NCERT Solutions Ganita Prakash (Part 1) Chapter 2 –452.5 A Pinch of History (chapter-end set) — Figure it Out

Book page 44 Updated on2026-09-05

Q1.

Find out the units digit in the value of 2²²⁴ ÷ 4³²? [Hint: 4 = 2²]

Answer

The units digit is 6.

4³² = (2²)³² = 2⁶⁴

2²²⁴ ÷ 4³² = 2²²⁴ ÷ 2⁶⁴
= 2^224–64
= 2¹⁶⁰

Now look at the units digits of the powers of 2. They repeat in a cycle of four:

Power	2¹	2²	2³	2⁴	2⁵	2⁶	2⁷	2⁸
Value	2	4	8	16	32	64	128	256
Units digit	2	4	8	6	2	4	8	6

160 ÷ 4 = 40, remainder 0
A remainder of 0 means we are at the end of a cycle, i.e. the same place as 2⁴
Units digit of 2¹⁶⁰ = 6

Why it happens: the units digit of a product depends only on the units digits of the factors, so multiplying by 2 sends 2 → 4 → 8 → 6 → 2 and can never escape those four. Because the cycle has length 4, the exponent only matters through its remainder on division by 4. This is why you can answer a question about a 49-digit number without ever writing it down.

Q2.

There are 5 bottles in a container. Every day, a new container is brought in. How many bottles would be there after 40 days?

Answer

200 bottles = 2 × 10².

Bottles added per day = 5
Days = 40
Total = 5 × 40 = 200 = 2 × 10²

Why it happens: this question is placed here as a deliberate contrast. Nothing is doubling or tripling — the same 5 bottles arrive each day, so the total adds and the growth is linear. After 40 days there are 200 bottles, not 5⁴⁰. Compare the magical pond, where 40 days of doubling would give 2⁴⁰ ≈ 1.1 × 10¹² lotuses. Read the situation before reaching for exponents.

Check it yourself: "5 bottles every day" is 5 × n. "5 times as many every day" would be 5ⁿ. One word changes the answer from 200 to a number with 28 digits.

Q3.

Write the given number as the product of two or more powers in three different ways. The powers can be any integers. (i) 64³ (ii) 192⁸ (iii) 32⁻⁵

Answer

Reduce each number to prime powers first; then any split of the exponents works.

(i) 64³ : 64 = 2⁶, so 64³ = (2⁶)³ = 2¹⁸
2¹⁰ × 2⁸
4⁵ × 4⁴ (since 4⁹ = 2¹⁸)
8³ × 8³ (since 8⁶ = 2¹⁸)

(ii) 192⁸ : 192 = 2⁶ × 3, so 192⁸ = 2⁴⁸ × 3⁸
2⁴⁸ × 3⁸
2⁴⁰ × 2⁸ × 3⁸
2⁴⁰ × 6⁸ (since 2⁸ × 3⁸ = 6⁸)

(iii) 32^–5 : 32 = 2⁵, so 32^–5 = (2⁵)^–5 = 2^–25
2^–10 × 2^–15
2^–5 × 2^–20
4^–12 × 2^–1 (since 4^–12 = 2^–24)

Why it happens: once a number is written as a single prime power, every way of writing the exponent as a sum gives a new factorisation — and there are unlimited ways, because the parts may be negative. For (iii) the exponents must add to –25, so 2^–30 × 2⁵ would work equally well. Checking is always the same test: do the exponents of each prime add up correctly?

Q4.

Examine each statement below and find out if it is ‘Always True’, ‘Only Sometimes True’, or ‘Never True’. Explain your reasoning. (i) Cube numbers are also square numbers. (ii) Fourth powers are also square numbers. (iii) The fifth power of a number is divisible by the cube of that number. (iv) The product of two cube numbers is a cube number. (v) q⁴⁶ is both a 4th power and a 6th power (q is a prime number).

Answer

	Verdict	Reasoning
(i)	Only Sometimes True	8 = 2³ is a cube but not a square. 64 = 4³ = 8² is both. A number is both exactly when it is a 6th power: n⁶ = (n²)³ = (n³)².
(ii)	Always True	n⁴ = (n²)², so every fourth power is the square of n². 4 is even, and that is all that is needed.
(iii)	Always True	n⁵ ÷ n³ = n^5–3 = n², a whole number. (For n = 0 both are 0 and the statement is vacuous, so it is stated for n ≠ 0.)
(iv)	Always True	a³ × b³ = (ab)³ by the rule m^a × n^a = (mn)^a. E.g. 8 × 27 = 216 = 6³.
(v)	Never True	q is prime, so q⁴⁶ can be a 4th power only if 4 divides 46, and a 6th power only if 6 divides 46. 46 = 4 × 11 + 2 and 46 = 6 × 7 + 4 — neither divides. So it is neither, let alone both.

