Large Numbers Around Us

Given: 1 year ≈ 365 days. We need to find y such that y × 365 is close to 1,00,000.

Let us choose y = 274.

Number of days = 274 × 365 = 1,00,010.

This is just 10 more than one lakh, so it is very close to one lakh.

Alternatively, choose y = 273:
Number of days = 273 × 365 = 99,645.
This is 1,00,000 − 99,645 = 355 less than one lakh.

So for y = 274, the number of days (1,00,010) is closest to one lakh — just 10 more than one lakh.

Given: Population = 75,000; One lakh = 1,00,000.

Concept: Subtraction.

$$1,00,000 - 75,000 = 25,000$$

∴ 75,000 is 25,000 less than one lakh.

Given: Population in 2024 = 1,06,000; One lakh = 1,00,000.

Concept: Subtraction.

$$1,06,000 - 1,00,000 = 6,000$$

∴ 1,06,000 is 6,000 more than one lakh.

Given: Population in 2011 = 75,000; Population in 2024 = 1,06,000.

Concept: Subtraction to find increase.

$$1,06,000 - 75,000 = 31,000$$

∴ The population increased by 31,000 from 2011 to 2024.

We need to find how many hundreds make 10,000.

$$10,000 ÷ 100 = 100$$

∴ 100 hundreds are required to make 10,000.

Fifty three thousand = 53,000.

$$53,000 ÷ 100 = 530$$

∴ 530 hundreds are required to make 53,000.

$$90,000 ÷ 100 = 900$$

∴ 900 hundreds are required to make 90,000.

$$97,600 ÷ 100 = 976$$

∴ 976 hundreds are required to make 97,600.

$$1,00,000 ÷ 100 = 1,000$$

∴ 1,000 hundreds are required to make one lakh.

Number = 582 × 100 = 58,200.

∴ The number is 58,200.

Ten thousand = 10,000.

$$10,000 ÷ 100 = 100$$

∴ 100 hundreds are required to make ten thousand.

One lakh = 1,00,000.

$$1,00,000 ÷ 100 = 1,000$$

∴ 1,000 hundreds are required to make one lakh.

Let us think carefully:

- Tedious Tens can show any multiple of 10 (10, 20, 30, …).
- Thoughtful Thousands can show any multiple of 1000 (1000, 2000, …).
- Handy Hundreds can show any multiple of 100 (100, 200, 300, …).

Now, every multiple of 1000 is also a multiple of 100 (e.g., 1000 = 10 × 100), and every multiple of 100 is also a multiple of 10 (e.g., 100 = 10 × 10).

So Tedious Tens can show ALL multiples of 100 (since 100 is itself a multiple of 10). Therefore, there is NO number that Handy Hundreds can show but Tedious Tens cannot.

However, Handy Hundreds CAN show numbers like 100, 200, 300, 500, 700 that Thoughtful Thousands cannot (since these are not multiples of 1000).

∴ The statement is partially true: Handy Hundreds can show numbers that Thoughtful Thousands cannot (e.g., 100, 200, 500), but Tedious Tens can show everything Handy Hundreds can. So the statement is true with respect to Thoughtful Thousands, but false with respect to Tedious Tens.

Method 1: Press +10 thirty-two times and +1 once.
$$32 \times 10 + 1 \times 1 = 320 + 1 = 321 ✓$$

Method 2: Press +100 two times, +10 twelve times, +1 once.
$$2 \times 100 + 12 \times 10 + 1 \times 1 = 200 + 120 + 1 = 321 ✓$$

Both methods correctly give 321.

We need to express 5072 in a different way using the available buttons.

Method 3: Press +1000 five times, +10 seven times, +1 two times.
$$5 \times 1000 + 7 \times 10 + 2 \times 1 = 5000 + 70 + 2 = 5072 ✓$$

Expression: $(5 \times 1000) + (7 \times 10) + (2 \times 1) = 5072$

Another Method: Press +100 forty-seven times, +10 twenty-two times, +1 two times.
$$47 \times 100 + 22 \times 10 + 2 \times 1 = 4700 + 220 + 2 = 4922 \neq 5072$$

Let us try: +1000 four times, +100 ten times, +1 two times:
$$4 \times 1000 + 10 \times 100 + 7 \times 10 + 2 \times 1 = 4000 + 1000 + 70 + 2 = 5072 ✓$$

Expression: $(4 \times 1000) + (10 \times 100) + (7 \times 10) + (2 \times 1) = 5072$

Way 1 (using thousands and hundreds):
$$8 \times 1000 + 3 \times 100 = 8000 + 300 = 8300$$
Expression: $(8 \times 1000) + (3 \times 100) = 8300$

Way 2 (using hundreds only):
$$83 \times 100 = 8300$$
Expression: $(83 \times 100) = 8300$

Way 3 (mixing differently):
$$5 \times 1000 + 33 \times 100 = 5000 + 3300 = 8300$$
Expression: $(5 \times 1000) + (33 \times 100) = 8300$

Way 1:
$$4 \times 10000 + 6 \times 100 + 2 \times 10 + 9 \times 1 = 40000 + 600 + 20 + 9 = 40629$$
Expression: $(4 \times 10000) + (6 \times 100) + (2 \times 10) + (9 \times 1) = 40629$

Way 2:
$$3 \times 10000 + 10 \times 1000 + 6 \times 100 + 2 \times 10 + 9 \times 1 = 30000 + 10000 + 600 + 20 + 9 = 40629$$
Expression: $(3 \times 10000) + (10 \times 1000) + (6 \times 100) + (2 \times 10) + (9 \times 1) = 40629$

Way 1:
$$5 \times 10000 + 6 \times 1000 + 3 \times 100 + 5 \times 10 + 4 \times 1 = 50000 + 6000 + 300 + 50 + 4 = 56354$$
Expression: $(5 \times 10000) + (6 \times 1000) + (3 \times 100) + (5 \times 10) + (4 \times 1) = 56354$

Way 2:
$$4 \times 10000 + 16 \times 1000 + 3 \times 100 + 5 \times 10 + 4 \times 1 = 40000 + 16000 + 300 + 50 + 4 = 56354$$
Expression: $(4 \times 10000) + (16 \times 1000) + (3 \times 100) + (5 \times 10) + (4 \times 1) = 56354$

Way 1:
$$6 \times 10000 + 6 \times 1000 + 6 \times 100 + 6 \times 10 + 6 \times 1 = 60000 + 6000 + 600 + 60 + 6 = 66666$$
Expression: $(6 \times 10000) + (6 \times 1000) + (6 \times 100) + (6 \times 10) + (6 \times 1) = 66666$

Way 2:
$$5 \times 10000 + 16 \times 1000 + 6 \times 100 + 6 \times 10 + 6 \times 1 = 50000 + 16000 + 600 + 60 + 6 = 66666$$
Expression: $(5 \times 10000) + (16 \times 1000) + (6 \times 100) + (6 \times 10) + (6 \times 1) = 66666$

Way 1:
$$3 \times 100000 + 6 \times 10000 + 7 \times 1000 + 8 \times 100 + 1 \times 10 + 3 \times 1$$
$$= 300000 + 60000 + 7000 + 800 + 10 + 3 = 367813$$
Expression: $(3 \times 100000) + (6 \times 10000) + (7 \times 1000) + (8 \times 100) + (1 \times 10) + (3 \times 1) = 367813$

Way 2:
$$2 \times 100000 + 16 \times 10000 + 7 \times 1000 + 8 \times 100 + 1 \times 10 + 3 \times 1$$
$$= 200000 + 160000 + 7000 + 800 + 10 + 3 = 367813$$
Expression: $(2 \times 100000) + (16 \times 10000) + (7 \times 1000) + (8 \times 100) + (1 \times 10) + (3 \times 1) = 367813$

Available buttons: +1, +10, +100, +1000, +10000, +100000, +1000000.
We need exactly 30 button presses and the result must be a 3-digit number (100–999).

Largest 3-digit number with 30 presses:
To maximise the number while keeping it 3-digit (≤ 999) and using exactly 30 presses:
- Press +100 nine times: 9 × 100 = 900 (9 presses used, 21 remaining)
- Press +10 nine times: 9 × 10 = 90 (18 more presses, 3 remaining)
- Press +1 three times: 3 × 1 = 3
- Total = 900 + 90 + 3 = 993, using 9 + 9 + 3 = 21 presses. Need 9 more.

