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NCERT Solutions · Class 7 Maths · Ganita Prakash Part 1 · Chapter 1

Chapter 1: Large Numbers Around Us (Numbers)

Step-by-step answers to every "Figure it Out" and in-text question of Chapter 1, Large Numbers Around Us (NCERT Class 7 Maths, Ganita Prakash Part 1, 2026-27): lakhs and crores, the Indian and American systems, button-click calculators, rounding and nearest neighbours, estimation, patterns in products, large-number facts and the Toothpick Digits puzzle. All 72 questions are answered, with the key answer highlighted.

1.1 A Lakh Varieties!

Think

If we tried a new variety of rice each day, would we come close to tasting all one lakh varieties in a lifetime of 100 years? What if we ate 2, or 3, varieties every day?

Solution

Number of days in 100 years (ignoring leap years) $= 365 \times 100 = 36, 500$ .

Varieties per day	Varieties tasted in 100 years	Reach 1 lakh?
1	36,500	No, far short
2	73,000	No
3	1,09,500	Yes (1 lakh is reached in about 91 years)

1 a day gives only 36,500; 2 a day gives 73,000; 3 a day gives 1,09,500, so only at 3 varieties a day can one taste all one lakh in 100 years.

Fill

Observe the pattern and fill in the boxes: the largest 3-digit number is 999; +1 gives the smallest 4-digit number, and so on, up to the smallest 6-digit number 1,00,000. Also complete the number line 99,995, 99,996, __, 99,998, __, __.

Solution

Largest 3-digit number 999 → +1 → smallest 4-digit number 1,000
Largest 4-digit number 9,999 → +1 → smallest 5-digit number 10,000
Largest 5-digit number 99,999 → +1 → smallest 6-digit number 1,00,000 (one lakh)

Number line: 99,995, 99,996, 99,997, 99,998, 99,999, 1,00,000.

1,000; 9,999; 10,000; 99,999; and the missing numbers on the line are 99,997, 99,999 and 1,00,000.

Choose y

Choose a number for

y

. How close to one lakh is the number of days in

y

years?

Solution

Days in $y$ years $= 365 \times y$ . For example:

$y = 100$ : $36, 500$ days, which is $63, 500$ less than one lakh.
$y = 250$ : $91, 250$ days, $8, 750$ less than one lakh.
$y = 274$ : $1, 00, 010$ days, just 10 more than one lakh.

So one lakh days is about 274 years.

E.g. 274 years gives 365 × 274 = 1,00,010 days, almost exactly one lakh.

Figure it Out (page 3)

According to the 2011 Census, the population of the town of Chintamani was about 75,000. How much less than one lakh is 75,000?

Solution

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

25,000 less than one lakh.

The estimated population of Chintamani in the year 2024 is 1,06,000. How much more than one lakh is 1,06,000?

Solution

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

6,000 more than one lakh.

By how much did the population of Chintamani increase from 2011 to 2024?

Solution

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

It increased by about 31,000.

Getting a Feel of Large Numbers

Somu is 1 metre tall. If each floor is about four times his height, what is the approximate height of the building (10 floors in the picture)? Which is taller, the Statue of Unity or this building, and by how much?

Solution

Each floor $\approx 4 \times 1 = 4$ m, so the 10-floor building $\approx 10 \times 4 = 40$ m.

The Statue of Unity (180 m) is taller by $180 - 40 = 140$ m.

The building is about 40 m tall; the Statue of Unity is about 140 m taller.

How much taller is the Kunchikal waterfall than Somu's building? How many floors should Somu's building have to be as high as the waterfall?

Solution

Waterfall − building $= 450 - 40 = 410$ m.

Floors needed $= 450 \div 4 = 112.5$ , so about 113 floors.

410 m taller; the building would need about 113 floors.

How do you view a lakh — is a lakh big or small?

Solution

It depends on what we are counting and what we compare it with. One lakh days (about 274 years) or one lakh people in a line (about 38 km) feels very large. But one lakh hairs on a head, one lakh fish eggs or one lakh people in a stadium fit in a small space, so there a lakh feels small. A number is "big" or "small" only in comparison with something.

Both: a lakh is large when counting days or people in a line, but small for hairs on a head or eggs of a fish; size depends on the context.

Reading and Writing Numbers

Write each number in words: (a) 3,00,600 (b) 5,04,085 (c) 27,30,000 (d) 70,53,138

Solution

(a) Three lakh six hundred

(b) Five lakh four thousand eighty-five

(d) Seventy lakh fifty-three thousand one hundred thirty-eight

(a) three lakh six hundred (b) five lakh four thousand eighty-five (c) twenty-seven lakh thirty thousand (d) seventy lakh fifty-three thousand one hundred thirty-eight

Write the number in the Indian place value system: (a) One lakh twenty three thousand four hundred and fifty six (b) Four lakh seven thousand seven hundred and four (c) Fifty lakhs five thousand and fifty (d) Ten lakhs two hundred and thirty five

Solution

(a) 1,23,456 (b) 4,07,704 (c) 50,05,050 (d) 10,00,235

1.2 Land of Tens

The Thoughtful Thousands only has a +1000 button. How many times should it be pressed to show: (a) Three thousand (b) 10,000 (c) Fifty three thousand (d) 90,000 (e) One lakh (f) ___ ? 153 times (g) How many thousands make one lakh?

Solution

Number of presses = number ÷ 1000.

(a) 3 times (b) 10 times (c) 53 times (d) 90 times (e) 100 times (f) $153 \times 1000 = 1, 53, 000$ (g) 100 thousands make one lakh.

(a) 3 (b) 10 (c) 53 (d) 90 (e) 100 (f) 1,53,000 (g) 100

The Tedious Tens only has a +10 button. How many times should it be pressed to show: (a) Five hundred (b) 780 (c) 1000 (d) 3700 (e) 10,000 (f) One lakh (g) ___ ? 435 times

Solution

Number of presses = number ÷ 10.

