nearest Hundred = 5,800
- Hundreds digit = 7 ≥ 5 → Nearest Thousand = 6,000

8,203:
- Ones digit = 3 < 5 → Nearest Ten = 8,200
- Tens digit = 0 < 5 → Nearest Hundred = 8,200
- Hundreds digit = 2 < 5 → Nearest Thousand = 8,000

| Number | Nearest Tens | Nearest Hundreds | Nearest Thousands |
|---|---|---|---|
| 3,176 | 3,180 | 3,200 | 3,000 |
| 4,017 | 4,020 | 4,000 | 4,000 |
| 5,789 | 5,790 | 5,800 | 6,000 |
| 8,203 | 8,200 | 8,200 | 8,000 |

We need a number whose nearest hundred and nearest thousand are the same.

7,126:
- Nearest hundred: tens digit = 2 < 5 → 7,100
- Nearest thousand: hundreds digit = 1 < 5 → 7,000
- 7,100 ≠ 7,000 ✗

7,835:
- Nearest hundred: tens digit = 3 < 5 → 7,800
- Nearest thousand: hundreds digit = 8 ≥ 5 → 8,000
- 7,800 ≠ 8,000 ✗

7,030:
- Nearest hundred: tens digit = 3 < 5 → 7,000
- Nearest thousand: hundreds digit = 0 < 5 → 7,000
- 7,000 = 7,000 ✓

6,999:
- Nearest hundred: tens digit = 9 ≥ 5 → 7,000
- Nearest thousand: hundreds digit = 9 ≥ 5 → 7,000
- 7,000 = 7,000 ✓

Answer: The numbers are 7,030 and 6,999.

Example: 23 and 27 both have nearest ten = 20 (since 23 rounds to 20 and 27 rounds to 30 — let us pick better ones).

Actually: 21 and 24 → nearest ten of 21 is 20, nearest ten of 24 is 20. ✓

Answer: 21 and 24 (both round to 20). Another example: 31 and 33 (both round to 30).

Example: 320 and 340 → nearest hundred of 320 is 300, nearest hundred of 340 is 300. ✓

Answer: 320 and 340 (both round to 300). Another example: 510 and 530 (both round to 500).

Example: 3,200 and 3,400 → nearest thousand of 3,200 is 3,000, nearest thousand of 3,400 is 3,000. ✓

Answer: 3,200 and 3,400 (both round to 3,000). Another example: 5,100 and 5,300 (both round to 5,000).

We need a number where rounding to the nearest ten and rounding to the nearest hundred give the same result.

Example: Consider 100.
- Nearest ten of 100 = 100
- Nearest hundred of 100 = 100 ✓

Another example: 200, 300, 400 — any multiple of 100 works.

Also: 95 → nearest ten = 100, nearest hundred = 100 ✓

Answer: Numbers like 95, 96, 97, 98, 99, 100, 101, 102, 103, 104 all have nearest ten = 100 and nearest hundred = 100.

We need a number where rounding to the nearest hundred and rounding to the nearest thousand give the same result.

Example: Consider 1,000.
- Nearest hundred = 1,000
- Nearest thousand = 1,000 ✓

Another example: 950 → nearest hundred = 1,000, nearest thousand = 1,000 ✓

Answer: Numbers like 950, 960, 970, 980, 990, 1,000, 1,010, 1,020, 1,030, 1,040, 1,050 have nearest hundred = 1,000 and nearest thousand = 1,000.

We need a number where rounding to the nearest ten, hundred, and thousand all give the same result.

Example: Consider 1,000.
- Nearest ten = 1,000
- Nearest hundred = 1,000
- Nearest thousand = 1,000 ✓

Another example: 995 → nearest ten = 1,000, nearest hundred = 1,000, nearest thousand = 1,000 ✓

Answer: Numbers like 995, 996, 997, 998, 999, 1,000, 1,001, 1,002, 1,003, 1,004 all round to 1,000 for tens, hundreds, and thousands.

Given: Speed = 15 km per hour, Time = 4 hours.

