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NCERT Solutions Ganita Prakash (Part 1) Chapter 1 –211.6 Did You Ever Wonder…? — Figure it Out

Book page 19 Updated on2026-09-05

Q1.
Using all digits from 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
Answer

(a) 98,76,54,32,10  (b) 1,02,34,56,798

(a) A multiple of 5 must end in 0 or 5.
Ending in 0 is better — it frees the 5 for a higher place.
Put the rest in decreasing order: 9,8,7,6,5,4,3,2,1
Largest = 9876543210

(b) An even number must end in 0, 2, 4, 6 or 8.
To be small, the front should read 1,0,2,3,4,…
Ending in 8 keeps the front smallest: 102345679 | 8
Smallest = 1023456798
Why it happens: In (b), why not end in 0? Because 0 is the most valuable digit for making the front small — it belongs right after the leading 1. Ending in 2, 4 or 6 would force a bigger digit into an earlier place. Ending in 8 leaves 0,1,2,3,4,5,6,7,9 to be arranged as 102345679, the smallest possible front.
Check it yourself: 1023456798 ÷ 2 = 511728399, so it is even ✓, and it uses each of 0–9 exactly once ✓.
Q2.
The number 10,30,285 in words is Ten lakhs thirty thousand two hundred eighty five, which has 42 letters. Give a 7-digit number name which has the maximum number of letters.
Answer

77,77,777 — “seventy seven lakh seventy seven thousand seven hundred seventy seven” — with 60 letters.

seventy (7) + seven (5) = 12
lakh (4) → 16
seventy (7) + seven (5) = 12 → 28
thousand (8) → 36
seven (5) → 41
hundred (7) → 48
seventy (7) + seven (5) = 12 → 60 letters
Why it happens: 60 is the maximum, and here is why. The longest two-digit name is “seventy-seven / seventy-three / seventy-eight”, all 12 letters. The longest single digit words are “three”, “seven” and “eight”, 5 letters each. So the best possible split is 12 + 4 (lakh) + 12 + 8 (thousand) + 5 + 7 (hundred) + 12 = 60. No 7-digit number can beat it.
Tip: Other 60-letter names exist — 73,78,373 gives “seventy three lakh seventy eight thousand three hundred seventy three”, also 60 letters. If you write “lakhs” instead of “lakh” the count becomes 61.
Q3.
Write a 9-digit number where exchanging any two digits results in a bigger number. How many such numbers exist?
Answer

The number is 123456789, and there is exactly one such number.

Swap the 1 and the 2 → 213456789 > 123456789 ✓
Swap the 1 and the 9 → 923456781 > 123456789 ✓
Swap the 8 and the 9 → 123456798 > 123456789 ✓
Every swap makes it bigger.
Why it happens: Suppose you swap the digits at two positions, an earlier one holding a and a later one holding b. The number grows only if b > a — a bigger digit has moved to the more valuable place. For every swap to grow the number, every digit must be smaller than every digit to its right, i.e. the digits must be strictly increasing. Also no two digits can be equal, since swapping equal digits leaves the number unchanged (not bigger). Nine strictly increasing digits, with 0 not allowed at the front, can only be 1,2,3,4,5,6,7,8,9 — so 123456789 is the one and only answer.
Note on the book's answer key: The appended key gives 987654312, but swapping its first two digits gives 897654312, which is smaller. So that answer does not satisfy “any two digits”. The correct answer is 123456789.
Q4.
Strike out 10 digits from the number 12345123451234512345 so that the remaining number is as large as possible.
Answer

The largest number left is 5534512345.

12345 12345 12345 12345 (20 digits, keep 10)

Step 1: for the 1st kept digit, scan positions 1–11 → biggest is 5 (position 5)
Step 2: for the 2nd, scan positions 6–12 → biggest is 5 (position 10)
Step 3: for the 3rd, scan positions 11–13 → biggest is 3 (position 13)
Step 4: only 7 digits are left (positions 14–20) and 7 are still needed → keep them all: 4512345

Result: 5 5 3 4512345 = 5534512345
Why it happens: The leftmost digit matters most, so grab the biggest digit you can while still leaving enough digits behind to finish the number. That is the whole strategy: at each step look only as far right as you can afford, and take the largest digit in that window.
Check it yourself: Count the struck-out digits — 1,2,3,4 (4 digits), then 1,2,3,4 (4 more), then 1,2 (2 more) = 10 ✓, and 10 digits remain ✓.
Q5.
The words ‘zero’ and ‘one’ share letters ‘e’ and ‘o’. The words ‘one’ and ‘two’ share a letter ‘o’, and the words ‘two’ and ‘three’ also share a letter ‘t’. How far do you have to count to find two consecutive numbers which do not share an English letter in common?
Answer

You would have to count for ever — no such pair exists, however far you go.