(a) 50 (b) 78 (c) 100 (d) 370 (e) 1,000 (f) 10,000 (g) $435 \times 10 = 4, 350$

(a) 50 (b) 78 (c) 100 (d) 370 (e) 1,000 (f) 10,000 (g) 4,350

The Handy Hundreds only has a +100 button. How many times should it be pressed to show: (a) Four hundred (b) 3,700 (c) 10,000 (d) Fifty three thousand (e) 90,000 (f) 97,600 (g) 1,00,000 (h) ___ ? 582 times (i) How many hundreds make ten thousand? (j) How many hundreds make one lakh? (k) Handy Hundreds says, "There are some numbers which Tedious Tens and Thoughtful Thousands can't show but I can." Is this true?

Solution

Number of presses = number ÷ 100.

(a) 4 (b) 37 (c) 100 (d) 530 (e) 900 (f) 976 (g) 1,000 (h) $582 \times 100 = 58, 200$ (i) 100 (j) 1,000

(k) Not true. Handy Hundreds can show only multiples of 100. Some of them (like 3,700 or 97,600) cannot be shown by Thoughtful Thousands, but every multiple of 100 is also a multiple of 10, so Tedious Tens can show all of them (3,700 = 370 tens). There is no number that both the others fail to show but Handy Hundreds can.

(a) 4 (b) 37 (c) 100 (d) 530 (e) 900 (f) 976 (g) 1,000 (h) 58,200 (i) 100 (j) 1,000 (k) False: every multiple of 100 can also be shown by Tedious Tens.

Creative Chitti has buttons +1, +10, +100, … To get 321, it presses +10 thirty-two times and +1 once. Will it get 321?

Solution

$32 \times 10 + 1 \times 1 = 320 + 1 = 321$ . Yes. (Also $2 \times 100 + 12 \times 10 + 1 = 200 + 120 + 1 = 321$ .)

Yes, 32 × 10 + 1 = 321.

Two ways to get 5072 are (50 × 100) + (7 × 10) + (2 × 1) and (3 × 1000) + (20 × 100) + (72 × 1). Find a different way to get 5072 and write an expression for it.

Solution

$(4 \times 1000) + (10 \times 100) + (7 \times 10) + (2 \times 1) = 4000 + 1000 + 70 + 2 = 5072$

(Another: $(5 \times 1000) + (6 \times 10) + (12 \times 1) = 5072$ .)

For example, (4 × 1000) + (10 × 100) + (7 × 10) + (2 × 1) = 5072.

Figure it Out (page 6)

For each number, write expressions for at least two different ways to obtain it through button clicks: (a) 8300 (b) 40629 (c) 56354 (d) 66666 (e) 367813

Solution

(a) $8300 = (8 \times 1000) + (3 \times 100)$ ; $8300 = (83 \times 100)$ ; $8300 = (7 \times 1000) + (13 \times 100)$

(b) $40629 = (4 \times 10000) + (6 \times 100) + (2 \times 10) + (9 \times 1)$ ; $40629 = (40 \times 1000) + (62 \times 10) + (9 \times 1)$

(c) $56354 = (5 \times 10000) + (6 \times 1000) + (3 \times 100) + (5 \times 10) + (4 \times 1)$ ; $56354 = (56 \times 1000) + (35 \times 10) + (4 \times 1)$

(d) $66666 = (6 \times 10000) + (6 \times 1000) + (6 \times 100) + (6 \times 10) + (6 \times 1)$ ; $66666 = (66 \times 1000) + (666 \times 1)$

(e) $367813 = (3 \times 100000) + (6 \times 10000) + (7 \times 1000) + (8 \times 100) + (1 \times 10) + (3 \times 1)$ ; $367813 = (36 \times 10000) + (78 \times 100) + (13 \times 1)$

Many answers are possible; e.g. 8300 = 8 × 1000 + 3 × 100 = 83 × 100, and 66666 = 66 × 1000 + 666 × 1.

(a) You have to make exactly 30 button presses. What is the largest 3-digit number you can make? What is the smallest? (b) 997 can be made using 25 clicks. Can you make 997 with a different number of clicks?

Solution

A useful fact: every button (+1, +10, +100, …) adds a number that is 1 more than a multiple of 9. So the number shown always leaves the same remainder on division by 9 as the number of clicks.

(a) 30 clicks leaves remainder 3 on division by 9, so the number must also leave remainder 3.

Largest 3-digit such number: 993 ( $993 = 110 \times 9 + 3$ ). Clicks: $(9 \times 100) + (8 \times 10) + (13 \times 1)$ , i.e. $9 + 8 + 13 = 30$ clicks.
Smallest 3-digit such number: 102. Clicks: $(8 \times 10) + (22 \times 1)$ , i.e. $8 + 22 = 30$ clicks.

(b) $997 = (9 \times 100) + (9 \times 10) + (7 \times 1)$ uses 25 clicks. Breaking one +100 into ten +10s adds 9 clicks: $(8 \times 100) + (19 \times 10) + (7 \times 1)$ uses 34 clicks. Similarly 43, 52, … clicks are possible (any number that is 25 plus a multiple of 9, up to 997 clicks of +1).

(a) Largest 993, smallest 102. (b) Yes, e.g. (8 × 100) + (19 × 10) + (7 × 1) = 997 in 34 clicks; possible counts are 25, 34, 43, … (each 9 more).

Systematic Sippy wants to use as few clicks as possible. How can we get (a) 5072 (b) 8300 with the fewest clicks? Is there a way to get 5072 using fewer than 23 clicks?

Solution

(a) $5072 = (5 \times 1000) + (7 \times 10) + (2 \times 1)$ , i.e. $5 + 7 + 2 = 14$ clicks (fewer than 23).