Let us try: Press +100 nine times (9 presses) = 900, press +10 nine times (9 presses) = 90, press +1 twelve times (12 presses) = 12. Total presses = 30. Number = 900 + 90 + 12 = 1002. This exceeds 3 digits.

Adjust: Press +100 nine times = 900 (9 presses), press +10 two times = 20 (2 presses), press +1 nineteen times = 19 (19 presses). Total = 30 presses. Number = 939. But can we do better?

Press +100 nine times = 900 (9 presses), press +10 nine times = 90 (9 presses), press +1 nine times = 9 (9 presses), press +10 three times = 30 (3 presses). Total = 30 presses. Number = 900 + 90 + 9 + 30 = 1029. Too big.

Best approach: We want the largest 3-digit number ≤ 999 using exactly 30 presses.
Press +100 nine times = 900 (9 presses), press +10 nine times = 90 (9 presses), press +1 nine times = 9 (9 presses). Total = 27 presses, number = 999. Need 3 more presses without changing the number — but every press adds at least 1. So we cannot use exactly 30 presses to get 999.

With 30 presses, each press adds at least 1, so minimum value = 30. We want a 3-digit number.
If we press +100 once (1 press) and +1 twenty-nine times (29 presses): 100 + 29 = 129. (30 presses)
If we press +100 nine times (9 presses) and +1 twenty-one times (21 presses): 900 + 21 = 921. (30 presses)
If we press +100 nine times (9 presses), +10 two times (2 presses), +1 nineteen times (19 presses): 900 + 20 + 19 = 939. (30 presses)
If we press +100 nine times (9 presses), +10 nine times (9 presses), +1 twelve times (12 presses): 900 + 90 + 12 = 1002. Too big.
If we press +100 nine times (9 presses), +10 eight times (8 presses), +1 thirteen times (13 presses): 900 + 80 + 13 = 993. (30 presses) ✓
If we press +100 nine times (9 presses), +10 eight times (8 presses), +1 thirteen times (13 presses): 993. Can we do better?
If we press +100 nine times (9 presses), +10 nine times (9 presses), +1 eleven times (11 presses): 900 + 90 + 11 = 1001. Too big.
So the largest 3-digit number = 993 (press +100 nine times, +10 eight times, +1 thirteen times).

Smallest 3-digit number with 30 presses:
We want the smallest number ≥ 100 using exactly 30 presses.
Press +100 once (1 press) and +1 twenty-nine times (29 presses): 100 + 29 = 129. (30 presses)
Can we get smaller? Press +100 once and +10 once and +1 twenty-eight times: 100 + 10 + 28 = 138. Bigger.
Press +100 once and +1 twenty-nine times = 129 is the smallest since we must press +100 at least once (to reach 3 digits) and each remaining press adds at least 1.

So the smallest 3-digit number = 129 (press +100 once, +1 twenty-nine times).

First, verify 25 clicks:
Standard way: 9 × 100 + 9 × 10 + 7 × 1 = 9 + 9 + 7 = 25 clicks. ✓

Different number of clicks:

With fewer clicks (using larger buttons):
Press +1000 once (1 click) — but 1000 > 997, so we'd need to subtract, which isn't possible.

So we cannot use +1000 directly. Let us try:
- Press +100 nine times (9 clicks) = 900
- Press +10 nine times (9 clicks) = 90
- Press +1 seven times (7 clicks) = 7
- Total = 25 clicks (same as before)

Alternative with more clicks:
- Press +100 eight times (8 clicks) = 800
- Press +10 nineteen times (19 clicks) = 190
- Press +1 seven times (7 clicks) = 7
- Total = 34 clicks, number = 997 ✓

Alternative with fewer clicks:
- Press +100 nine times = 900 (9 clicks)
- Press +10 eight times = 80 (8 clicks)
- Press +1 seventeen times = 17 (17 clicks)
- Total = 34 clicks, number = 997 ✓

For fewer than 25: We need to use bigger denominations. Since +1000 overshoots, we cannot reduce below 25 clicks with the given buttons for 997.

∴ Yes, 997 can be made with 34 clicks (or other numbers greater than 25). The minimum is 25 clicks.

The minimum number of clicks equals the sum of all digits of the number (since each digit tells us how many times to press that place-value button).

(a) 8300:
Digits: 8, 3, 0, 0
Minimum clicks = 8 + 3 = 11
Expression: $(8 \times 1000) + (3 \times 100) = 8300$
Press +1000 eight times and +100 three times.

(b) 40629:
Digits: 4, 0, 6, 2, 9
Minimum clicks = 4 + 0 + 6 + 2 + 9 = 21
Expression: $(4 \times 10000) + (6 \times 100) + (2 \times 10) + (9 \times 1) = 40629$

(c) 56354:
Digits: 5, 6, 3, 5, 4
Minimum clicks = 5 + 6 + 3 + 5 + 4 = 23
Expression: $(5 \times 10000) + (6 \times 1000) + (3 \times 100) + (5 \times 10) + (4 \times 1) = 56354$

(d) 66666:
Digits: 6, 6, 6, 6, 6
Minimum clicks = 6 + 6 + 6 + 6 + 6 = 30
Expression: $(6 \times 10000) + (6 \times 1000) + (6 \times 100) + (6 \times 10) + (6 \times 1) = 66666$

(e) 367813:
Digits: 3, 6, 7, 8, 1, 3
Minimum clicks = 3 + 6 + 7 + 8 + 1 + 3 = 28
Expression: $(3 \times 100000) + (6 \times 10000) + (7 \times 1000) + (8 \times 100) + (1 \times 10) + (3 \times 1) = 367813$

Yes! The minimum number of button clicks = sum of all the digits of the number.

For example:
- 8300 → 8 + 3 + 0 + 0 = 11 clicks
- 56354 → 5 + 6 + 3 + 5 + 4 = 23 clicks

This is because each digit tells us exactly how many times to press the corresponding place-value button, and pressing it that many times is the most efficient way.

In the Indian place value system, every number is written as a sum of its digits multiplied by their respective place values:
$$N = d_n \times 10^n + d_{n-1} \times 10^{n-1} + \cdots + d_1 \times 10 + d_0 \times 1$$

For example: $56354 = 5 \times 10000 + 6 \times 1000 + 3 \times 100 + 5 \times 10 + 4 \times 1$

This is exactly what Systematic Sippy does with minimum clicks — it presses each place-value button exactly as many times as the digit in that place. So the expression for minimum clicks directly mirrors the place value expansion of the number. This is why the two are the same.

Hundred thousand = 1,00,000.

Counting the zeros: 1-0-0-0-0-0 → 5 zeros.

(a) 4050678
Indian notation: 40,50,678
Indian name: Forty lakh fifty thousand six hundred seventy-eight
American notation: 4,050,678
American name: Four million fifty thousand six hundred seventy-eight

(b) 48121620
Indian notation: 4,81,21,620
Indian name: Four crore eighty-one lakh twenty-one thousand six hundred twenty
American notation: 48,121,620
American name: Forty-eight million one hundred twenty-one thousand six hundred twenty

(c) 20022002
Indian notation: 2,00,22,002
Indian name: Two crore twenty-two thousand two
American notation: 20,022,002
American name: Twenty million twenty-two thousand two

(d) 246813579
Indian notation: 24,68,13,579
Indian name: Twenty-four crore sixty-eight lakh thirteen thousand five hundred seventy-nine
American notation: 246,813,579
American name: Two hundred forty-six million eight hundred thirteen thousand five hundred seventy-nine

(e) 345000543
Indian notation: 34,50,00,543
Indian name: Thirty-four crore fifty lakh five hundred forty-three
American notation: 345,000,543
American name: Three hundred forty-five million five hundred forty-three

(f) 1020304050
Indian notation: 1,02,03,04,050
Indian name: One arab two crore three lakh four thousand fifty
American notation: 1,020,304,050
American name: One billion twenty million three hundred four thousand fifty

(a) One crore one lakh one thousand ten:
= 1,00,00,000 + 1,00,000 + 1,000 + 10
= 1,01,01,010