PairShared letter(s)PairShared letter(s)
one, twoosix, sevens
two, threetseven, eighte
three, fourreight, ninee, i
four, fivefnine, tenn, e
five, sixiten, elevene, n
Why it happens: Split all numbers into three ranges.
Below 100: check all 99 consecutive pairs by hand — every single one shares at least one letter (the table shows the first ten).
From 100 to 999: both numbers contain the word hundred, so they share h, u, n, d, r, e.
1000 and beyond: both numbers contain thousand, or million, or billion, or trillion — and every one of those words contains the letters n and o. So the two names always share n and o.
The two boundary cases behave the same way: 99/100 share n and e, and 999/1000 share n, e, h, u, d.
Did you know? This is a nice example of a question whose honest answer is “never”. You cannot settle it by testing examples — you have to argue about all numbers at once, which is what the three cases above do.
Q6.
Suppose you write down all the numbers 1, 2, 3, 4, …, 9, 10, 11, ... The tenth digit you write is ‘1’ and the eleventh digit is ‘0’, as part of the number 10. (a) What would the 1000th digit be? At which number would it occur? (b) What number would contain the millionth digit? (c) When would you have written the digit ‘5’ for the 5000th time?
Answer

(a) The 1000th digit is 3, the first digit of the number 370.

Numbers 1–9: 9 numbers × 1 digit = 9 digits (running total 9)
Numbers 10–99: 90 numbers × 2 digits = 180 digits (running total 189)
Digits still needed = 1000 − 189 = 811
811 ÷ 3 = 270 remainder 1
270 three-digit numbers: 100 to 369 → uses 810 digits (total 999)
The next digit, the 1000th, is the 1st digit of 370, which is 3

(b) The millionth digit lies in the number 1,85,185.

1–9: 9 digits (total 9)
10–99: 180 digits (total 189)
100–999: 2,700 digits (total 2,889)
1000–9999: 36,000 digits (total 38,889)
10,000–99,999: 4,50,000 digits (total 4,88,889)
Digits still needed = 10,00,000 − 4,88,889 = 5,11,111
5,11,111 ÷ 6 = 85,185 remainder 1
85,185 six-digit numbers: 1,00,000 to 1,85,184 → 5,11,110 digits (total 9,99,999)
The next digit, the millionth, is the 1st digit of 1,85,185

(c) You write the digit ‘5’ for the 5000th time at the number 13,495 — it is the units digit of 13,495.

Count of 5's written from 1 up to 13,494 = 4,999
13,495 contains one 5 (in the units place)
4,999 + 1 = 5,000 ✓
Why it happens: The trick in all three parts is the same — count digits in blocks (1-digit numbers, 2-digit numbers, 3-digit numbers, …) instead of one at a time, then divide to see how far into the next block you land. The remainder tells you which digit of which number you have reached.
Note on the book's answer key: The appended key gives 13,995 for part (c). Counting carefully, the 5,000th ‘5’ actually appears in 13,495; by 13,995 you have already written the digit 5 more than 5,000 times.
Q7.
A calculator has only ‘+10,000’ and ‘+100’ buttons. Write an expression describing the number of button clicks to be made for the following numbers: (a) 20,800 (b) 92,100 (c) 1,20,500 (d) 65,30,000 (e) 70,25,700
Answer

Take as many ten-thousands as you can, then make up the rest in hundreds.