(b) $8300 = (8 \times 1000) + (3 \times 100)$ , i.e. $8 + 3 = 11$ clicks.

5072 in 14 clicks and 8300 in 11 clicks.

Figure it Out (page 7)

For the numbers in the previous exercise, find how to get each number by making the smallest number of button clicks and write the expression.

Solution

Number	Fewest clicks	Clicks
8300	(8 × 1000) + (3 × 100)	8 + 3 = 11
40629	(4 × 10000) + (6 × 100) + (2 × 10) + (9 × 1)	4 + 6 + 2 + 9 = 21
56354	(5 × 10000) + (6 × 1000) + (3 × 100) + (5 × 10) + (4 × 1)	5 + 6 + 3 + 5 + 4 = 23
66666	(6 × 10000) + (6 × 1000) + (6 × 100) + (6 × 10) + (6 × 1)	30
367813	(3 × 100000) + (6 × 10000) + (7 × 1000) + (8 × 100) + (1 × 10) + (3 × 1)	3 + 6 + 7 + 8 + 1 + 3 = 28

8300: 11, 40629: 21, 56354: 23, 66666: 30, 367813: 28 clicks.

Do you see any connection between each number and the corresponding smallest number of button clicks?

Solution

Yes. The smallest number of clicks equals the sum of the digits of the number. For example, $4 + 0 + 6 + 2 + 9 = 21$ .

The fewest clicks = the sum of the digits.

The expressions for the least button clicks also give the Indian place value notation of the numbers. Think about why this is so.

Solution

To use the fewest clicks, we should never press a button 10 or more times, because 10 presses of one button can be replaced by 1 press of the next bigger button (saving 9 clicks). So each button is pressed 0 to 9 times, and the number of presses of +1000, +100, +10 and +1 are exactly the digits in the thousands, hundreds, tens and ones places. That is precisely the place value expansion of the number.

Ten presses of a button can always be replaced by one press of the next button, so in the least-click way each button is pressed 0–9 times; these counts are exactly the digits in each place.

What if we press the +10,00,000 button ten times? What number will come up? How many zeroes will it have? What should we call it?

Solution

$10 \times 10, 00, 000 = 1, 00, 00, 000$ : 100 lakh, called one crore. It has 7 zeroes.

1,00,00,000 (one crore), with 7 zeroes.

1.3 Of Crores and Crores!

How many zeros does a thousand lakh have? How many zeros does a hundred thousand have?

Solution

A thousand lakh $= 1000 \times 1, 00, 000 = 10, 00, 00, 000$ (ten crore): 8 zeros.

A hundred thousand $= 100 \times 1000 = 1, 00, 000$ (one lakh): 5 zeros.

A thousand lakh has 8 zeros; a hundred thousand has 5 zeros.

Figure it Out (page 9)

Read the following numbers in Indian place value notation and write their number names in both the Indian and American systems: (a) 4050678 (b) 48121620 (c) 20022002 (d) 246813579 (e) 345000543 (f) 1020304050

Solution

	Indian system	American system
(a)	40,50,678: forty lakh fifty thousand six hundred seventy-eight	4,050,678: four million fifty thousand six hundred seventy-eight
(b)	4,81,21,620: four crore eighty-one lakh twenty-one thousand six hundred twenty	48,121,620: forty-eight million one hundred twenty-one thousand six hundred twenty
(c)	2,00,22,002: two crore twenty-two thousand two	20,022,002: twenty million twenty-two thousand two
(d)	24,68,13,579: twenty-four crore sixty-eight lakh thirteen thousand five hundred seventy-nine	246,813,579: two hundred forty-six million eight hundred thirteen thousand five hundred seventy-nine
(e)	34,50,00,543: thirty-four crore fifty lakh five hundred forty-three	345,000,543: three hundred forty-five million five hundred forty-three
(f)	1,02,03,04,050: one arab two crore three lakh four thousand fifty (or one hundred two crore three lakh four thousand fifty)	1,020,304,050: one billion twenty million three hundred four thousand fifty

See the table: e.g. (a) 40,50,678 = forty lakh fifty thousand six hundred seventy-eight = 4,050,678 = four million fifty thousand six hundred seventy-eight.

Write in Indian place value notation: (a) One crore one lakh one thousand ten (b) One billion one million one thousand one (c) Ten crore twenty lakh thirty thousand forty (d) Nine billion eighty million seven hundred thousand six hundred

Solution

(a) 1,01,01,010

(b) 1 billion = 100 crore and 1 million = 10 lakh, so the number is 1,00,10,01,001.

(d) 9 billion = 900 crore, 80 million = 8 crore, 700 thousand = 7 lakh, so the number is 9,08,07,00,600.

(a) 1,01,01,010 (b) 1,00,10,01,001 (c) 10,20,30,040 (d) 9,08,07,00,600

Compare and write '<', '>' or '=': (a) 30 thousand __ 3 lakhs (b) 500 lakhs __ 5 million (c) 800 thousand __ 8 million (d) 640 crore __ 60 billion

Solution

(a) 30 thousand = 30,000 and 3 lakh = 3,00,000, so 30 thousand < 3 lakhs.

(b) 500 lakh = 5 crore = 50 million, so 500 lakhs > 5 million.

(d) 640 crore = 6.4 billion, so 640 crore < 60 billion.

(a) < (b) > (c) < (d) <

1.4 Exact and Approximate Values

"1 lakh people visited the book fair." "If I had not gone, they would have written '99,999 people visited'." What do you think of this conversation?

Solution

The number "1 lakh" in a headline is an approximation (a rounded number), not an exact count. The actual number might have been 98,640 or 1,02,315. One person more or less makes no difference to a rounded figure, so the joke is that the speaker takes a rounded number as exact. Newspapers often use such rounded figures: "about 5 crore people watched the match", "nearly 2 lakh devotees".