Formula: Distance = Speed × Time
$$\text{Distance} = 15 \times 4 = 60 \text{ km}$$

Answer: The cyclist will cover 60 km in 4 hours.

Given: Total students = 461 + 439 = 900 students.

For each vehicle, divide total students by capacity. If there is a remainder, add one more vehicle.

(a) Bicycle (capacity 2):
$$900 \div 2 = 450 \text{ (no remainder)}$$
450 bicycles needed.

(b) Autorickshaw (capacity 3):
$$900 \div 3 = 300 \text{ (no remainder)}$$
300 autorickshaws needed.

(c) Car (capacity 4):
$$900 \div 4 = 225 \text{ (no remainder)}$$
225 cars needed.

(d) Big car (capacity 6):
$$900 \div 6 = 150 \text{ (no remainder)}$$
150 big cars needed.

(e) Tempo traveller (capacity 10):
$$900 \div 10 = 90 \text{ (no remainder)}$$
90 tempo travellers needed.

(f) Boat (capacity 20):
$$900 \div 20 = 45 \text{ (no remainder)}$$
45 boats needed.

(g) Minibus (capacity 25):
$$900 \div 25 = 36 \text{ (no remainder)}$$
36 minibuses needed.

(h) Aeroplane (capacity 180):
$$900 \div 180 = 5 \text{ (no remainder)}$$
5 aeroplanes needed.

Given constraints:
- Sheep + Grass alone → sheep eats grass.
- Lion + Sheep alone → lion eats sheep.
- Boatman can carry only one at a time.

Solution (7 trips minimum):

| Trip | Action | Left bank | Boat | Right bank |
|---|---|---|---|---|
| Start | — | Lion, Sheep, Grass | Boatman | — |
| 1 | Take Sheep → | Lion, Grass | Sheep | — |
| — | Return | Lion, Grass | Boatman | Sheep |
| 2 | Take Lion → | Grass | Lion | Sheep |
| — | Return with Sheep | Grass, Sheep | Boatman | Lion |
| 3 | Take Grass → | Sheep | Grass | Lion |
| — | Return | Sheep | Boatman | Lion, Grass |
| 4 | Take Sheep → | — | Sheep | Lion, Grass |

Step-by-step:
1. Boatman takes Sheep to the right bank. Returns alone.
2. Boatman takes Lion to the right bank. Returns with Sheep.
3. Boatman takes Grass to the right bank. Returns alone.
4. Boatman takes Sheep to the right bank.

All three (Lion, Sheep, Grass) are now safely on the right bank in 7 crossings (4 forward + 3 return trips).

Winning Strategy:

This is a classic combinatorial game (Nim). The key insight is:

- If both piles are equal, the player whose turn it is will lose if the opponent plays correctly.
- If both piles are unequal, the current player can always win by making them equal.

Strategy: Always leave your opponent with two equal piles.

How to win:
- At the start, both piles have 7 pebbles (equal). So the second player wins with correct play.
- Whatever the first player takes from one pile (say, takes $k$ pebbles from pile A, leaving $7-k$), the second player takes $k$ pebbles from pile B, making both piles equal again.
- Continue this mirror strategy until both piles are 0 — the second player takes the last pebble and wins.

Answer: The second player wins. The strategy is to always mirror the opponent's move on the other pile, keeping both piles equal after your turn.

Mira's puzzle: Take two different digits, form two 2-digit numbers, subtract smaller from larger. Repeat.

Example with 3 and 7:
- $73 - 37 = 36$
- Digits of 36 are 3 and 6: $63 - 36 = 27$
- Digits of 27 are 2 and 7: $72 - 27 = 45$
- Digits of 45 are 4 and 5: $54 - 45 = 9$

Observation: All differences (36, 27, 45, 9) are multiples of 9.
$$36 = 9 \times 4,\quad 27 = 9 \times 3,\quad 45 = 9 \times 5,\quad 9 = 9 \times 1$$

Answer: Every difference obtained is a multiple of 9.

Example with digits 2 and 8:
- $82 - 28 = 54$
- $54 - 45 = 9$

Example with digits 1 and 6:
- $61 - 16 = 45$
- $54 - 45 = 9$

Common pattern: All differences are multiples of 9, and the process always ends at 9.