Why it happens: every one of these is really a statement about the exponent. "Is a square" means the exponent is a multiple of 2; "is a cube" means a multiple of 3; "is both" means a multiple of 6. Once you translate the question into divisibility of exponents, the answers follow without any large arithmetic. Part (v) needs q to be prime, because for a composite base such as q = 4 the number 4⁴⁶ = 2⁹² is a 4th power after all.

Q5.

Simplify and write these in the exponential form. (i) 10⁻² × 10⁻⁵ (ii) 5⁷ ÷ 5⁴ (iii) 9⁻⁷ ÷ 9⁴ (iv) (13⁻²)⁻³ (v) m⁵n¹²(mn)⁹

Answer

(i) 10^–2 × 10^–5 = 10^–2+(–5) = 10^–7 ( = 1/10⁷)

(ii) 5⁷ ÷ 5⁴ = 5^7–4 = 5³ ( = 125)

(iii) 9^–7 ÷ 9⁴ = 9^–7–4 = 9^–11 ( = 1/9¹¹)

(iv) (13^–2)^–3 = 13^{(–2)×(–3)} = 13⁶

Part (v) has two bases, so handle each separately.

(v) m⁵n¹²(mn)⁹
= m⁵n¹² × m⁹n⁹ (since (mn)⁹ = m⁹n⁹)
= (m⁵ × m⁹) × (n¹² × n⁹)
= m¹⁴n²¹

Why it happens: (iii) is the one to watch. Dividing by 9⁴ subtracts 4, so –7 – 4 = –11 — it is easy to write –3 by mistake. (iv) shows two negatives producing a positive exponent, exactly as (–2) × (–3) = 6 for ordinary integers, which is no accident: the power-of-a-power rule turns the exponents into a plain multiplication.

Q6.

If 12² = 144 what is (i) (1.2)² (ii) (0.12)² (iii) (0.012)² (iv) 120²

Answer

Write each number as 12 times a power of 10, then square both parts.

(i) 1.2 = 12 × 10^–1
(1.2)² = 12² × 10^–2 = 144 × 0.01 = 1.44

(ii) 0.12 = 12 × 10^–2
(0.12)² = 144 × 10^–4 = 0.0144

(iii) 0.012 = 12 × 10^–3
(0.012)² = 144 × 10^–6 = 0.000144

(iv) 120 = 12 × 10¹
120² = 144 × 10² = 14400

Number	Decimal places	Square	Decimal places in square
1.2	1	1.44	2
0.12	2	0.0144	4
0.012	3	0.000144	6

Why it happens: the digits 144 never change, only the position of the decimal point does — because moving the point is multiplying by a power of 10, and squaring simply doubles that power's exponent. Hence the neat rule visible in the table: squaring doubles the number of decimal places.

Q7.

Circle the numbers that are the same — 2⁴ × 3⁶ 6⁴ × 3² 6¹⁰ 18² × 6² 6²⁴

Answer

Bring every expression down to powers of the primes 2 and 3.

Expression	In primes	Value
2⁴ × 3⁶	2⁴ × 3⁶	16 × 729 = 11664
6⁴ × 3²	(2×3)⁴ × 3² = 2⁴ × 3⁴⁺² = 2⁴ × 3⁶	1296 × 9 = 11664
6¹⁰	2¹⁰ × 3¹⁰	6,04,66,176
18² × 6²	(2×3²)² × (2×3)² = 2²⁺² × 3⁴⁺² = 2⁴ × 3⁶	324 × 36 = 11664
6²⁴	2²⁴ × 3²⁴	about 4.7 × 10¹⁸

So the three that are the same are 2⁴ × 3⁶, 6⁴ × 3² and 18² × 6², each equal to 11664.

Why it happens: two numbers are equal exactly when their prime factorisations match, so reducing everything to 2s and 3s turns the whole comparison into checking two pairs of exponents. There is no need to evaluate 6²⁴ at all — its exponents (24, 24) obviously differ from (4, 6). Also worth noticing: 11664 = 108², since 2⁴ × 3⁶ = (2² × 3³)².