The 1 lakh is a rounded figure, not an exact count; one person missing would not change it to 99,999.

Think and share situations where it is appropriate to (a) round up, (b) round down, (c) either is okay and (d) exact numbers are needed.

Solution

(a) Round up: ordering food or sweets for a function, buying tiles or paint, arranging chairs or buses for a trip, keeping money for a journey, so we do not fall short.

(b) Round down: telling how much money we can surely spend, the number of full boxes we can pack, a weight limit we must not cross (e.g. lift capacity), a selling price said casually ("around ₹450").

(c) Either: the population of a city, the distance between two towns, the number of people at a fair, the height of a mountain.

(d) Exact numbers needed: phone numbers, bank account numbers and money in a bank, medicine doses, exam marks, train times, measurements for scientific experiments.

Round up when we must not fall short (food, buses); round down when we must not exceed (money to spend, weight limits); either for general sizes like populations and distances; exact for phone numbers, bank amounts and medicine doses.

Nearest Neighbours

Write the five nearest neighbours for these numbers: (a) 3,87,69,957 (b) 29,05,32,481

Solution

Nearest	(a) 3,87,69,957	(b) 29,05,32,481
thousand	3,87,70,000	29,05,32,000
ten thousand	3,87,70,000	29,05,30,000
lakh	3,88,00,000	29,05,00,000
ten lakh	3,90,00,000	29,10,00,000
crore	4,00,00,000	29,00,00,000

(a) 3,87,70,000; 3,87,70,000; 3,88,00,000; 3,90,00,000; 4,00,00,000 (b) 29,05,32,000; 29,05,30,000; 29,05,00,000; 29,10,00,000; 29,00,00,000

I have a number for which all five nearest neighbours are 5,00,00,000. What could the number be? How many such numbers are there?

Solution

If the nearest thousand is 5,00,00,000, then the nearest ten thousand, lakh, ten lakh and crore are also 5,00,00,000. So the number must be within 500 of 5,00,00,000: from 4,99,99,500 to 5,00,00,499 (taking 500 to round up).

Examples: 4,99,99,750, 5,00,00,000, 5,00,00,321. There are $500 + 500 =$ 1,000 such numbers.

Any number from 4,99,99,500 to 5,00,00,499, e.g. 5,00,00,321; there are 1,000 such numbers.

Estimation

4,63,128 + 4,19,682. Roxie: "The sum is near 8,00,000 and is more than 8,00,000." Estu: "The sum is near 9,00,000 and is less than 9,00,000." (a) Are these estimates correct? Whose is closer? (b) Will the sum be greater or less than 8,50,000? (c) Greater or less than 8,83,128? (d) Find the exact value.

Solution

(a) Both are correct: $4, 63, 128 > 4$ lakh and $4, 19, 682 > 4$ lakh, so the sum is more than 8 lakh; and both are less than 4.5 lakh, so the sum is less than 9 lakh. The exact sum is 8,82,810, which is 17,190 from 9,00,000 but 82,810 from 8,00,000, so Estu's estimate is closer.

(b) Greater than 8,50,000: $4, 63, 128 > 4, 50, 000$ and $4, 19, 682 > 4, 00, 000$ , so the sum $> 8, 50, 000$ .

(d) $4, 63, 128 + 4, 19, 682 =$ 8,82,810

(a) Both correct; Estu is closer. (b) Greater. (c) Less. (d) 8,82,810

14,63,128 − 4,90,020. Roxie: "The difference is near 10,00,000 and is less than 10,00,000." Estu: "The difference is near 9,00,000 and is more than 9,00,000." (a) Are these correct? Whose is closer? (b) Greater or less than 9,50,000? (c) Greater or less than 9,63,128? (d) Find the exact value.

Solution

(a) Both are correct: subtracting a little less than 5 lakh from about 14.6 lakh gives a little less than 10 lakh but more than 9 lakh. The exact difference 9,73,108 is 26,892 from 10 lakh but 73,108 from 9 lakh, so Roxie's estimate is closer.

(b) Greater than 9,50,000: $14, 63, 128 - 5, 00, 000 = 9, 63, 128$ already exceeds 9,50,000, and we subtract less than 5 lakh.

(c) Greater than 9,63,128, because 4,90,020 is less than 5,00,000, so we take away less than in $14, 63, 128 - 5, 00, 000 = 9, 63, 128$ .

(d) $14, 63, 128 - 4, 90, 020 =$ 9,73,108

(a) Both correct; Roxie is closer. (b) Greater. (c) Greater. (d) 9,73,108

Populations of Cities

What is your general observation about this data?

Solution

The population of almost every city grew from 2001 to 2011; only Kolkata's fell slightly (45,72,876 to 44,86,679).
Mumbai and New Delhi are the only cities above 1 crore.
Some cities like Bengaluru, Hyderabad, Surat and Vadodara grew very fast (almost doubling), while Chennai and Kanpur grew slowly.
The ranking is by the 2011 population, from the largest (Mumbai) to the twentieth (Patna).

Nearly all cities grew (Kolkata fell slightly); only Mumbai and Delhi crossed 1 crore; Bengaluru, Hyderabad, Surat and Vadodara nearly doubled.

What is an appropriate title for the above table?

Solution

"The 20 most populous cities of India: population in 2001 and 2011"

"Population of the 20 largest Indian cities, 2001 and 2011".

How much is the population of Pune in 2011? Approximately, by how much has it increased compared to 2001?

Solution

Pune in 2011: 31,15,431 (about 31 lakh). In 2001 it was about 25 lakh, so the increase is about $31 - 25 = 6$ lakh (exactly 5,76,958).

31,15,431; it increased by about 6 lakh.

Which city's population increased the most between 2001 and 2011?

