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NCERT Solutions · Class 8 Maths · Ganita Prakash Part 1 · Chapter 2

Chapter 2: Power Play (Exponents)

Step-by-step answers to every "Figure it Out" and in-text question of Chapter 2, Power Play (NCERT Class 8 Maths, Ganita Prakash Part 1, 2026-27): paper folding and exponential growth, exponential notation, the laws of exponents, zero and negative exponents, power lines, powers of 10, scientific notation, combinations and estimation thought experiments. All 43 questions are answered, with the key answer highlighted.

2.1 Experiencing the Power Play

A sheet 0.001 cm thick doubles in thickness with every fold. Complete the tables of thickness after folds 21 to 46.

Solution

Thickness after n folds $= 0.001 \times 2^{n}$ cm.

Fold	Thickness	Fold	Thickness	Fold	Thickness
21	≈ 21 m	30	≈ 10.7 km	39	≈ 5,498 km
22	≈ 42 m	31	≈ 21.5 km	40	≈ 10,995 km
23	≈ 84 m	32	≈ 43 km	41	≈ 21,990 km
24	≈ 168 m	33	≈ 86 km	42	≈ 43,980 km
25	≈ 336 m	34	≈ 172 km	43	≈ 87,961 km
26	≈ 671 m	35	≈ 344 km	44	≈ 1,75,922 km
27	≈ 1.3 km	36	≈ 687 km	45	≈ 3,51,844 km
28	≈ 2.7 km	37	≈ 1,374 km	46	≈ 7,03,687 km
29	≈ 5.4 km	38	≈ 2,749 km

After 46 folds the thickness (≈ 7 lakh km) is more than the distance to the Moon (3,84,400 km).

Thickness = 0.001 × 2ⁿ cm: about 10.7 km after 30 folds and about 7,03,687 km after 46.

2.2 Exponential Notation and Operations

Which expression gives the thickness after 10 folds, if the initial thickness is v? (i) 10v (ii) 10 + v (iii) 2 × 10 × v (iv) 2¹⁰ (v) 2¹⁰v (vi) 10²v

Solution

Each fold multiplies by 2, so 10 folds multiply by $2^{10}$ : (v) $2^{10} v$ .

(v) 2¹⁰v

What is (−1)⁵? Is it positive or negative? What about (−1)⁵⁶? Is (−2)⁴ = 16? What are 0², 0⁵ and 0ⁿ?

Solution

$(- 1)^{5} = - 1$ (negative: an odd number of −1s); $(- 1)^{56} = 1$ (positive: an even number).

$(- 2)^{4} = (- 2) (- 2) (- 2) (- 2) = 16$ ✓.

$0^{2} = 0$ , $0^{5} = 0$ , and $0^{n} = 0$ for every counting number n.

(−1)⁵ = −1, (−1)⁵⁶ = 1; yes, (−2)⁴ = 16; 0ⁿ = 0.

Figure it Out (page 22)

Express in exponential form: (i) 6 × 6 × 6 × 6 (ii) y × y (iii) b × b × b × b (iv) 5 × 5 × 7 × 7 × 7 (v) 2 × 2 × a × a (vi) a × a × a × c × c × c × c × d

Solution

(i) $6^{4}$ (ii) $y^{2}$ (iii) $b^{4}$ (iv) $5^{2} \times 7^{3}$ (v) $2^{2} a^{2}$ (or $(2 a)^{2}$ ) (vi) $a^{3} c^{4} d$

(i) 6⁴ (ii) y² (iii) b⁴ (iv) 5² × 7³ (v) 2²a² (vi) a³c⁴d

Express as a product of powers of prime factors: (i) 648 (ii) 405 (iii) 540 (iv) 3600

Solution

(i) $648 = 2^{3} \times 3^{4}$ (ii) $405 = 3^{4} \times 5$ (iii) $540 = 2^{2} \times 3^{3} \times 5$ (iv) $3600 = 2^{4} \times 3^{2} \times 5^{2}$

(i) 2³ × 3⁴ (ii) 3⁴ × 5 (iii) 2² × 3³ × 5 (iv) 2⁴ × 3² × 5²

Find the value: (i) 2 × 10³ (ii) 7² × 2³ (iii) 3 × 4⁴ (iv) (−3)² × (−5)² (v) 3² × 10⁴ (vi) (−2)⁵ × (−10)⁶

Solution

(i) 2000 (ii) $49 \times 8 = 392$ (iii) $3 \times 256 = 768$ (iv) $9 \times 25 = 225$ (v) $9 \times 10000 = 90000$ (vi) $- 32 \times 1 0^{6} = - 3, 20, 00, 000$

(i) 2000 (ii) 392 (iii) 768 (iv) 225 (v) 90,000 (vi) −3,20,00,000

The Stones that Shine

Three daughters, each with three baskets, each with three keys, … each necklace with three diamonds. How many rooms? How many diamonds? Why is 3⁷ also 3² × 3⁵?

Solution

Daughters 3, baskets $3^{2}$ , keys $3^{3}$ , rooms $3^{4} =$ 81, tables $3^{5}$ , necklaces $3^{6}$ , diamonds $3^{7} = 3^{4} \times 3^{3} = 81 \times 27 =$ 2187.

$3^{7}$ is seven 3s multiplied; group them as two 3s and five 3s: $(3 \times 3) \times (3 \times 3 \times 3 \times 3 \times 3) = 3^{2} \times 3^{5}$ .

81 rooms; 3⁷ = 2187 diamonds; seven 3s can be grouped as 2 + 5.