(b) One billion one million one thousand one:
1 billion = 1,00,00,00,000
1 million = 10,00,000
1 thousand = 1,000
1 = 1
= 1,00,00,00,000 + 10,00,000 + 1,000 + 1
= 1,00,10,01,001

(c) Ten crore twenty lakh thirty thousand forty:
= 10,00,00,000 + 20,00,000 + 30,000 + 40
= 10,20,30,040

(d) Nine billion eighty million seven hundred thousand six hundred:
9 billion = 9,00,00,00,000
80 million = 8,00,00,000
7 hundred thousand = 7,00,000
600 = 600
= 9,00,00,00,000 + 8,00,00,000 + 7,00,000 + 600
= 9,08,07,00,600

(a) 30 thousand vs 3 lakhs:
30 thousand = 30,000
3 lakhs = 3,00,000
30,000 < 3,00,000
30 \text{ thousand} &lt; 3 \text{ lakhs}

(b) 500 lakhs vs 5 million:
500 lakhs = 500 × 1,00,000 = 5,00,00,000 = 5 crore
5 million = 50 lakh = 50,00,000
5,00,00,000 > 50,00,000
500 \text{ lakhs} &gt; 5 \text{ million}

(c) 800 thousand vs 8 million:
800 thousand = 8,00,000
8 million = 80,00,000
8,00,000 < 80,00,000
800 \text{ thousand} &lt; 8 \text{ million}

(d) 640 crore vs 60 billion:
640 crore = 640 × 1,00,00,000 = 6,40,00,00,000
60 billion = 60 × 1,00,00,00,000 = 6,00,00,00,000
6,40,00,00,000 > 6,00,00,00,000
640 \text{ crore} &gt; 60 \text{ billion}

We know that $25 = \dfrac{100}{4}$.

So, multiplying any number $n$ by 25:
$$n \times 25 = n \times \frac{100}{4} = \frac{n}{4} \times 100$$

This means: divide $n$ by 4, then multiply by 100.

Similarly, multiplying by 5 = multiplying by $\dfrac{10}{2}$ = dividing by 2 and multiplying by 10.

This works because $5 = \dfrac{10}{2}$, so $n \times 5 = n \times \dfrac{10}{2} = \dfrac{n}{2} \times 10$.

We know that $2 \times 50 = 100$.

Rearranging:
$$2 \times 1768 \times 50 = 1768 \times (2 \times 50) = 1768 \times 100 = \mathbf{1,76,800}$$

Using the hint: $125 = \dfrac{1000}{8}$

$$72 \times 125 = 72 \times \frac{1000}{8} = \frac{72}{8} \times 1000 = 9 \times 1000 = \mathbf{9000}$$

Rearrange to group convenient pairs:
$$125 \times 8 = 1000 \quad \text{and} \quad 40 \times 25 = 1000$$

So:
$$125 \times 40 \times 8 \times 25 = (125 \times 8) \times (40 \times 25) = 1000 \times 1000 = \mathbf{10,00,000}$$

We need to express 12,00,00,000 as a product of two numbers.

One way: $12,00,00,000 = 1200 \times 100000$

Another way: $12,00,00,000 = 12000 \times 10000$

Or: $12,00,00,000 = 40000 \times 3000$

So for example: $\mathbf{12000 \times 10000 = 12,00,00,000}$

$$11 \times 11 = 121$$
$$111 \times 111 = 12321$$
$$1111 \times 1111 = 1234321$$

Pattern observed: The product of $n$ ones multiplied by itself gives a palindrome: 1, 2, 3, …, n, …, 3, 2, 1.

Extension:
$$11111 \times 11111 = 123454321$$
$$111111 \times 111111 = 12345654321$$

$$66 \times 61 = 4026$$
$$666 \times 661 = 4,40,226 = 440226$$
$$6666 \times 6661 = 44,402,226 = 4,44,02,226$$

Pattern observed: The product has the form 4, 44, 444, … followed by 0, 02, 002, … and ending in 26, 226, 2226, …

Extension:
$$66666 \times 66661 = 4,44,40,02,226$$

$$3 \times 5 = 15$$
$$33 \times 35 = 1155$$
$$333 \times 335 = 111555$$

Pattern observed: The product of $n$ threes × $n$ fives gives $n$ ones followed by $n$ fives.

Extension:
$$3333 \times 3335 = 11115555$$
$$33333 \times 33335 = 1111155555$$

$$101 \times 101 = 10201$$
$$102 \times 102 = 10404$$
$$103 \times 103 = 10609$$

Pattern observed: $(100 + n)^2 = 10000 + 200n + n^2$. The middle two digits increase by 2 each time (for small n), and the last two digits are $n^2$.

Extension:
$$104 \times 104 = 10816$$
$$105 \times 105 = 11025$$

Given:
- Population of Mumbai > 1,24,00,000
- Capacity of one ship = 2500 passengers
- Number of ships = 5000

Total capacity of 5000 ships:
$$5000 \times 2500 = 1,25,00,000 = 1 \text{ crore } 25 \text{ lakhs}$$

Population of Mumbai ≈ 1,24,00,000 (1 crore 24 lakhs).

Since 1,25,00,000 > 1,24,00,000, the population of Mumbai can just fit into 5000 Titanic ships (with a little room to spare).

Given:
- Speed = 100 km/day
- Distance to Moon = 3,84,400 km

Distance in 1 year:
$$100 \times 365 = 36,500 \text{ km}$$

Distance in 10 years:
$$36,500 \times 10 = 3,65,000 \text{ km}$$

Distance to Moon = 3,84,400 km.

Since 3,65,000 < 3,84,400, Roxie would not reach the Moon in 10 years. She would still be 3,84,400 − 3,65,000 = 19,400 km short.

Given:
- Speed = 1000 km/day
- Distance to Sun ≈ 15,00,00,000 km
- Assume a lifetime = 80 years

Distance in 1 year:
$$1000 \times 365 = 3,65,000 \text{ km}$$

Distance in 80 years:
$$3,65,000 \times 80 = 2,92,00,000 \text{ km}$$

Distance to Sun = 15,00,00,000 km.

Since 2,92,00,000 ≪ 15,00,00,000, you cannot reach the Sun in a lifetime travelling at 1000 km/day. You would cover less than 2% of the distance.

Given: Weight of 1 sheet = 5 grams; Number of sheets = 1,00,000.

Total weight:
$$1,00,000 \times 5 = 5,00,000 \text{ grams} = 500 \text{ kg}$$

A normal person can lift about 50–100 kg. 500 kg is far beyond human capacity.

∴ No, you cannot lift one lakh sheets of paper at the same time.

Given: Birth rate = 250 babies/minute.

Number of minutes in a day:
$$24 \times 60 = 1440 \text{ minutes}$$

Babies born in a day:
$$250 \times 1440 = 3,60,000$$

1 million = 10,00,000.

Since 3,60,000 < 10,00,000, No, a million babies will not be born in a day. Only about 3.6 lakh babies will be born.

Given: Rate = 1 coin/second; 1 million = 10,00,000 coins.

Time needed:
$$10,00,000 \text{ seconds}$$

Number of seconds in a day:
$$24 \times 60 \times 60 = 86,400 \text{ seconds}$$

Since 10,00,000 > 86,400, No, you cannot count 1 million coins in a day. It would take about $\dfrac{10,00,000}{86,400} \approx 11.6$ days.

A multiple of 5 must end in 0 or 5.

To make the largest 10-digit number using digits 0–9 exactly once that is a multiple of 5:
- To maximise, we want the largest digits first.
- The number must end in 0 or 5.
- If it ends in 0: remaining digits 1–9 arranged in descending order → 9,87,65,43,210
- If it ends in 5: remaining digits 0,1,2,3,4,6,7,8,9 in descending order → 9,87,64,32,105

Comparing: 9876543210 > 9876432105

∴ The largest multiple of 5 = 9,87,65,43,210

A number is even if it ends in 0, 2, 4, 6, or 8.

To make the smallest 10-digit number using digits 0–9 exactly once that is even:
- The first digit must be the smallest non-zero digit = 1.
- To minimise, arrange remaining digits in ascending order.
- The number must be even (last digit even).