Solution

Bengaluru: $84, 25, 970 - 43, 01, 326 = 41, 24, 644$ , about 41 lakh. (Next is Hyderabad, about 32 lakh.)

Bengaluru, by about 41 lakh.

Are there cities whose population has almost doubled? Which are they?

Solution

Yes: Bengaluru (43 lakh → 84 lakh), Hyderabad (36 lakh → 68 lakh), Surat (24 lakh → 45 lakh) and Vadodara (17 lakh → 36 lakh; more than doubled).

Bengaluru, Hyderabad, Surat and Vadodara.

By what number should we multiply Patna's population to get a number close to that of Mumbai?

Solution

Patna ≈ 17 lakh and Mumbai ≈ 124 lakh. $124 \div 17 \approx 7.3$ . Check: $17 \times 7 = 119$ lakh, which is close to 124 lakh.

By about 7 (more exactly about 7.4).

1.5 Patterns in Products

Math Talk

Using the meaning of multiplication and division, can you explain why multiplying by 5 is the same as dividing by 2 and multiplying by 10?

Solution

$5 = \frac{10}{2}$ . Taking 5 groups of a number is half of taking 10 groups of it. So $n \times 5 = n \times \frac{10}{2} = \frac{n}{2} \times 10$ . For example, $116 \times 5 = 58 \times 10 = 580$ .

Since 5 is half of 10, 5 groups of a number = half of 10 groups; so n × 5 = (n ÷ 2) × 10.

Figure it Out (page 14)

Find quick ways to calculate: (a) 2 × 1768 × 50 (b) 72 × 125 (c) 125 × 40 × 8 × 25

Solution

(a) $2 \times 1768 \times 50 = (2 \times 50) \times 1768 = 100 \times 1768 =$ 1,76,800

(b) $72 \times 125 = 72 \times \frac{1000}{8} = \frac{72}{8} \times 1000 = 9 \times 1000 =$ 9,000

(a) 1,76,800 (b) 9,000 (c) 10,00,000

Calculate quickly: (a) 25 × 12 (b) 25 × 240 (c) 250 × 120 (d) 2500 × 12 (e) __ × __ = 120000000

Solution

Use $25 = \frac{100}{4}$ , $250 = \frac{1000}{4}$ , $2500 = \frac{10000}{4}$ .

(a) $25 \times 12 = 100 \times \frac{12}{4} = 100 \times 3 =$ 300

(b) $25 \times 240 = 100 \times 60 =$ 6,000

(d) $2500 \times 12 = 10000 \times 3 =$ 30,000

(e) $120000000 = 12, 00, 00, 000$ (12 crore), e.g. 12,000 × 10,000, or $2500 \times 48, 000$ , or $1200 \times 1, 00, 000$ .

(a) 300 (b) 6,000 (c) 30,000 (d) 30,000 (e) e.g. 12,000 × 10,000

How Long is the Product?

Evaluate the multiplications and extend the patterns: 11 × 11, 111 × 111, 1111 × 1111; 66 × 61, 666 × 661, 6666 × 6661; 3 × 5, 33 × 35, 333 × 335; 101 × 101, 102 × 102, 103 × 103.

Solution

Pattern 1	Pattern 2	Pattern 3	Pattern 4
11 × 11 = 121	66 × 61 = 4026	3 × 5 = 15	101 × 101 = 10201
111 × 111 = 12321	666 × 661 = 440226	33 × 35 = 1155	102 × 102 = 10404
1111 × 1111 = 1234321	6666 × 6661 = 44402226	333 × 335 = 111555	103 × 103 = 10609
11111 × 11111 = 123454321	66666 × 66661 = 4444022226	3333 × 3335 = 11115555	104 × 104 = 10816

121, 12321, 1234321 → 123454321; 4026, 440226, 44402226 → 4444022226; 15, 1155, 111555 → 11115555; 10201, 10404, 10609 → 10816.

Is there any connection between the number of digits in the numbers being multiplied and in their product? Roxie says the product of two 2-digit numbers can only be a 3- or 4-digit number. Is she correct? Should we try all possible multiplications?

Solution

The product has either the sum of the numbers of digits, or one less than that. (2-digit × 2-digit → 3 or 4 digits.)

Roxie is correct, and we need not try all products. The smallest product is $10 \times 10 = 100$ (3 digits); every product of two 2-digit numbers is less than $100 \times 100 = 10, 000$ (so at most 4 digits). Checking just the smallest and largest cases is enough.

Digits in the product = sum of the digits of the factors, or one less. Roxie is correct: 10 × 10 = 100 and every product is below 100 × 100 = 10,000.

Can multiplying a 3-digit number with another 3-digit number give a 4-digit number? Can multiplying a 4-digit number with a 2-digit number give a 5-digit number?

Solution

No. The smallest such product is $100 \times 100 = 10, 000$ , which already has 5 digits.
Yes. For example, $1000 \times 10 = 10, 000$ or $1200 \times 20 = 24, 000$ (5 digits).

No (the least is 100 × 100 = 10,000); yes (e.g. 1000 × 10 = 10,000).

Complete: 5-digit × 5-digit = __ or __; 8-digit × 3-digit = __ or __; 12-digit × 13-digit = __ or __.

Solution

5-digit × 5-digit = 9-digit or 10-digit
8-digit × 3-digit = 10-digit or 11-digit
12-digit × 13-digit = 24-digit or 25-digit

9 or 10; 10 or 11; 24 or 25 digits.

Fascinating Facts about Large Numbers

1250 × 380 is the number of kīrtanas composed by Purandaradāsa according to legends. Find the number. If he composed 4,75,000 songs, how many songs per year did he have to compose?

Solution

$1250 \times 380 = 4, 75, 000$ : four lakh seventy-five thousand (Indian) = 475 thousand (American).