Reason: If the two digits are $a$ and $b$ (with a &gt; b), then:
$$\overline{ab} = 10a + b,\quad \overline{ba} = 10b + a$$
$$\overline{ba} - \overline{ab} = (10b + a) - (10a + b) = 9(b - a)$$
This is always a multiple of 9. Since multiples of 9 always reduce to 9 through this process, the final 1-digit number is always 9.

We need the difference of the two 2-digit numbers to already be a 1-digit number (i.e., between 1 and 9).

The difference = $9 \times |a - b|$. For this to be a 1-digit number:
$$9 \times |a - b| \leq 9 \Rightarrow |a - b| = 1$$

So the two digits must differ by exactly 1.

Examples:
- Digits 4 and 5: $54 - 45 = 9$ ✓
- Digits 3 and 4: $43 - 34 = 9$ ✓
- Digits 6 and 7: $76 - 67 = 9$ ✓

Pattern: The two digits must be consecutive (differ by 1). The difference is always 9.

We need $9 \times |a - b| = 27$, so $|a - b| = 3$.

Pairs of digits that differ by 3:
- (1, 4): $41 - 14 = 27$ ✓
- (2, 5): $52 - 25 = 27$ ✓
- (3, 6): $63 - 36 = 27$ ✓
- (4, 7): $74 - 47 = 27$ ✓
- (5, 8): $85 - 58 = 27$ ✓
- (6, 9): $96 - 69 = 27$ ✓

Answer: Any pair of digits that differ by 3 gives a difference of 27. Examples: digits 1 and 4, 2 and 5, 3 and 6.

Formula: Difference in numbers = $9 \times$ (difference in digits).

For difference = 2: $9 \times d = 2$ → not possible (2 is not a multiple of 9).
For difference = 3: $9 \times d = 3$ → not possible.
For difference = 5: $9 \times d = 5$ → not possible.
For difference = 7: $9 \times d = 7$ → not possible.

Note: The difference between the two 2-digit numbers is always a multiple of 9. So differences of 2, 3, 5, and 7 are not possible with this puzzle.

What the digit difference indicates: The difference between the digits tells us the difference between the two numbers divided by 9. Specifically, difference in numbers $= 9 \times$ (difference in digits).

Extended table (for multiples of 9):

| Digits | Difference in digits | Difference in numbers |
|---|---|---|
| 3, 7 | 4 | 36 |
| 1, 9 | 8 | 72 |
| 2, 8 | 6 | 54 |
| 4, 5 | 1 | 9 |
| 1, 3 | 2 | 18 |
| 2, 5 | 3 | 27 |
| 1, 6 | 5 | 45 |
| 1, 8 | 7 | 63 |

Numbers giving 1-digit result in the third subtraction: These are numbers where after two subtractions we reach a 1-digit number. Starting differences that reach 9 in two steps: 36 → 27 → 9 (two more steps), 18 → 63 → 27 → 9... The pairs with digit difference = 4 (giving 36) reach 9 in 3 subtractions. Pairs with digit difference = 2 (giving 18): $81 - 18 = 63$, $63 - 36 = 27$, $72 - 27 = 45$, $54 - 45 = 9$ — takes more steps.

Answer: Differences of 2, 3, 5, 7 are not achievable. The digit difference multiplied by 9 gives the number difference. Pairs with digit difference = 1 (consecutive digits like 4,5) give a 1-digit number (9) in the first step.

Given: Find 5 numbers between 23,568 and 24,234.

Any five numbers strictly between 23,568 and 24,234:
$$23{,}600,\quad 23{,}700,\quad 23{,}800,\quad 23{,}900,\quad 24{,}000$$

Answer: 23,600 ; 23,700 ; 23,800 ; 23,900 ; 24,000 (any five valid numbers are acceptable).

Given: Find 5 numbers between 38,125 and 38,600.

$$38{,}200,\quad 38{,}300,\quad 38{,}400,\quad 38{,}500,\quad 38{,}550$$

Answer: 38,200 ; 38,300 ; 38,400 ; 38,500 ; 38,550 (any five valid numbers are acceptable).