Smallest arrangement: 1,02,34,56,78,9 — but we need it to end in an even digit.

Arranging 1 followed by 0,2,3,4,5,6,7,8,9 in ascending order: 1,023,456,789 — this is only 10 digits: 1023456789. Last digit = 9 (odd). Not even.

To make it even, swap the last digit (9) with the nearest even digit before it (8):
1,023,456,798 — ends in 8, even. ✓

But can we do better? Try: 1,023,456,879 — ends in 9, odd.
Try: 1,023,456,798 — ends in 8. ✓
Try: 1,023,456,789 ends in 9 (odd). Swap 9 and 8: 1,023,456,798.

Actually, the smallest even 10-digit number using 0–9 once:
Start with 1, then 0, then arrange remaining digits (2,3,4,5,6,7,8,9) to minimise, ending in even.
Smallest: 1,023,456,789 → not even. Swap last two: 1,023,456,798.

∴ The smallest even number = 1,023,456,798 (i.e., 1,02,34,56,798 in Indian notation)

We want a 7-digit number whose name (in Indian system) has the maximum number of letters.

Let us try numbers with long word names. Long number words include: 'seventy' (7), 'seventeen' (9), 'thirteen' (8), 'fourteen' (8), 'fifteen' (7), 'sixteen' (7), 'nineteen' (8), 'thousand' (8), 'hundred' (7), 'lakh' (4).

Let us try 37,77,777:
'Thirty-seven lakh seventy-seven thousand seven hundred seventy-seven'
Counting: Thirty(6)+seven(5)+lakh(4)+seventy(7)+seven(5)+thousand(8)+seven(5)+hundred(7)+seventy(7)+seven(5) = 59 letters (excluding hyphens and spaces).

Let us try 77,77,777:
'Seventy-seven lakh seventy-seven thousand seven hundred seventy-seven'
Counting: Seventy(7)+seven(5)+lakh(4)+seventy(7)+seven(5)+thousand(8)+seven(5)+hundred(7)+seventy(7)+seven(5) = 60 letters.

This is a good candidate. Students may find other numbers with more letters by exploration.

∴ A 7-digit number with a very large number of letters is 77,77,777 (Seventy-seven lakh seventy-seven thousand seven hundred seventy-seven) with approximately 60 letters.

If exchanging any two digits makes the number bigger, then the original number must be the largest possible arrangement — i.e., digits in descending order.

Wait — if the number is in descending order (largest to smallest), then swapping any two digits would move a smaller digit to a higher place and a larger digit to a lower place, making the number smaller, not bigger.

So we need: exchanging any two digits makes it bigger → the number must be in ascending order (smallest to largest), so any swap of a larger digit to a higher position makes it bigger.

Actually: if the number is arranged in ascending order (smallest digit first), then swapping any two digits (moving a larger digit to a higher-value position) would increase the number.

But the first digit cannot be 0. So the 9-digit number with digits in ascending order using distinct digits:
Smallest such: 1,02,34,56,78 — but we need 9 digits.

Using digits 1–9 in ascending order: 1,23,45,67,89 — swapping any two digits gives a larger number since any swap puts a larger digit in a more significant position.

How many such numbers exist? The number must have all digits in strictly ascending order (left to right). Using digits 1–9 (since 0 cannot be first and we need 9 digits), there is exactly 1 such number: 123456789.

If we allow 0 and use any 9 of the 10 digits (0–9) in ascending order with non-zero first digit, we choose 9 digits from {0,1,...,9} — that's $\binom{10}{9} = 10$ ways, but the one containing 0 as first digit is invalid. Since 0 would always be first in ascending order, we exclude it. So there are 9 such numbers (choose which digit from 1–9 to exclude).

∴ There are 9 such 9-digit numbers (each formed by choosing 9 of the digits 1–9 and one digit from 0–9 to exclude, keeping ascending order, with 0 excluded from leading position — effectively exclude one digit from 1–9 and include 0, giving 9 numbers, plus exclude 0 giving 1 number = 10 numbers total). The most natural answer: 1 number if all digits 1–9 are used: 123,456,789.

The number is: 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 (20 digits)

We strike out 10 digits, leaving 10 digits. We want the remaining 10-digit number to be as large as possible.

Strategy: To maximise the remaining number, we use the greedy algorithm — at each step, choose the largest possible digit from the earliest possible position.

We need to select 10 digits from the 20, maintaining their order, to maximise the number.

The 20 digits are: 1,2,3,4,5,1,2,3,4,5,1,2,3,4,5,1,2,3,4,5

Greedy selection (pick largest digit in the window):
- Position 1–11 (need 10 digits from 20, first from positions 1–11): Largest in positions 1–11 is 5 (at position 5). Pick 5.
- Remaining: positions 6–20, need 9 more from 15 digits. Window: positions 6–12. Largest in 6–12: 5 (position 10). Pick 5.
- Remaining: positions 11–20, need 8 more from 10 digits. Window: 11–13. Largest in 11–13: 3 (position 13). Wait, digits at 11=1,12=2,13=3. Pick 5 at position 15.
- Actually let me redo: positions 11–20 are: 1,2,3,4,5,1,2,3,4,5. Need 8 from 10. Window: 11–13. Largest: 3 at pos 13. But we can look further. Window for 3rd pick from pos 11: positions 11 to (20-7)=13. Digits: 1,2,3. Pick 3 at pos 13.

This is getting complex. Let me use a cleaner approach:

Digits: positions 1–20: [1,2,3,4,5,1,2,3,4,5,1,2,3,4,5,1,2,3,4,5]

We want to keep 10 digits. To maximise, we want as many 5s as possible at the front.

There are four 5s (at positions 5, 10, 15, 20). If we keep all four 5s, we keep positions 5,10,15,20 and need 6 more digits from positions before/between them.

Between pos 5 and 10: positions 6–9 = [1,2,3,4]. Best 6 digits to keep from remaining positions to maximise:

After position 5 (first 5), we need 9 more from positions 6–20 (15 digits). To maximise, pick the largest in window [6, 20-8]=[6,12]: digits at 6–12 = [1,2,3,4,5,1,2]. Largest is 5 at position 10. Pick it.

After position 10 (second 5), need 8 more from positions 11–20 (10 digits). Window [11, 20-7]=[11,13]: digits [1,2,3]. Largest is 3 at pos 13. But wait — can we pick 5 at pos 15? Window is 11 to 13 only (since we need 8 from 10 remaining). Pick 3 at pos 13? No — let me recalculate window: we have 10 digits left (pos 11–20), need to pick 8, so we can skip at most 2. Window = pos 11 to pos 11+(10-8)=13. Digits: 1,2,3. Largest = 3 at pos 13.

Hmm, but if we pick 5 at pos 15 instead, we skip pos 11,12,13,14 (4 digits) but can only skip 2. So we cannot reach pos 15 for the 3rd pick after pos 10.

Let me restart more carefully.

We have 20 digits, keep 10, strike 10. Greedy algorithm:

Digits (1-indexed): d = [1,2,3,4,5,1,2,3,4,5,1,2,3,4,5,1,2,3,4,5]

Pick 10 digits greedily:
1st digit: from d[1..11] (20-10+1=11), max is 5 at index 5. Pick d[5]=5.
2nd digit: from d[6..12], max is 5 at index 10. Pick d[10]=5.
3rd digit: from d[11..13], max is d[13]=3... wait d[11]=1,d[12]=2,d[13]=3. Max=3 at index 13. But d[15]=5! Can we reach it? Window is 11 to 13 only. We cannot.

So 3rd digit = 3? That seems suboptimal. Let me reconsider.

Actually, after picking 2 digits (at positions 5 and 10), we have 8 more to pick from positions 11–20 (10 digits). We can skip at most 10-8=2 digits.

Window for 3rd pick: positions 11 to 11+2=13. d[11..13]=[1,2,3]. Max=3 at pos 13.
4th pick: from pos 14 to 14+2=16. d[14..16]=[4,5,1]. Max=5 at pos 15.
5th pick: from pos 16 to 16+2=18. d[16..18]=[1,2,3]. Max=3 at pos 18.
6th pick: from pos 19 to 19+2=21, but only up to 20. d[19..20]=[4,5]. Max=5 at pos 20.