Purandaradāsa lived about 80 years (c. 1484–1564). If he composed for about 50 years: $4, 75, 000 \div 50 = 9, 500$ songs a year, or about $9, 500 \div 365 \approx 26$ songs every day!

4,75,000 (475 thousand); over about 50 years that is 9,500 songs a year, about 26 a day.

2100 × 70,000 is the approximate distance in km between the Earth and the Sun. 6400 × 62,500 is the average number of litres of water the Amazon discharges into the Atlantic every second.

Solution

$2100 \times 70, 000 = 14, 70, 00, 000$ km: fourteen crore seventy lakh = 147 million km.

$6400 \times 62, 500 = 40, 00, 00, 000$ litres: forty crore = 400 million litres every second.

14,70,00,000 km (147 million km); 40,00,00,000 litres (400 million litres) per second.

Divide to uncover the facts: 13,95,000 ÷ 150 (km, longest single-train journey); 10,50,00,000 ÷ 700 (kg, weight of an adult blue whale); 52,00,00,00,000 ÷ 130 (tonnes of plastic waste generated in 2021).

Solution

$13, 95, 000 \div 150 = 9, 300$ km: nine thousand three hundred (same in both systems).

$10, 50, 00, 000 \div 700 = 1, 50, 000$ kg: one lakh fifty thousand = 150 thousand kg.

$52, 00, 00, 00, 000 \div 130 = 40, 00, 00, 000$ tonnes: forty crore = 400 million tonnes.

9,300 km; 1,50,000 kg (150 thousand); 40,00,00,000 tonnes (400 million).

1.6 Did You Ever Wonder…?

The RMS Titanic carried about 2500 passengers. Can the population of Mumbai fit into 5000 such ships?

Solution

$5000 \times 2500 = 1, 25, 00, 000$ (1 crore 25 lakh). Mumbai's population (2011) is 1,24,42,373, which is a little less. So yes, just barely (with about 57,600 places to spare).

Yes, just: 5000 ships hold 1,25,00,000 people, slightly more than 1,24,42,373.

If Roxie could travel 100 km every day, could she reach the Moon (3,84,400 km away) in 10 years? How far would she travel in a year? In 10 years?

Solution

In a year: $100 \times 365 = 36, 500$ km. In 10 years: $36, 500 \times 10 = 3, 65, 000$ km.

This is $3, 84, 400 - 3, 65, 000 = 19, 400$ km short. No, she would need about 10½ years.

36,500 km in a year and 3,65,000 km in 10 years, so she would fall 19,400 km short of the Moon.

Find out if you can reach the Sun in a lifetime if you travel 1000 km every day.

Solution

Distance to the Sun ≈ 14,70,00,000 km. Days needed $= 14, 70, 00, 000 \div 1000 = 1, 47, 000$ days. Years $\approx 1, 47, 000 \div 365 \approx 403$ years.

No. It would take about 400 years, far longer than a lifetime.

No: it would take about 1,47,000 days, i.e. about 403 years.

Make reasonable assumptions and answer: (a) If a sheet of paper weighs 5 g, could you lift one lakh sheets together? (b) If 250 babies are born every minute, will a million babies be born in a day? (c) Can you count 1 million coins in a day at 1 coin per second?

Solution

(a) $1, 00, 000 \times 5$ g $= 5, 00, 000$ g $= 500$ kg. No, a person cannot lift 500 kg.

(b) Babies in a day $= 250 \times 60 \times 24 = 3, 60, 000$ , which is less than 10,00,000 (1 million). No.

(c) Seconds in a day $= 60 \times 60 \times 24 = 86, 400$ , so at most 86,400 coins (even without sleep). No; 1 million coins would take about $10, 00, 000 \div 86, 400 \approx 11.6$ days of non-stop counting.

(a) No, they weigh 500 kg. (b) No, only 3,60,000 a day. (c) No, only 86,400 a day.

Figure it Out (page 19)

Using all digits 0–9 exactly once (the first digit cannot be 0) to create a 10-digit number, write the (a) largest multiple of 5 (b) smallest even number.

Solution

(a) A multiple of 5 ends in 0 or 5. Arranging the rest in decreasing order with 0 at the end gives the largest: 98,76,54,32,10 (9876543210).

(b) The smallest arrangement 1023456789 is odd. Keep the smallest start and put the largest even digit (8) last: 10,23,45,67,98 (1023456798).

(a) 9876543210 (b) 1023456798

The number 10,30,285 in words, "Ten lakhs thirty thousand two hundred eighty five", has 42 letters. Give a 7-digit number name which has the maximum number of letters.

Solution

To have the most letters, each part should use the longest words: "seventy" (7 letters) for the tens and a 5-letter word (three, seven, eight) for the ones, giving 12 letters for each two-digit part; and a 5-letter word before "hundred".

77,77,777: seventy seven lakhs seventy seven thousand seven hundred seventy seven, which has 61 letters (60 if we write "lakh"). Numbers like 73,73,373 or 78,78,878 also have the same number of letters.

E.g. 77,77,777 (seventy seven lakhs seventy seven thousand seven hundred seventy seven) with 61 letters.

Write a 9-digit number where exchanging any two digits results in a bigger number. How many such numbers exist?

Solution

If an exchange of two digits always makes the number bigger, then every digit must be smaller than every digit after it (if a bigger digit came first, swapping it with a smaller later digit would make the number smaller; equal digits give the same number). So the digits must be different and strictly increasing. A 9-digit number cannot start with 0, so the only choice is the digits 1 to 9 in order:

12,34,56,789 (123456789). There is only one such number.

123456789; only one such number exists.

Strike out 10 digits from the number 12345123451234512345 so that the remaining number is as large as possible.

Solution

We keep 10 of the 20 digits and want the biggest possible digit at each place from the left.