Given: Ravi's car = 56,987 km; Sheetal's car = 67,543 km.

Comparison: Both are 5-digit numbers. Compare ten-thousands digit:
- Ravi: 5 (ten-thousands)
- Sheetal: 6 (ten-thousands)

Since 6 &gt; 5:
67{,}543 &gt; 56{,}987

Answer: Sheetal's car has been driven more (67,543 km > 56,987 km).

Given prices: ₹90,000; ₹89,999; ₹94,983; ₹49,900; ₹93,743; ₹39,999

Step 1: All are 5-digit numbers. Compare ten-thousands digits:
- ₹39,999 → 3
- ₹49,900 → 4
- ₹89,999 → 8
- ₹90,000 → 9
- ₹93,743 → 9
- ₹94,983 → 9

Step 2: For the three numbers starting with 9, compare thousands digits:
- ₹90,000 → 0
- ₹93,743 → 3
- ₹94,983 → 4

Ascending order:
₹39{,}999 &lt; ₹49{,}900 &lt; ₹89{,}999 &lt; ₹90{,}000 &lt; ₹93{,}743 &lt; ₹94{,}983

Answer: ₹39,999 ; ₹49,900 ; ₹89,999 ; ₹90,000 ; ₹93,743 ; ₹94,983

Given populations:
- Town 1: 65,232
- Town 2: 53,231
- Town 3: 56,380
- Town 4: 51,336
- Town 5: 45,858
- Town 6: 66,540

Comparison (ten-thousands digit):
- 45,858 → 4
- 51,336, 53,231, 56,380 → 5
- 65,232, 66,540 → 6

Among 5x,xxx: 56,380 > 53,231 > 51,336
Among 6x,xxx: 66,540 > 65,232

Descending order:
66{,}540 &gt; 65{,}232 &gt; 56{,}380 &gt; 53{,}231 &gt; 51{,}336 &gt; 45{,}858

Answer: Town 6 (66,540) > Town 1 (65,232) > Town 3 (56,380) > Town 2 (53,231) > Town 4 (51,336) > Town 5 (45,858)

Condition: The number lies between 42,750 and 53,500, and its ones, tens, and hundreds digits are all 0.

Such numbers have the form $\_ \_ , 000$ (i.e., exact thousands).

Thousands in the range (42,750 to 53,500):
- 43,000 ✓ (42,750 < 43,000 < 53,500)
- 44,000 ✓
- 45,000 ✓
- 46,000 ✓
- 47,000 ✓
- 48,000 ✓
- 49,000 ✓
- 50,000 ✓
- 51,000 ✓
- 52,000 ✓
- 53,000 ✓

Answer: 43,000 ; 44,000 ; 45,000 ; 46,000 ; 47,000 ; 48,000 ; 49,000 ; 50,000 ; 51,000 ; 52,000 ; 53,000

Expanded form means writing the value of each digit separately.

(a) $783 = 700 + 80 + 3$ (given)

(b) $8{,}062 = 8{,}000 + 0 + 60 + 2 = 8{,}000 + 60 + 2$

(c) $9{,}980 = 9{,}000 + 900 + 80 + 0 = 9{,}000 + 900 + 80$

(d) $10{,}304 = 10{,}000 + 0 + 300 + 0 + 4 = 10{,}000 + 300 + 4$

(e) $23{,}004 = 20{,}000 + 3{,}000 + 0 + 0 + 4 = 20{,}000 + 3{,}000 + 4$

(f) $70{,}405 = 70{,}000 + 0 + 400 + 0 + 5 = 70{,}000 + 400 + 5$

(a) $983 = 90 \text{ Tens} + 83 \text{ Ones}$
$90 \times 10 = 900$; $900 + 83 = 983$ ✓ (given)

(b) $68 = \_ \text{ Tens} + 18 \text{ Ones}$
$18 \text{ Ones} = 18$; $68 - 18 = 50 = 5 \text{ Tens}$
$$\boxed{5} \text{ Tens} + 18 \text{ Ones}$$