NumberExpressionTotal clicks
(a) 20,800(2 × 10,000) + (8 × 100)2 + 8 = 10
(b) 92,100(9 × 10,000) + (21 × 100)9 + 21 = 30
(c) 1,20,500(12 × 10,000) + (5 × 100)12 + 5 = 17
(d) 65,30,000(653 × 10,000)653 + 0 = 653
(e) 70,25,700(702 × 10,000) + (57 × 100)702 + 57 = 759
Check (e): 702 × 10,000 = 70,20,000
57 × 100 = 5,700
70,20,000 + 5,700 = 70,25,700 ✓
Why it happens: With only these two buttons, every number you can reach must be a multiple of 100. Split the number as (number ÷ 10,000) ten-thousands plus whatever hundreds are left. In (b), 92,100 = 90,000 + 2,100, so 9 clicks of +10,000 and 21 clicks of +100.
Try This: Could this calculator show 20,850? No — 50 is not a whole number of hundreds, so that number is out of reach.
Q8.
How many lakhs make a billion?
Answer

10,000 lakhs make one billion.

1 billion = 1,000,000,000 = 109
1 lakh = 1,00,000 = 105
Number of lakhs = 109 ÷ 105
= 104
= 10,000 lakhs
Why it happens: Just count zeroes. A billion has 9 zeroes, a lakh has 5, so the answer has 9 − 5 = 4 zeroes: 10,000. In Indian terms, 1 billion = 100 crore = 1 arab, and 100 crore = 10,000 lakh.
Tip: The same method answers “how many lakhs make a million?” — 6 − 5 = 1 zero, so 10 lakhs.
Q9.
You are given two sets of number cards numbered from 1 – 9. Place a number card in each box below to get the (a) largest possible sum (b) smallest possible difference of the two resulting numbers. (The boxes make a 7-digit number above a 5-digit number.)
Answer

(a) Largest sum = 1,00,75,308   (b) Smallest difference = 10,22,447

(a) 99,87,654
   +     87,654
   = 1,00,75,308

(b) 11,22,334
   −     99,887
   = 10,22,447
Why it happens: There are two of every card, 1 to 9.
For the largest sum, list the place values of all 12 boxes: 106 and 105 occur once each (only the 7-digit number reaches that far), while 104, 103, 102, 101 and 100 occur twice each. Now give the biggest cards to the biggest place values: 9 to 106, 9 to 105, both 8s to the two 104 places, both 7s to the 103 places, and so on.
For the smallest difference, make the 7-digit number as small as possible (11,22,334, using the two 1s, two 2s, two 3s and one 4) and the 5-digit number as large as possible from what is left (99,887).
Note on the book's answer key: The key prints the sum as 1,00,54,320. Adding the cards it shows — 9,9,8,8,7,7,6,6,5,5,4,4 — actually gives 99,87,654 + 87,654 = 1,00,75,308. The difference answer, 10,22,447, is correct.
Q10.
You are given some number cards; 4000, 13000, 300, 70000, 150000, 20, 5. Using the cards get as close as you can to the numbers below using any operation you want. Each card can be used only once for making a particular number. (a) 1,10,000 (b) 2,00,000 (c) 5,80,000 (d) 12,45,000 (e) 20,90,800
Answer

Four of the five targets can be hit exactly.

TargetExpressionValue
(a) 1,10,000(13,000 × 20) − 1,50,0001,10,000 — exact
(b) 2,00,000(70,000 × 5) − 1,50,0002,00,000 — exact
(c) 5,80,000(70,000 × 5) + 1,50,000 + (4,000 × 20)5,80,000 — exact
(d) 12,45,000(13,000 + 70,000) × 300 ÷ 2012,45,000 — exact
(e) 20,90,80070,000 + 4,000 × (5 + 1,50,000 ÷ 300) + (13,000 ÷ 20)20,90,650 — off by 150
Check (a): 13,000 × 20 = 2,60,000; 2,60,000 − 1,50,000 = 1,10,000 ✓
Check (c): 3,50,000 + 1,50,000 + 80,000 = 5,80,000 ✓
Check (d): 83,000 × 300 = 2,49,00,000; ÷ 20 = 12,45,000 ✓
Check (e): 1,50,000 ÷ 300 = 500; 500 + 5 = 505; 4,000 × 505 = 20,20,000;
+ 70,000 = 20,90,000; 13,000 ÷ 20 = 650; total = 20,90,650
Why it happens: The useful moves are dividing to make small handy numbers (1,50,000 ÷ 300 = 500, 13,000 ÷ 20 = 650) and multiplying to jump up fast. Once you spot 13,000 × 20 = 2,60,000 or 70,000 × 5 = 3,50,000, the rest is a short adjustment.
Try This: The book's own sample for (a) — 4000 × (20 + 5) + 13000 = 1,13,000 — is 3,000 away. Can you find any exact expression for 20,90,800? The closest known is 70,000 − (5 − 300) × ((1,50,000 − 13,000) ÷ 20) = 20,90,750, only 50 away.
Q11.
Find out how many coins should be stacked to match the height of the Statue of Unity. Assume each coin is 1 mm thick.
Answer

1,80,000 coins — one lakh eighty thousand.