1st digit: choose from the first 11 digits (10 must remain after it): the largest is the first 5.
2nd digit: the largest of the next digits that leaves 8 after it: the second 5.
3rd digit: from "1 2 3", the largest is 3; now only 7 digits are left and all must be kept: 4 5 1 2 3 4 5.

Largest number: 5534512345

5534512345

The words 'zero' and 'one' share letters 'e' and 'o'; 'one' and 'two' share 'o'; 'two' and 'three' share 't'. How far do you have to count to find two consecutive numbers which do not share an English letter in common?

Solution

You will never find such a pair: every two consecutive numbers share a letter. Here is why:

From zero to twenty, check each pair: zero–one (e, o), one–two (o), two–three (t), three–four (r), four–five (f), five–six (i), six–seven (s), seven–eight (e), eight–nine (e, i, n), nine–ten (e, n), ten–eleven (e, n), eleven–twelve (e, l, v), and all the "-teen" numbers share t, e, n; nineteen–twenty (t, e, n).
After twenty, two consecutive numbers have the same beginning words (twenty one, twenty two, …) and so share letters, unless the first ends in "nine". Then the next number starts a new tens: twenty nine–thirty (t, y), thirty nine–forty (t, r, y), forty nine–fifty (f, t, y), fifty nine–sixty (i, t, y), sixty nine–seventy (s, e, n, t, y), seventy nine–eighty (e, i, t, y), eighty nine–ninety (n, i, e, t, y), ninety nine–one hundred (n, e).
For larger numbers, a number ending in "…nine" is followed by one with "hundred", "thousand", "lakh", "million", etc. along with the same earlier words, which again share letters (e.g. "nine" and "hundred" share n, e).

(A computer check of all numbers up to 30 lakh, in both systems, confirms this.)

Never: every pair of consecutive numbers shares at least one letter.

Write down 1, 2, 3, …, 9, 10, 11, … The tenth digit is '1' and the eleventh is '0' (in 10). (a) What would the 1000th digit be, and in which number? (b) What number would contain the millionth digit? (c) When would you have written the digit '5' for the 5000th time?

Solution

(a) Digits used: 1–9 use 9 digits; 10–99 use $90 \times 2 = 180$ digits; total 189. Remaining $1000 - 189 = 811 = 270 \times 3 + 1$ . The 270 three-digit numbers 100 to 369 use 810 digits (up to the 999th digit). The 1000th digit is the first digit of the next number, 370: it is 3.

(b) Count the digits used:

Numbers	Digits used	Total so far
1–9	9	9
10–99	180	189
100–999	2,700	2,889
1,000–9,999	36,000	38,889
10,000–99,999	4,50,000	4,88,889

Remaining $10, 00, 000 - 4, 88, 889 = 5, 11, 111 = 85, 185 \times 6 + 1$ . The 85,185 six-digit numbers from 1,00,000 to 1,85,184 finish the 9,99,999th digit; so the millionth digit is the first digit ('1') of 1,85,185.

(c) Count the 5s: from 1 to 9,999, each of the 4 places has a 5 in one-tenth of the numbers: $4 \times 1000 = 4000$ fives. Each block of a thousand after that (10,000–10,999, 11,000–11,999, 12,000–12,999) has 300 fives, giving $4000 + 900 = 4900$ by 12,999. From 13,000 to 13,499 there are 50 fives in the tens place and 50 in the ones place: 100 more. The last of these is in 13,495, where the 5000th '5' is written.

(a) The digit 3, in 370. (b) 1,85,185. (c) In the number 13,495.

A calculator has only '+10,000' and '+100' buttons. Write an expression for the button clicks for: (a) 20,800 (b) 92,100 (c) 1,20,500 (d) 65,30,000 (e) 70,25,700

Solution

(a) $20, 800 = (2 \times 10, 000) + (8 \times 100)$ : 10 clicks

(b) $92, 100 = (9 \times 10, 000) + (21 \times 100)$ : 30 clicks

(d) $65, 30, 000 = (653 \times 10, 000)$ : 653 clicks

(e) $70, 25, 700 = (702 \times 10, 000) + (57 \times 100)$ : 759 clicks

(a) 2 × 10,000 + 8 × 100 (b) 9 × 10,000 + 21 × 100 (c) 12 × 10,000 + 5 × 100 (d) 653 × 10,000 (e) 702 × 10,000 + 57 × 100

How many lakhs make a billion?

Solution

1 billion $= 1, 00, 00, 00, 000 = 100$ crore $= 100 \times 100$ lakh $=$ 10,000 lakh.

10,000 lakhs.

You are given two sets of number cards 1–9. Place a card in each box (a 7-digit number above a 5-digit number, right-aligned) to get (a) the largest possible sum (b) the smallest possible difference.

Solution

(a) For the largest sum, the two highest places (only in the 7-digit number) get 9 and 9. Each of the remaining five places has one card from each number, so give the biggest pairs to the highest places: 8 and 8, then 7 and 7, 6 and 6, 5 and 5, 4 and 4.

$99, 87, 654 + 87, 654 1, 00, 75, 308$

Largest sum = 1,00,75,308.

(b) For the smallest difference, make the 7-digit number as small as possible and the 5-digit number as large as possible: $11, 22, 334 - 99, 887 =$ 10,22,447.

(a) 99,87,654 + 87,654 = 1,00,75,308 (b) 11,22,334 − 99,887 = 10,22,447

Number cards: 4000, 13000, 300, 70000, 150000, 20, 5. Using any operations (each card at most once), get as close as you can to (a) 1,10,000 (b) 2,00,000 (c) 5,80,000 (d) 12,45,000 (e) 20,90,800.