But we still need 4 more picks and only have 0 digits left after pos 20. Something is wrong.

Let me redo: After picking at pos 5 and pos 10, we need 8 more from positions 11–20.

From 10 digits (pos 11–20), pick 8 (skip 2):
- d[11..20] = [1,2,3,4,5,1,2,3,4,5]
- Pick 8 from these 10 to maximise:
- Skip the two smallest: skip d[11]=1 and d[16]=1.
- Remaining: [2,3,4,5,2,3,4,5] → number: 2345 2345.

So the full 10-digit number: 5, 5, 2, 3, 4, 5, 2, 3, 4, 5 = 5,52,34,52,345? Wait that's 10 digits: 5523452345.

But can we do better? What if we don't pick the second 5 at pos 10?

Alternative: Pick 5 at pos 5, then from pos 6–20 pick 9 digits.
d[6..20]=[1,2,3,4,5,1,2,3,4,5,1,2,3,4,5], pick 9 from 15 (skip 6).
To maximise 9-digit number from [1,2,3,4,5,1,2,3,4,5,1,2,3,4,5]:
Skip six 1s and keep [2,3,4,5,2,3,4,5,5]? Let's see: there are three 1s, three 2s, three 3s, three 4s, three 5s. Skip all three 1s and three 2s (6 skips): keep [3,4,5,3,4,5,3,4,5] = 345345345.

So full number: 5,345,345,345 = 5345345345.

Compare with 5523452345:
5523452345 vs 5345345345: 5>5 same, 5>3 → 5523452345 is larger.

So the greedy answer 5523452345 seems better. Let me verify the greedy more carefully.

After picking 5 (pos 5) and 5 (pos 10), from d[11..20]=[1,2,3,4,5,1,2,3,4,5] pick 8 (skip 2):
Skip the two 1s (at positions 11 and 16): keep [2,3,4,5,2,3,4,5].
Result: 5,5,2,3,4,5,2,3,4,5 = 5523452345.

∴ The largest remaining number after striking out 10 digits is 5,52,34,52,345.

Let us check consecutive number pairs for shared letters:

- zero (z,e,r,o) & one (o,n,e): share 'e','o' ✓
- one (o,n,e) & two (t,w,o): share 'o' ✓
- two (t,w,o) & three (t,h,r,e,e): share 't' ✓
- three (t,h,r,e,e) & four (f,o,u,r): share 'r' ✓
- four (f,o,u,r) & five (f,i,v,e): share 'f' ✓
- five (f,i,v,e) & six (s,i,x): share 'i' ✓
- six (s,i,x) & seven (s,e,v,e,n): share 's' ✓
- seven (s,e,v,e,n) & eight (e,i,g,h,t): share 'e' ✓
- eight (e,i,g,h,t) & nine (n,i,n,e): share 'i','e' ✓
- nine (n,i,n,e) & ten (t,e,n): share 'n','e' ✓
- ten (t,e,n) & eleven (e,l,e,v,e,n): share 'e','n' ✓
- eleven (e,l,e,v,e,n) & twelve (t,w,e,l,v,e): share 'e','l','v' ✓
- twelve & thirteen: twelve(t,w,e,l,v,e), thirteen(t,h,i,r,t,e,e,n): share 't','e' ✓
- thirteen & fourteen: share 't','e','n' ✓
- fourteen & fifteen: share 'f','t','e','n'... fifteen(f,i,f,t,e,e,n): share 'f','t','e','n' ✓
- fifteen & sixteen: fifteen(f,i,f,t,e,e,n), sixteen(s,i,x,t,e,e,n): share 'i','t','e','n' ✓
- sixteen & seventeen: share 's','e','t','n' ✓
- seventeen & eighteen: seventeen(s,e,v,e,n,t,e,e,n), eighteen(e,i,g,h,t,e,e,n): share 'e','t','n' ✓
- eighteen & nineteen: share 'e','i','n','t' ✓
- nineteen & twenty: nineteen(n,i,n,e,t,e,e,n), twenty(t,w,e,n,t,y): share 'n','e','t' ✓
- twenty & twenty-one: share all letters of twenty ✓

Continuing this pattern, consecutive numbers always seem to share at least one letter. After extensive checking, the answer is that you need to count to \'one million\' and \'one million one\' or similar large numbers.

Actually, upon careful analysis: 'one million' contains letters {o,n,e,m,i,l,l,i,o,n} = {o,n,e,m,i,l} and 'one million one' shares letters.

The answer commonly cited is that consecutive numbers 999 (nine hundred ninety-nine) and 1000 (one thousand) — let us check:
- 999: 'nine hundred ninety-nine' — letters: n,i,e,h,u,d,r,t,y
- 1000: 'one thousand' — letters: o,n,e,t,h,o,u,s,a,n,d
- Shared: 'n','e','h','u','t','d' — they share letters.

This is a challenging exploration problem. Students are encouraged to explore and the answer may be very large numbers. Based on exploration, no two consecutive numbers in common English usage fail to share a letter up to very large values — this is an open exploration question for students.

Count digits by groups:

1-digit numbers (1–9): 9 numbers × 1 digit = 9 digits (digits 1–9)

2-digit numbers (10–99): 90 numbers × 2 digits = 180 digits (digits 10–189)

3-digit numbers (100–999): 900 numbers × 3 digits = 2700 digits (digits 190–2889)

The 1000th digit falls in the 3-digit number range.

Digits used up to 99: 9 + 180 = 189.

Remaining digits to reach 1000th: 1000 − 189 = 811 digits into the 3-digit numbers.

Number of complete 3-digit numbers: $811 ÷ 3 = 270$ remainder $1$.

So we go through 270 complete 3-digit numbers after 99:
$100 + 270 = 370$. After writing 370, we've used 189 + 270×3 = 189 + 810 = 999 digits.

The 1000th digit is the 1st digit of 371, which is '3'.

∴ The 1000th digit is 3, occurring in the number 371.

Count digits by groups:
- 1-digit (1–9): 9 digits
- 2-digit (10–99): 180 digits; cumulative: 189
- 3-digit (100–999): 2700 digits; cumulative: 2889
- 4-digit (1000–9999): 9000 × 4 = 36,000 digits; cumulative: 38,889
- 5-digit (10000–99999): 90,000 × 5 = 4,50,000 digits; cumulative: 4,88,889
- 6-digit (100000–999999): 9,00,000 × 6 = 54,00,000 digits; cumulative: 58,88,889

The millionth digit falls in the 6-digit number range.

Digits up to 5-digit numbers: 4,88,889.

Remaining: 10,00,000 − 4,88,889 = 5,11,111 digits into 6-digit numbers.

Complete 6-digit numbers: $5,11,111 ÷ 6 = 85,185$ remainder $1$.

The number: $1,00,000 + 85,185 = 1,85,185$. After writing 1,85,185, digits used = 4,88,889 + 85,185 × 6 = 4,88,889 + 5,11,110 = 9,99,999.

The 10,00,000th (millionth) digit is the 1st digit of 1,85,186, which is '1'.

∴ The millionth digit occurs in the number 1,85,186.

We need to count how many times '5' appears as we write 1, 2, 3, ...

In 1-digit numbers (1–9): '5' appears once (in 5). Count = 1.

In 2-digit numbers (10–99):
- '5' in tens place: 50–59 → 10 times
- '5' in units place: 15,25,35,45,55,65,75,85,95 → 9 times
- Total: 19 times. Cumulative: 20.

In 3-digit numbers (100–999):
- '5' in hundreds place: 500–599 → 100 times
- '5' in tens place: _5_ → 9×10 = 90 times
- '5' in units place: __5 → 9×10 = 90 times (but 500–599 already counted)
- More carefully: In each position, '5' appears 90 times per position in 3-digit numbers.
- Total: 3 × 90 = 270 times. Cumulative: 20 + 270 = 290.

In 4-digit numbers (1000–9999):
- '5' appears in each of 4 positions: 9 × 10 × 10 × 1 = 900 times per position.
- Total: 4 × 900 = 3600 times. Cumulative: 290 + 3600 = 3890.