Q8.

Identify the greater number in each of the following — (i) 4³ or 3⁴ (ii) 2⁸ or 8² (iii) 100² or 2¹⁰⁰

Answer

(i) 4³ = 64 and 3⁴ = 81 → 3⁴ is greater

(ii) 2⁸ = 256 and 8² = 64 → 2⁸ is greater

(iii) 100² = 10,000 = 10⁴
2¹⁰⁰ = (2¹⁰)¹⁰ ≈ (10³)¹⁰ = 10³⁰
→ 2¹⁰⁰ is greater, by a factor of about 10²⁶

The exact value of 2¹⁰⁰ is 1,267,650,600,228,229,401,496,703,205,376 — about 1.27 × 10³⁰.

Why it happens: when the base and the exponent are swapped, the larger exponent usually wins, because the exponent controls how many times the multiplication happens. 2⁸ beats 8² comfortably, and 2¹⁰⁰ annihilates 100². The one place this fails is among the very smallest numbers: 2³ = 8 is less than 3² = 9, and 2⁴ = 4² exactly.

Tip: for part (iii) you never need the 31-digit number. 2¹⁰ = 1024 ≈ 10³ lets you estimate any power of 2 in one step — a habit worth keeping.

Q9.

A dairy plans to produce 8.5 billion packets of milk in a year. They want a unique ID (identifier) code for each packet. If they choose to use the digits 0–9, how many digits should the code consist of?

Answer

10 digits.

Packets in a year = 8.5 billion = 8.5 × 10⁹ = 8,50,00,00,000
Each slot holds one of 10 digits, so an n-digit code gives 10ⁿ different codes

Need 10ⁿ ≥ 8.5 × 10⁹
10⁹ = 1 × 10⁹ < 8.5 × 10⁹ — not enough
10¹⁰ = 10 × 10⁹ > 8.5 × 10⁹ — enough
So n = 10

A 10-digit code gives 1000 crore codes for 850 crore packets — comfortably more than needed, with room to spare.

Why it happens: the answer must be a whole number, so the question is really "what is the smallest n with 10ⁿ ≥ 8.5 × 10⁹?" Nine digits fall short by a factor of 8.5, and there is no such thing as a 9½-digit code — you must round up. Note also how little the extra digit costs: one more slot, ten times the capacity.

Q10.

64 is a square number (8²) and a cube number (4³). Are there other numbers that are both squares and cubes? Is there a way to describe such numbers in general?

Answer

Yes — infinitely many. They are exactly the sixth powers, n⁶.

n⁶ = (n³)² → a square (of n³)
n⁶ = (n²)³ → a cube (of n²)

n	n⁶	As a square	As a cube
1	1	1²	1³
2	64	8²	4³
3	729	27²	9³
4	4096	64²	16³
5	15625	125²	25³

Why it happens: a number is a square when every exponent in its prime factorisation is a multiple of 2, and a cube when every exponent is a multiple of 3. To be both, every exponent must be a multiple of both 2 and 3 — that is, of their LCM, 6. So the numbers that are both are precisely the 6th powers, and since there is an n⁶ for every n, there are infinitely many.

Try This: what numbers are squares, cubes and fourth powers at once? The LCM of 2, 3 and 4 is 12, so they are the twelfth powers: 1, 4096, 5,31,441, …

Q11.

A digital locker has an alphanumeric (it can have both digits and letters) passcode of length 5. Some example codes are G89P0, 38098, BRJKW, and 003AZ. How many such codes are possible?

Answer

36⁵ = 6,04,66,176 — about 6 crore, or 6.05 × 10⁷.

Characters available = 26 letters + 10 digits = 36
Slots = 5, each independently filled

Total codes = 36 × 36 × 36 × 36 × 36 = 36⁵
36² = 1296
36⁴ = 1296² = 16,79,616
36⁵ = 16,79,616 × 36 = 6,04,66,176

Compare: a 5-digit numeric code gives only 10⁵ = 1,00,000. Allowing letters as well makes the locker about 600 times harder to open.

Why it happens: the examples in the question confirm that repeats are allowed (38098 repeats the 8) and that any mix is allowed, so every slot really does have all 36 choices, independently. That independence is what licenses the multiplication. If, say, the first character had to be a letter, the count would fall to 26 × 36⁴ = 4,36,70,016.

Q12.