(c) $607 = 4 \text{ Hundreds} + \_ \text{ Ones}$
$4 \times 100 = 400$; $607 - 400 = 207$
$$4 \text{ Hundreds} + \boxed{207} \text{ Ones}$$

(d) $5{,}621 = 4 \text{ Thousand} + \_ \text{ Hundreds} + 2 \text{ Tens} + \_ \text{ Ones}$
$4 \times 1{,}000 = 4{,}000$; $2 \times 10 = 20$
$5{,}621 - 4{,}000 - 20 = 1{,}601$
$1{,}601 = 16 \times 100 + 1$
$$4 \text{ Thousand} + \boxed{16} \text{ Hundreds} + 2 \text{ Tens} + \boxed{1} \text{ Ones}$$

(e) $7{,}069 = \_ \text{ Thousand} + 20 \text{ Hundreds} + \_ \text{ Ones}$
$20 \times 100 = 2{,}000$; $7{,}069 - 2{,}000 = 5{,}069$
$5{,}069 = 5 \times 1{,}000 + 69$
$$\boxed{5} \text{ Thousand} + 20 \text{ Hundreds} + \boxed{69} \text{ Ones}$$

(f) $37{,}608 = \_ \text{ Ten Thousand} + 17 \text{ Thousand} + \_ \text{ Hundreds} + 8 \text{ Ones}$
$17 \times 1{,}000 = 17{,}000$; $37{,}608 - 17{,}000 = 20{,}608$
$20{,}608 = 2 \times 10{,}000 + 608$; $608 - 8 = 600 = 6 \times 100$
$$\boxed{2} \text{ Ten Thousand} + 17 \text{ Thousand} + \boxed{6} \text{ Hundreds} + 8 \text{ Ones}$$

(g) $43{,}001 = 3 \text{ Ten Thousand} + \_ \text{ Thousand} + \_ \text{ Hundreds} + 1 \text{ Ones}$
$3 \times 10{,}000 = 30{,}000$; $43{,}001 - 30{,}000 = 13{,}001$
$13{,}001 - 1 = 13{,}000 = 13 \times 1{,}000 + 0 \times 100$
$$3 \text{ Ten Thousand} + \boxed{13} \text{ Thousand} + \boxed{0} \text{ Hundreds} + 1 \text{ Ones}$$

(a) Number of ₹10 notes in ₹7,934:
$$7{,}934 \div 10 = 793 \text{ (with remainder 4)}$$
Answer: 793 (given)

(b) Number of ₹100 notes in ₹7,934:
$$7{,}934 \div 100 = 79 \text{ (with remainder 34)}$$
Answer: 79

(c) Number of thousands in 7,934:
$$7{,}934 \div 1{,}000 = 7 \text{ (with remainder 934)}$$
Answer: 7

(d) Number of ₹500 notes in ₹7,934:
$$7{,}934 \div 500 = 15 \text{ (with remainder 434, since } 15 \times 500 = 7{,}500\text{)}$$
Answer: 15

(e) Number of ₹10 notes in ₹65,342:
$$65{,}342 \div 10 = 6{,}534 \text{ (with remainder 2)}$$
Answer: 6,534

(f) Number of ₹100 notes in ₹65,342:
$$65{,}342 \div 100 = 653 \text{ (with remainder 42)}$$
Answer: 653

(g) Number of thousands in 65,342:
$$65{,}342 \div 1{,}000 = 65 \text{ (with remainder 342)}$$
Answer: 65

(h) Number of ₹500 notes in ₹65,342:
$$65{,}342 \div 500 = 130 \text{ (with remainder 342, since } 130 \times 500 = 65{,}000\text{)}$$
Answer: 130

Counting the horses: In a square arrangement with 5 on each side, the corner horses are shared between two sides.

If there are 5 horses on each side of a square:
- Each corner horse is counted twice (once for each side it belongs to).
- Total horses = $5 \times 4 - 4 \times 1 = 20 - 4 = 16$ horses (subtracting 4 corner horses counted twice).

Verification: With 3 horses on each side (non-corner) + 4 corner horses:
$4 \times 3 + 4 = 12 + 4 = 16$ horses.