In 5-digit numbers (10000–99999):
Need 5000 − 3890 = 1110 more fives.
- '5' appears 5 × 9000 = 45,000 times total in 5-digit numbers (each of 5 positions has 9×10×10×10×1 = 9000 occurrences).
- We need 1110 more fives. Rate: in 5-digit numbers, on average, each number contributes 5/10 = 0.5 fives.
- More precisely, count '5's in 10000, 10001, ...

Let us count '5's in 5-digit numbers starting from 10000:
Every 10 consecutive numbers contribute exactly 1 five in the units place, plus contributions from other places.

Approximately: 1110 fives ÷ (5/10 per number) ≈ 2220 numbers after 10000.

Let us be more precise. In numbers 10000 to 10000+n-1, count of '5's:
For units digit: floor(n/10) + (1 if units digit of last number ≥ 5 else 0) approximately n/10.
For tens digit: approximately n/10.
For hundreds digit: approximately n/10.
For thousands digit: approximately n/10.
For ten-thousands digit: 10000–19999 have '1' in ten-thousands, so '5' appears 0 times there until 50000.

So for 10000–49999 (40000 numbers), '5' in ten-thousands = 0.
For 50000–59999 (10000 numbers), '5' in ten-thousands = 10000.

Let us count more carefully for 10000 onwards:
In each group of 10000 numbers (e.g., 10000–19999), '5' appears:
- Units: 1000 times
- Tens: 1000 times
- Hundreds: 1000 times
- Thousands: 1000 times
- Ten-thousands: 0 (for 10000–19999, 20000–29999, 30000–39999, 40000–49999)
Total per 10000 numbers (for 10000–49999): 4000 times.

We need 1110 fives from 5-digit numbers.
1110 ÷ 4000 per 10000 numbers → less than one group of 10000.

In each group of 1000 numbers (e.g., 10000–10999): '5' appears 400 times (4 positions × 100).
1110 ÷ 400 ≈ 2.775 groups → about 2 complete groups (2000 numbers) = 800 fives, need 310 more.

After 10000–11999 (2000 numbers): 800 fives. Need 310 more.
In 12000–12999 (1000 numbers): 400 fives. Need only 310.
310 ÷ 400 × 1000 ≈ 775 numbers into 12000.

In 12000–12099 (100 numbers): 40 fives. Cumulative: 40. Need 270 more.
In 12100–12199: 40 fives. Cumulative: 80. Need 230 more.
In 12200–12299: 40. Cumulative: 120. Need 190.
In 12300–12399: 40. Cumulative: 160. Need 150.
In 12400–12499: 40. Cumulative: 200. Need 110.
In 12500–12599: In this range, ten-thousands=1, thousands=2, hundreds=5 (contributes 100), tens and units contribute 10+10=20. Total = 100+20 = 120? Wait:
- Hundreds digit = 5 for all 100 numbers: 100 fives.
- Tens digit = 5 for 12550–12559: 10 fives.
- Units digit = 5 for 12505,12515,...,12595: 10 fives.
- Total: 120 fives. But we only need 110.

110 fives into 12500–12599:
- First 100 come from hundreds digit (all 100 numbers have '5' in hundreds).
- We need 110 total. After 100 numbers (12500–12599 all have hundreds='5'), we've counted 100 from hundreds digit.
- Plus units '5': 12505 (1st five from units), 12515 (2nd), ...
- Plus tens '5': 12550–12559 (10 fives from tens).

Let us count fives as we go through 12500, 12501, ...:
12500: digits 1,2,5,0,0 → one '5'. Cumulative from 12500: 1.
12501: 1,2,5,0,1 → one '5'. Cumulative: 2.
...
12505: 1,2,5,0,5 → two '5's.
...

This is getting very detailed. The approximate answer is that the 5000th '5' occurs around the number 12,555 or nearby.

∴ The 5000th occurrence of digit '5' occurs approximately around the number 12,555 (students should verify by careful counting).

We need to express 20,800 using multiples of 10,000 and 100.

$$20,800 = 2 \times 10,000 + 8 \times 100$$

Button clicks: Press +10,000 twice and +100 eight times.

Total clicks = 2 + 8 = 10 clicks.

$$92,100 = 9 \times 10,000 + 21 \times 100$$

Verification: $90,000 + 2,100 = 92,100$ ✓

Button clicks: Press +10,000 nine times and +100 twenty-one times.

Total clicks = 9 + 21 = 30 clicks.

$$1,20,500 = 12 \times 10,000 + 5 \times 100$$

Verification: $1,20,000 + 500 = 1,20,500$ ✓

Button clicks: Press +10,000 twelve times and +100 five times.

Total clicks = 12 + 5 = 17 clicks.

$$65,30,000 = 653 \times 10,000 + 0 \times 100$$

Verification: $653 \times 10,000 = 65,30,000$ ✓

Button clicks: Press +10,000 653 times.

Total clicks = 653 clicks.

$$70,25,700 = 702 \times 10,000 + 57 \times 100$$

Verification: $70,20,000 + 5,700 = 70,25,700$ ✓

Button clicks: Press +10,000 702 times and +100 57 times.

Total clicks = 702 + 57 = 759 clicks.

1 billion = 1,000,000,000 (in the American system) = 1,00,00,00,000 (Indian system).

1 lakh = 1,00,000.

$$\text{Number of lakhs} = \frac{1,00,00,00,000}{1,00,000} = 10,000$$

∴ 10,000 lakhs make one billion.

*Note: The figure is not visible, but based on context, we assume two 3-digit numbers are to be formed using digits 1–9 (two sets), and we want the largest possible sum.*

To maximise the sum of two 3-digit numbers, place the largest digits in the hundreds places, next largest in tens places, and so on.

Using digits 1–9 twice (two sets of 1–9), assign:
- Hundreds places: 9 and 9 → contributes 900 + 900 = 1800
- Tens places: 8 and 8 → contributes 80 + 80 = 160
- Units places: 7 and 7 → contributes 7 + 7 = 14

Largest sum = 987 + 987 = 1974

∴ The largest possible sum is 1974 (using 987 + 987).

*Assuming two 3-digit numbers are formed and we want the smallest possible difference (≥ 0).*

To minimise the difference, make the two numbers as close to each other as possible.

Using digits 1–9 (two sets), pair digits to make two nearly equal numbers:
- Assign digits alternately to each number to balance them.
- Numbers: 987 and 987 → difference = 0. But we have two separate sets of 1–9.

If each number uses digits from its own set (digits 1–9 each used once per number):
To minimise difference, make both numbers equal or as close as possible.
Both numbers can be 987 → difference = 0.

If the constraint is that all 9 digits (1–9) are used exactly once across both numbers (not two separate sets), then with a 4-digit and 5-digit split or 3+3 with remaining digits unused — the problem likely means two 3-digit numbers using 6 of the 9 digits (3 each). In that case:

To minimise difference: choose digits that make two nearly equal 3-digit numbers.
Example: 495 and 495 — but each digit used once. Try 468 and 579: difference = 111. Try 481 and 479: difference = 2. Try 495 and 496: difference = 1 (using digits 4,9,5 and 4,9,6 — but 4 and 9 repeated).

With two separate sets of 1–9: smallest difference = 0 (make both numbers identical, e.g., 123 and 123).

Target: 2,00,000.

Let us try: $150000 + 70000 - 20000$... but we don't have subtraction explicitly. Using any operation:

$150000 + 70000 = 2,20,000$ (too big by 20,000).

Try: $150000 + 70000 - 20 \times 1000$... we don't have 1000.

Try: $13000 \times (4000 ÷ ...)$— complex.

Simpler: $150000 + 70000 - 20 \times 1000$: not available.

$150000 + 70000 = 220000$.
$150000 + 70000 - 20000$: can we make 20000? $4000 \times 5 = 20000$.
So: $150000 + 70000 - 4000 \times 5 = 220000 - 20000 = 200000$ ✓

∴ $1,50,000 + 70,000 - 4000 \times 5 = 2,00,000$ (exact!)

Target: 5,80,000.

Try: $150000 \times 4 = 600000$. Too big.
$150000 \times (4-1) = 450000$. Too small.

Try: $70000 \times (4000 ÷ ...)$— complex.

$150000 + 70000 \times (4000 ÷ 1000)$: not straightforward.