Solution

(a) $(13000 \times 20) - 150000 = 2, 60, 000 - 1, 50, 000 =$ 1,10,000 (exact; better than the book's 1,13,000)

(b) $(70000 \times 5) - 150000 = 3, 50, 000 - 1, 50, 000 =$ 2,00,000 (exact)

(d) $(70000 + 13000) \div 20 \times 300 = 83, 000 \div 20 \times 300 = 4, 150 \times 300 =$ 12,45,000 (exact)

(e) $(300 - 5) \times (150000 - 13000) \div 20 + 70000 = 295 \times 1, 37, 000 \div 20 + 70, 000 = 20, 20, 750 + 70, 000 =$ 20,90,750 (only 50 away; a computer search shows 20,90,800 itself cannot be made)

(a) 13000 × 20 − 150000 = 1,10,000 (b) 70000 × 5 − 150000 = 2,00,000 (c) 70000 × 5 + 150000 + 4000 × 20 = 5,80,000 (d) (70000 + 13000) ÷ 20 × 300 = 12,45,000 (e) (300 − 5) × (150000 − 13000) ÷ 20 + 70000 = 20,90,750

How many coins should be stacked to match the height of the Statue of Unity? Assume each coin is 1 mm thick.

Solution

Height $= 180$ m $= 180 \times 1000$ mm $= 1, 80, 000$ mm. Each coin is 1 mm, so 1,80,000 coins are needed.

1,80,000 coins.

Albatrosses can cover about 900–1000 km in a day. One of the longest single trips recorded is about 12,000 km. How many days would such a trip take?

Solution

At 1000 km a day: $12, 000 \div 1000 = 12$ days. At 900 km a day: $12, 000 \div 900 \approx 13.3$ days.

About 12 to 13 days.

A bar-tailed godwit travelled 13,560 km non-stop from Alaska to Australia in about 11 days. Find the approximate distance it covered every day and every hour.

Solution

Per day: $13, 560 \div 11 \approx 1, 233$ km (roughly 1,200 km).

Per hour: $1, 233 \div 24 \approx 51$ km.

About 1,233 km a day and about 51 km an hour.

Bald eagles fly as high as 4500–6000 m; Mount Everest is about 8850 m high; aeroplanes fly at 10,000–12,800 m. How many times bigger are these heights compared to Somu's building?

Solution

Somu's building ≈ 40 m.

Height	÷ 40	Times the building
Bald eagle: 4500–6000 m	4500 ÷ 40 = 112.5; 6000 ÷ 40 = 150	about 112 to 150 times
Mount Everest: 8850 m	8850 ÷ 40 ≈ 221	about 221 times
Aeroplanes: 10,000–12,800 m	10,000 ÷ 40 = 250; 12,800 ÷ 40 = 320	about 250 to 320 times

Eagles: about 112–150 times; Everest: about 221 times; aeroplanes: about 250–320 times.

Puzzle Time: Toothpick Digits

Sticks needed for each digit (as in the picture):

Digit	0	1	2	3	4	5	6	7	8	9
Sticks	6	2	5	5	4	5	6	3	7	6

Write or make the number 5108. How many sticks are required?

Solution

$5 + 2 + 6 + 7 =$ 20 sticks (5 → 5, 1 → 2, 0 → 6, 8 → 7).

20 sticks.

42,019 requires exactly 23 sticks. Starting with 42,019, add two more sticks and make a bigger number (e.g. 42,078). What other numbers bigger than 42,019 can you make this way?

Solution

Adding a stick can change 0 → 8, 1 → 7, 9 → 8; adding two sticks can change 1 → 4, 2 → 8 or 4 → 9. The bigger numbers possible are:

92,019 (4 → 9), 48,019 (2 → 8), 42,879 (0 → 8, 1 → 7), 42,818 (0 → 8, 9 → 8), 42,049 (1 → 4) and the given 42,078 (1 → 7, 9 → 8). The biggest is 92,019.

92,019; 48,019; 42,879; 42,818; 42,049 (and 42,078).

Preetham wants to insert the digit '1' somewhere among the digits 4, 2, 0, 1, 9. Where should he place it to get the biggest number? What other numbers can he make?

Solution

Possible numbers: 1,42,019, 4,12,019, 4,21,019, 4,20,119 (1 before or after the existing 1) and 4,20,191.

The biggest is 4,21,019: place the 1 right after the 2 (between 2 and 0), since 4,21,… beats 4,20,… and 4,12,….

Between 2 and 0, giving 4,21,019; others: 1,42,019; 4,12,019; 4,20,119; 4,20,191.

63,890 uses 30 sticks. Rearrange exactly four sticks and make a bigger number (e.g. 88,078). What other numbers bigger than 63,890 can you make?

Solution

Moving 4 sticks keeps the total at 30. Some possibilities (each changes the sticks so that exactly 4 are removed and placed elsewhere):

99,985, 99,982, 99,906, 99,805, 88,078, 66,208, 66,066

For example, 63,890 → 99,985: take one stick from the 6, one from the 8 and two from the 0 (4 sticks), and use them to turn the 6 and the 3 into 9s, the 9 into an 8 and the 0 into a 5. The largest number that can be made this way is 99,985.

E.g. 99,985 (the largest), 99,982, 99,906, 99,805, 66,208 and 66,066, each made by moving exactly four sticks.

Make any number using exactly 24 sticks. What is the biggest and the smallest number that can be made using 24 sticks?

Solution

Example: 51,084 uses 5 + 2 + 6 + 7 + 4 = 24 sticks.
Biggest: more digits make a bigger number, and 1 uses the fewest sticks (2). So use 12 ones: 1,11,11,11,11,111 (111111111111).
Smallest: fewest digits. 3 digits use at most $3 \times 7 = 21$ sticks, so we need 4 digits. The first digit cannot be 1 (then 22 sticks for 3 digits is too many), so it is 2 (5 sticks); then 0 (6), 0 (6) and 8 (7): $5 + 6 + 6 + 7 = 24$ . Smallest: 2,008.

E.g. 51,084; biggest 111111111111 (twelve 1s); smallest 2,008.

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