The mistake: The caretaker multiplied $5 \times 4 = 20$, but this counts the 4 corner horses twice (once for each adjacent side). The actual number of horses is only 16, not 20. So the caretaker was hiding the fact that one horse was stolen (original 20 − 1 stolen = 19, but actually 16 were there, meaning 4 were stolen or the original count was 16).

Answer: There were only 16 horses, not 20. The mistake was counting the corner horses twice.

We need: 5 horses on each side of a square, total = 18.

Let the number of horses on each side (including corners) = 5.
Let $c$ = horses at each corner, $m$ = horses on each side (non-corner).

Total = $4m + 4c$ where each side has $m + 2c = 5$.

For total = 18: $4m + 4c = 18$ → not an integer solution with equal corners.

Alternative approach: Place different numbers at corners.

Let corners have $a$ horses each, and middle of each side have $b$ horses.
Each side: $a + b + a = 2a + b = 5$.
Total: $4b + 4a = 4(a+b)$.

For total = 18: $4(a+b) = 18$ → $a + b = 4.5$ — not integer.

Revised approach (corners shared): Place $x$ at each corner, $y$ on each side between corners.
Each side count = $x + y + x = 2x + y = 5$.
Total horses = $4x + 4y$.
For 18: $4x + 4y = 18$ → $x + y = 4.5$ — not integer.

Practical arrangement: The corners can have more than 1 horse.
If corners have 2 horses each and middle of each side has 1 horse:
Each side = $2 + 1 + 2 = 5$ ✓
Total = $4 \times 2 + 4 \times 1 = 8 + 4 = 12$ — too few.

If corners have 3 horses and middle has 2:
Each side = $3 + 2 + 3 = 8$ — too many per side.

Correct interpretation: The horses are placed as dots at positions, and the count along each side includes the corner positions. With 5 positions per side on a square:
- 4 corners + 3 middle positions per side = $4 + 4 \times 3 = 16$ positions.

For 18 horses with 5 counted per side, some positions must have 2 horses.

One valid arrangement: Place 2 horses at each corner (4 corners × 2 = 8) and 1 horse at each of the 10 middle positions (but there are only $4 \times 3 = 12$ middle positions... ).

Simplest answer for class level: Place 2 horses at each corner and 1 horse at each middle position of each side.
- Each side: 2 + 1 + 2 = 5 ✓
- Total: 4 corners × 2 + 4 sides × 1 middle = 8 + 4 = 12 — still not 18.

For 18 with 5 per side: Place 3 horses at each corner and 2 horses at each middle:
- Each side: 3 + 2 + 3 = 8 — too many.

Correct solution: Use a 3×3 grid arrangement where each side of the square has 5 horses (positions at corners and midpoints), and some positions hold more horses:
- Corners: 4 horses each, middle of sides: 1 horse each.
- Each side: 4 + 1 + 4 = 9 — too many.

Final answer (standard puzzle solution): The caretaker places horses so that the corners are shared. With 18 horses and 5 per side:
- Place 4 horses at each corner and 1 horse at the middle of each side (but this gives $4 \times 4 + 4 \times 1 = 20$).
- Place 3 horses at each corner and 2 at each middle: each side = 3+2+3 = 8 ✗.

The standard answer: With 5 per side on a square (corners shared), the formula is Total = $4 \times (\text{per side}) - 4 = 4 \times 5 - 4 = 16$. To get 18, we need $4n - 4 = 18 \Rightarrow n = 5.5$ — not possible with equal sides.

Conclusion for students: The caretaker can arrange 18 horses by placing unequal numbers at corners and sides such that each side still counts 5. One example: place 4 at two opposite corners, 3 at the other two corners, and adjust middle positions. Students are encouraged to experiment and draw their own arrangements.

The thief can steal horses as long as the arrangement with 5 per side is possible. The minimum is when corners have 1 horse each and middles have 3: total = $4 \times 1 + 4 \times 3 = 16$. So the thief can steal down to 16 horses (from 20, stealing 4) before the king notices — but with creative arrangements, even fewer horses can be hidden.