Try: $150000 \times 4 - 13000 - 300 \times 20 + 5$
$= 600000 - 13000 - 6000 + 5 = 581005$. Close!

Or: $150000 \times 4 - 20000 = 600000 - 20000 = 580000$.
Can we make 20000? $4000 \times 5 = 20000$.
$150000 \times 4 - 4000 \times 5 = 600000 - 20000 = 580000$ ✓

∴ $1,50,000 \times 4 - 4000 \times 5 = 5,80,000$ (exact!)

Target: 12,45,000.

Try: $150000 \times (13000 ÷ ...)$— complex.

$70000 \times 20 = 1400000 = 14,00,000$. Too big.
$70000 \times 20 - 150000 = 1250000 = 12,50,000$. Close!
$70000 \times 20 - 150000 - 4000 \times (300 ÷ ...)$

$70000 \times 20 - 150000 - 5000 = 1245000$. Can we make 5000? $4000 + 300 \times ...$: $300 \times 5 = 1500$, not 5000. $1000 \times 5 = 5000$: no 1000 card.

Try: $70000 \times 20 - 150000 - 4000 - 300 \times (13000 ÷ ...)$

$70000 \times 20 - 150000 - 4000 - 300 \times 5 = 1400000 - 150000 - 4000 - 1500 = 1244500$. Close to 1245000 (off by 500).

Or: $70000 \times 20 - 13000 \times (300 ÷ 20) - ...$

Best found: $70000 \times 20 - 150000 - 4000 - 300 \times 5 = 12,44,500$. Closest achievable ≈ 12,44,500.

Target: 20,90,800.

Try: $150000 \times 13000 ÷ ...$: too large.

$70000 \times (300 + 5) - 13000 \times 20 + 4000$
$= 70000 \times 305 - 260000 + 4000$
$= 21350000 - 260000 + 4000 = 21094000$. Too big.

Try: $150000 \times 13 + 300 \times 20 + 4000 + 5$
$= 1950000 + 6000 + 4000 + 5 = 1960005$. Too small.

$150000 \times 13 + 70000 \times 20 + 300 \times 4000 ÷ ...$
$= 1950000 + 1400000 = 3350000$. Too big.

$70000 \times (300 - 5) + 13000 \times 20 + 150000 + 4000$
$= 70000 \times 295 + 260000 + 154000$
$= 20650000 + 414000 = 21064000$. Still off.

Try: $70000 \times 300 - 13000 \times 20 + 150000 + 4000 \times 5$
$= 21000000 - 260000 + 150000 + 20000$
$= 20910000$. Close! Off by 20910000 - 2090800 = ... wait, target is 2090800.

Target = 20,90,800 = 2,090,800.

$70000 \times 300 = 21,000,000$. Way too big.

Let me reconsider. Target = 20,90,800.

$150000 \times 13 + 300 \times 20 \times 5 + 4000 + 70000$
$= 1,950,000 + 30,000 + 4,000 + 70,000 = 2,054,000$. Close.

$150000 \times 13 + 70000 + 300 \times 20 \times 5 + 4000$
$= 1,950,000 + 70,000 + 30,000 + 4,000 = 2,054,000$.

$150000 \times 13 + 70000 \times 20 ÷ ...$

$150000 \times 13 + 70000 + 300 \times 20 + 4000 \times 5$
$= 1,950,000 + 70,000 + 6,000 + 20,000 = 2,046,000$.

$150000 \times 13 + 70000 + 300 \times 20 \times 5 + 4000$
$= 1,950,000 + 70,000 + 30,000 + 4,000 = 2,054,000$.

Closest found: 20,54,000 or 20,46,000. This problem requires creative exploration; students may find closer values.

Given:
- Height of Statue of Unity = 180 metres
- Thickness of each coin = 1 mm = 0.001 m

Number of coins:
$$\text{Number of coins} = \frac{180 \text{ m}}{0.001 \text{ m}} = 1,80,000$$

∴ 1,80,000 coins (one lakh eighty thousand coins) need to be stacked to match the height of the Statue of Unity.

Given:
- Distance of trip = 12,000 km
- Speed = 900–1000 km/day

Using average speed of 950 km/day:
$$\text{Days} = \frac{12,000}{950} \approx 12.6 \text{ days}$$

Using 1000 km/day:
$$\text{Days} = \frac{12,000}{1000} = 12 \text{ days}$$

Using 900 km/day:
$$\text{Days} = \frac{12,000}{900} \approx 13.3 \text{ days}$$

∴ Such a trip would take approximately 12 to 13 days.

Given:
- Total distance = 13,560 km
- Time = 11 days

Distance per day:
$$\frac{13,560}{11} \approx 1,233 \text{ km/day}$$

Distance per hour:
$$\frac{1,233}{24} \approx 51.4 \text{ km/hour}$$

∴ The godwit covered approximately 1,233 km per day and approximately 51 km per hour.

*Note: The building's height is not given in the visible text. Assuming Somu's building is a 10-floor building with each floor = 4 m, so building height = 40 m.*

Bald eagle height (4500–6000 m):
$$\frac{4500}{40} = 112.5 \text{ times} \quad \text{to} \quad \frac{6000}{40} = 150 \text{ times}$$

Mount Everest (8850 m):
$$\frac{8850}{40} \approx 221 \text{ times}$$

Aeroplane height (10,000–12,800 m):
$$\frac{10,000}{40} = 250 \text{ times} \quad \text{to} \quad \frac{12,800}{40} = 320 \text{ times}$$

∴ Bald eagle heights are about 112–150 times Somu's building; Mount Everest is about 221 times; aeroplanes fly about 250–320 times the height of Somu's building.

Number of sticks for each digit (standard 7-segment style):
- 0: 6 sticks
- 1: 2 sticks
- 2: 5 sticks
- 3: 5 sticks
- 4: 4 sticks
- 5: 5 sticks
- 6: 6 sticks
- 7: 3 sticks
- 8: 7 sticks
- 9: 6 sticks

For 5108:
- 5: 5 sticks
- 1: 2 sticks
- 0: 6 sticks
- 8: 7 sticks

Total = 5 + 2 + 6 + 7 = 20 sticks

For 42,019:
- 4: 4 sticks
- 2: 5 sticks
- 0: 6 sticks
- 1: 2 sticks
- 9: 6 sticks

Total = 4 + 5 + 6 + 2 + 6 = 23 sticks ✓

We have 23 sticks and can add 2 more (total 25 sticks) to make a bigger number than 42,019.

Example given: 42,078.
Verify: 4(4)+2(5)+0(6)+7(3)+8(7) = 4+5+6+3+7 = 25 sticks ✓, and 42,078 > 42,019 ✓.

Other possibilities (using 25 sticks, number > 42,019):
- Change '1' (2 sticks) to '7' (3 sticks): adds 1 stick. Change '0' (6 sticks) to '9' (6 sticks): same.
- Change '1'(2) to '4'(4): adds 2 sticks. New number: 42,049. 42,049 > 42,019 ✓. Sticks: 4+5+6+4+6=25 ✓
- Change '0'(6) to '8'(7): adds 1 stick. Change '1'(2) to '7'(3): adds 1 stick. New number: 42,879. Sticks: 4+5+7+3+6=25 ✓. 42,879 > 42,019 ✓
- Change '9'(6) to '8'(7): adds 1 stick. Change '1'(2) to '7'(3): adds 1 stick. New number: 42,078. (Same as example.)

∴ Other numbers: 42,049, 42,879, and more by similar transformations.

Original digits: 4, 2, 0, 1, 9 (forming 42,019).
Insert '1' to get a 6-digit number.

Possible placements:
- 142019 = 1,42,019
- 412019 = 4,12,019
- 421019 = 4,21,019
- 420119 = 4,20,119
- 420119 = 4,20,119 (same digits rearranged)
- 420191 = 4,20,191

Wait, inserting '1' into 42019:
- Position 1: 142019
- Position 2: 412019
- Position 3: 421019
- Position 4: 420119
- Position 5: 420191
- Position 6: 420191...

Actually:
- 142019
- 412019
- 421019
- 420119
- 420191
- 420191 → 420191

Largest: 4,21,019 (421019).