The worldwide population of sheep (2024) is about 10⁹, and that of goats is also about the same. What is the total population of sheep and goats? (i) 20⁹ (ii) 10¹¹ (iii) 10¹⁰ (iv) 10¹⁸ (v) 2 × 10⁹ (vi) 10⁹ + 10⁹

Answer

(vi) 10⁹ + 10⁹ — and this equals (v) 2 × 10⁹, so both express the correct total.

Total = sheep + goats = 10⁹ + 10⁹
= 2 × 10⁹
= 2,00,00,00,000 (200 crore / 2 billion)

Why the others are wrong:

20⁹ wrongly doubles the base. 20⁹ = 2⁹ × 10⁹ = 512 × 10⁹ — 512 times too big.
10¹⁰ and 10¹¹ wrongly raise the exponent. Adding does not change the exponent; 10¹⁰ is 5 times the true answer.
10¹⁸ comes from multiplying the two populations instead of adding them — that would be 10⁹ × 10⁹, and it answers a different question altogether.

Why it happens: this is the trap the whole chapter has been building towards. The rules of exponents are rules for multiplication and division; there is no rule for adding powers. To add, you must factor out the common power: 10⁹ + 10⁹ = (1 + 1) × 10⁹ = 2 × 10⁹. The answer is a number with a new coefficient, not a new exponent.

Check it yourself: options (v) and (vi) are the same number written two ways, so both are correct — the book marks (vi), which is the form that shows where the total came from.

Q13.

Calculate and write the answer in scientific notation: (i) If each person in the world had 30 pieces of clothing, find the total number of pieces of clothing. (ii) There are about 100 million bee colonies in the world. Find the number of honeybees if each colony has about 50,000 bees. (iii) The human body has about 38 trillion bacterial cells. Find the bacterial population residing in all humans in the world. (iv) Total time spent eating in a lifetime in seconds.

Answer

(i) World population ≈ 8.2 × 10⁹, clothing per person = 30 = 3 × 10¹
Total = (8.2 × 10⁹) × (3 × 10¹)
= 24.6 × 10¹⁰
= 2.46 × 10¹¹ pieces

(ii) Colonies = 100 million = 10⁸, bees per colony = 50,000 = 5 × 10⁴
Total = 10⁸ × 5 × 10⁴
= 5 × 10¹² honeybees

(iii) Bacteria per person = 38 trillion = 3.8 × 10¹³
People ≈ 8.2 × 10⁹
Total = (3.8 × 10¹³) × (8.2 × 10⁹)
= 31.16 × 10²²
= 3.116 × 10²³ bacterial cells

(iv) Assume a lifetime of 70 years and 1 hour a day spent eating
1 hour = 3600 s
Total = 3600 × 365 × 70
= 3600 × 25,550
= 9,19,80,000 s
= 9.198 × 10⁷ seconds (about 2.9 years)

Why it happens: parts (i) and (iii) both needed the final tidy-up — 24.6 × 10¹⁰ and 31.16 × 10²² are correct values but not standard form, since the coefficient must stay below 10. Part (iv) is different in kind: it has no data at all, only assumptions, so the honest answer is "about 10⁸ seconds", and anyone assuming 2 hours a day would get 1.8 × 10⁸ and be equally right.

Did you know? Part (iii) says there are more bacterial cells living on the world's people than there are stars in our galaxy — 3 × 10²³ against about 10¹¹.

Q14.

What was the date 1 arab/1 billion seconds ago?

Answer

Convert seconds to days, then to years.

1 billion = 1 arab = 10⁹ seconds
1 day = 86,400 s = 8.64 × 10⁴ s

Days = 10⁹ ÷ (8.64 × 10⁴)
= 0.11574 × 10⁵
= 11,574 days

Years = 11,574 ÷ 365.25
= about 31.7 years (31 years and roughly 8½ months)

So subtract 31 years and about 8 months and 20 days from today's date. Worked from 31 August 2026, one billion seconds ago was 23 December 1994. Work it from your own date and you will land in the same season, roughly 31 years and 8 months earlier.

Why it happens: the answer is startling because a billion feels like a number and 31 years feels like a life stage — and yet they are the same quantity, seen through a very small unit. This is the closing lesson of the chapter: a power of 10 has no meaning by itself, only when attached to a unit and compared with something we already know.

Check it yourself: a million seconds ago was less than a fortnight ago; a billion seconds ago you were not born; a trillion seconds ago (10¹² s ≈ 31,700 years) the last ice age had not yet peaked.

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