Height of Statue of Unity = 180 m
1 m = 100 cm and 1 cm = 10 mm
So 180 m = 180 × 100 × 10 mm
= 1,80,000 mm

Thickness of one coin = 1 mm
Number of coins = 1,80,000 ÷ 1
= 1,80,000 coins
Why it happens: The whole question is really a unit conversion. Once the height is in millimetres, the division by 1 mm is trivial — but forgetting that 1 m = 1000 mm would make the answer a thousand times too small.
Did you know? 1,80,000 one-rupee coins weigh roughly 700 kg — about as much as a blue whale's heart!
Q12.
Grey-headed albatrosses have a roughly 7-feet wide wingspan. They are known to migrate across several oceans. 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 to cross the Pacific Ocean approximately?
Answer

About 12 to 14 days.

At 1000 km a day: 12,000 ÷ 1000 = 12 days
At 900 km a day: 12,000 ÷ 900 = 13.3 → about 14 days
(13 full days cover only 11,700 km, so a 14th day is needed)

So the trip takes roughly 12 to 14 days
Why it happens: The speed is given as a range, so the answer must be a range too. Using the fastest speed gives the shortest time and the slowest speed gives the longest time — that pair of calculations brackets the true answer.
Did you know? With a 7-foot wingspan, a grey-headed albatross can glide for hours without flapping, riding the wind over the Southern Ocean.
Q13.
A bar-tailed godwit holds the record for the longest recorded non-stop flight. It travelled 13,560 km from Alaska to Australia without stopping. Its journey started on 13 October 2022 and continued for about 11 days. Find out the approximate distance it covered every day. Find out the approximate distance it covered every hour.
Answer

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

Distance per day = 13,560 ÷ 11
= 1232.7…
≈ 1,233 km per day

Hours in the journey = 11 × 24 = 264
Distance per hour = 13,560 ÷ 264
= 51.36…
≈ 51 km per hour

(Or: 1,233 ÷ 24 ≈ 51 km per hour)
Why it happens: The bird never lands, so it is flying all 24 hours of each day — that is why we divide by 264 hours and not by, say, 12 daylight hours. 51 km/h non-stop for eleven days is roughly the speed of a car in city traffic, kept up without a single break.
Did you know? The godwit doubles its body weight before setting off and burns almost all of it on the way — it cannot eat, drink or sleep properly for eleven days.
Q14.
Bald eagles are known to fly as high as 4500 – 6000 m above the ground level. Mount Everest is about 8850 m high. Aeroplanes can fly as high as 10,000 – 12,800 m. How many times bigger are these heights compared to Somu’s building?
Answer

Somu's building is 40 m tall, so divide every height by 40.

HeightWorkingTimes Somu's building
Bald eagle (lower) 4,500 m4500 ÷ 40112.5 ≈ 112 times
Bald eagle (upper) 6,000 m6000 ÷ 40150 times
Mount Everest 8,850 m8850 ÷ 40221.25 ≈ 221 times
Aeroplane (lower) 10,000 m10000 ÷ 40250 times
Aeroplane (upper) 12,800 m12800 ÷ 40320 times
So: bald eagles fly about 112 to 150 times the height of the building,
Mount Everest is about 221 times as tall,
and aeroplanes fly about 250 to 320 times as high.
Why it happens: This is the “compare it with something familiar” idea from page 3, used one last time. Saying “an aeroplane flies 12,800 m up” means little; saying “that is more than 300 ten-storey buildings stacked one on another” makes it real.
Check it yourself: A bald eagle at 6,000 m is flying above the top of Somu's building 150 times over — and even Mount Everest is lower than a cruising aeroplane.
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