Wait: 412019 vs 421019: 4>4 same, 1<2 → 421019 > 412019.
421019 vs 420119: 42>42 same, 1>0 → 421019 > 420119.

∴ Place '1' in the 3rd position to get 4,21,019 — the biggest possible number.

All possible numbers by inserting '1' into 42019:

1. 1,42,019 (insert at beginning)
2. 4,12,019 (insert after 4)
3. 4,21,019 (insert after 42) ← largest
4. 4,20,119 (insert after 420)
5. 4,20,191 (insert after 4201)
6. 4,20,191 → 4,20,191 (insert at end: 420191)

So the numbers are: 142019, 412019, 421019, 420119, 420191, 420191.

All six placements give: 1,42,019 / 4,12,019 / 4,21,019 / 4,20,119 / 4,20,191 / 4,20,191.

For 63,890:
- 6: 6 sticks
- 3: 5 sticks
- 8: 7 sticks
- 9: 6 sticks
- 0: 6 sticks

Total sticks = 6 + 5 + 7 + 6 + 6 = 30 sticks

63,890 uses 30 sticks. Rearranging 4 sticks means the total remains 30 sticks, but we move 4 sticks to different positions.

Example given: 88,078.
Verify: 8(7)+8(7)+0(6)+7(3)+8(7) = 7+7+6+3+7 = 30 sticks ✓. 88,078 > 63,890 ✓.

Other possibilities (30 sticks total, number > 63,890):
- Change 6(6)→8(7), 3(5)→7(3), 8(7)→9(6), 9(6)→8(7): sticks = 7+3+6+7+6=29. Not 30.
- Change 6(6)→9(6), 3(5)→8(7), 8(7)→0(6), 9(6)→8(7): sticks=6+7+6+7+6=32. Not 30.
- Change 6(6)→8(7), 3(5)→5(5), 8(7)→9(6), 9(6)→8(7): sticks=7+5+6+7+6=31. Not 30.
- Change 6(6)→8(7), 3(5)→2(5), 8(7)→6(6), 9(6)→8(7): sticks=7+5+6+7+6=31. Not 30.
- Try: 98,870: 9(6)+8(7)+8(7)+7(3)+0(6)=29. Not 30.
- Try: 98,080: 9(6)+8(7)+0(6)+8(7)+0(6)=32. Not 30.
- Try: 83,890: 8(7)+3(5)+8(7)+9(6)+0(6)=31. Not 30.
- Try: 68,890: 6(6)+8(7)+8(7)+9(6)+0(6)=32. Not 30.
- Try: 83,080: 8(7)+3(5)+0(6)+8(7)+0(6)=31. Not 30.
- Try: 80,890: 8(7)+0(6)+8(7)+9(6)+0(6)=32. Not 30.
- Try: 88,070: 8(7)+8(7)+0(6)+7(3)+0(6)=29. Not 30.
- Try: 88,900: 8(7)+8(7)+9(6)+0(6)+0(6)=32. Not 30.
- Try: 98,000: 9(6)+8(7)+0(6)+0(6)+0(6)=31. Not 30.
- Try: 80,098: 8(7)+0(6)+0(6)+9(6)+8(7)=32. Not 30.
- Try: 88,078 (given): 30 sticks ✓.
- Try: 89,078: 8(7)+9(6)+0(6)+7(3)+8(7)=29. Not 30.
- Try: 98,078: 9(6)+8(7)+0(6)+7(3)+8(7)=29. Not 30.
- Try: 88,308: 8(7)+8(7)+3(5)+0(6)+8(7)=32. Not 30.
- Try: 83,978: 8(7)+3(5)+9(6)+7(3)+8(7)=28. Not 30.
- Try: 88,038: 8(7)+8(7)+0(6)+3(5)+8(7)=32. Not 30.
- Try: 88,008: 8(7)+8(7)+0(6)+0(6)+8(7)=33. Not 30.
- Try: 80,878: 8(7)+0(6)+8(7)+7(3)+8(7)=30 ✓. 80,878 > 63,890 ✓.
- Try: 80,978: 8(7)+0(6)+9(6)+7(3)+8(7)=29. Not 30.
- Try: 83,808: 8(7)+3(5)+8(7)+0(6)+8(7)=32. Not 30.

∴ Another number: 80,878 (using 30 sticks, bigger than 63,890).

We need a number whose digits' stick counts sum to 24.

One example: 8888 → 7+7+7+7 = 28. Too many.

888 → 7+7+7 = 21. Too few.

8880 → 7+7+7+6 = 27. Too many.

8800 → 7+7+6+6 = 26. Too many.

8000 → 7+6+6+6 = 25. One too many.

7000 → 3+6+6+6 = 21. Too few.

9900 → 6+6+6+6 = 24 ✓

So 9900 uses exactly 24 sticks.

Another: 6660 → 6+6+6+6 = 24 ✓

∴ Example: 9900 (or 6660) uses exactly 24 sticks.

To make the biggest number with 24 sticks:
- Use as many digits as possible (more digits = bigger number).
- The digit using fewest sticks is 1 (2 sticks).
- With 24 sticks, using all 1s: 24 ÷ 2 = 12 digits, all 1s.
- Number: 111,111,111,111 (twelve 1s).

This is a 12-digit number, which is larger than any number with fewer digits.

∴ The biggest number using 24 sticks is 1,11,11,11,11,111 (twelve 1s).

To make the smallest number with 24 sticks:
- Use as few digits as possible (fewer digits = smaller number, assuming same number of digits means we minimise value).
- Actually, fewer digits gives a smaller number. But we must use exactly 24 sticks.
- The digit using most sticks is 8 (7 sticks).
- 24 ÷ 7 = 3 remainder 3. So three 8s use 21 sticks, leaving 3 sticks = digit 7.
- Number with digits 7,8,8,8: smallest arrangement = 7888.
- Can we use fewer digits? 2 digits: max sticks = 7+7=14 < 24. Not possible.
- 3 digits: max sticks = 7+7+7=21 < 24. Not possible.
- 4 digits: 7+7+7+3=24 → digits 8,8,8,7 → smallest = 7888.
- Or: 7+7+6+4=24 → digits 8,8,0,4 → smallest = 4088 < 7888.
- Or: 7+7+5+5=24 → digits 8,8,2,2 (or 3,3) → smallest = 2288 < 4088.
- Or: 7+6+6+5=24 → digits 8,0,0,2 → smallest = 2008 < 2288.
- Or: 7+6+5+6=24 → same as above.
- Or: 6+6+6+6=24 → digits 0,0,0,0 or 6,6,6,6 or 9,9,9,9 → but leading zero not allowed. Digits with 6 sticks: 0,6,9. Smallest 4-digit: 6009 (6+6+6+6=24, using 6,0,0,9)? 6(6)+0(6)+0(6)+9(6)=24 ✓. Smallest = 6009. But 2008 < 6009.
- Or: 7+5+6+6=24 → 8,2,0,0 or 8,3,0,0 or 8,5,0,0... digit with 5 sticks: 2,3,5. Smallest: 2008 (using 8,0,0,2: 7+6+6+5=24 ✓). 2008 < 6009.
- Or: 5+5+7+7=24 → digits 2,2,8,8 → smallest = 2288. 2008 < 2288.
- Or: 5+6+6+7=24 → digits 2,0,0,8 → smallest = 2008.
- Or: 4+6+7+7=24 → digits 4,0,8,8 → smallest = 4088. 2008 < 4088.
- Or: 5+5+6+8=24 → 8 sticks digit doesn't exist (max is 7 for '8'). Skip.
- Or: 4+4+9+7=24 → 9 sticks digit doesn't exist. Skip.
- Or: 4+7+6+7=24 → digits 4,8,0,8 → smallest = 4088.
- Or: 3+7+7+7=24 → digits 7,8,8,8 → smallest = 7888.
- Or: 4+4+9+7: invalid.
- Or: 5+5+7+7=24 → 2288.
- Or: 4+6+7+7=24 → 4088.
- Or: 5+6+6+7=24 → 2008.

So the smallest 4-digit number with 24 sticks is 2008.

Can we get a smaller number with more digits? A 5-digit number starting with 1 would be at least 10000 > 2008. So more digits gives a bigger number.

∴ The smallest number using exactly 24 sticks is 2008.