Use

n^{a} \times n^{b} = n^{a + b}

to compute (i) 2⁹ (ii) 5⁷ (iii) 4⁶. Write as a power of a power in two ways: (i) 8⁶ (ii) 7¹⁵ (iii) 9¹⁴ (iv) 5⁸.

Solution

(i) $2^{9} = 2^{4} \times 2^{5} = 16 \times 32 = 512$ (ii) $5^{7} = 5^{3} \times 5^{4} = 125 \times 625 = 78125$ (iii) $4^{6} = 4^{3} \times 4^{3} = 64 \times 64 = 4096$

Powers of powers:

(i) $8^{6} = (8^{2})^{3} = (8^{3})^{2}$ (also $= (2^{3})^{6} = 2^{18}$ )

(ii) $7^{15} = (7^{3})^{5} = (7^{5})^{3}$

(iii) $9^{14} = (9^{2})^{7} = (9^{7})^{2}$ (also $= 3^{28}$ )

(iv) $5^{8} = (5^{2})^{4} = (5^{4})^{2}$

512, 78125, 4096; e.g. 8⁶ = (8²)³ = (8³)², 7¹⁵ = (7³)⁵ = (7⁵)³, 9¹⁴ = (9²)⁷ = (9⁷)², 5⁸ = (5²)⁴ = (5⁴)²

Magical Pond

Lotuses double every day and cover the pond after 30 days (starting from 1). Write the number of lotuses when the pond was fully covered and half covered. Compute 2⁵ × 5⁵ and simplify 10⁴ / 5⁴.

Solution

Fully covered (day 30): $2^{30}$ lotuses; half covered (day 29): $2^{29}$ .

$2^{5} \times 5^{5} = (2 \times 5)^{5} = 1 0^{5} = 1, 00, 000$ . $\frac{1 0 ^{4}}{5 ^{4}} = (\frac{10}{5})^{4} = 2^{4} = 16$ .

2³⁰ and 2²⁹; 2⁵ × 5⁵ = 10⁵; 10⁴/5⁴ = 2⁴ = 16

How Many Combinations

Roxie has 7 dresses, 2 hats and 3 pairs of shoes: how many outfits? How many passwords for a 6-slot lock with letters A to Z?

Solution

Outfits: $7 \times 2 \times 3 =$ 42.

Lock: 26 choices in each of 6 slots: $2 6^{6} =$ 30,89,15,776 passwords.

42 outfits; 26⁶ = 30,89,15,776 passwords.

Think about the number of possible PIN codes, mobile numbers and vehicle registration numbers.

Solution

PIN codes: 6 digits, but the first digit is 1–9 (it names the postal region), so at most $9 \times 1 0^{5} = 9$ lakh codes; far fewer are actually used.
Mobile numbers: 10 digits, starting with 6, 7, 8 or 9 in India: up to $4 \times 1 0^{9}$ numbers.
Vehicle numbers (like MH 12 AB 1234): state code, 2-digit district, up to 2 letters and 4 digits: each district can issue about $2 6^{2} \times 1 0^{4} \approx 67.6$ lakh numbers.

PIN codes ≤ 9 × 10⁵; mobile numbers ≈ 4 × 10⁹; about 26² × 10⁴ vehicle numbers per district series.

2.3 The Other Side of Powers

What is 2¹⁰⁰ ÷ 2²⁵? Why can't n be 0 in

n^{a} \div n^{b} = n^{a - b}

? Can a and b be any integers?

Solution

$2^{100} \div 2^{25} = 2^{75}$ .

n cannot be 0 because $0^{b} = 0$ , and division by 0 is not defined.

Yes: with $n^{- a} = \frac{1}{n ^{a}}$ and $n^{0} = 1$ , all three rules hold for any integers a and b (e.g. $2^{- 3} \times 2^{5} = \frac{2 ^{5}}{2 ^{3}} = 2^{2}$ ).

2⁷⁵; dividing by 0ᵇ = 0 is impossible; yes, the rules work for all integer exponents.

Write equivalent forms: (i) 2⁻⁴ (ii) 10⁻⁵ (iii) (−7)⁻² (iv) (−5)⁻³ (v) 10⁻¹⁰⁰

Solution

(i) $\frac{1}{2 ^{4}} = \frac{1}{16}$ (ii) $\frac{1}{1 0 ^{5}}$ (iii) $\frac{1}{( - 7 ) ^{2}} = \frac{1}{49}$ (iv) $\frac{1}{( - 5 ) ^{3}} = - \frac{1}{125}$ (v) $\frac{1}{1 0 ^{100}}$

1/16, 1/10⁵, 1/49, −1/125, 1/10¹⁰⁰

Simplify in exponential form: (i) 2⁻⁴ × 2⁷ (ii) 3² × 3⁻⁵ × 3⁶ (iii) p³ × p⁻¹⁰ (iv) 2⁴ × (−4)⁻² (v) 8ᵖ × 8^q

Solution

(i) $2^{3}$ (ii) $3^{3}$ (iii) $p^{- 7}$ (iv) $\frac{2 ^{4}}{4 ^{2}} = \frac{2 ^{4}}{2 ^{4}} = 2^{0} = 1$ (v) $8^{p + q}$

(i) 2³ (ii) 3³ (iii) p⁻⁷ (iv) 1 (v) 8^(p+q)

Power Lines

How many times larger than 4⁻² is 4²? Use the power line for 7 to find: 2401 × 49, 49³, 343 × 2401, 16807 / 49, 7 / 343, 16807 / 823543, 117649 × 1/343, 1/343 × 1/343.

Solution

$4^{2} \div 4^{- 2} = 4^{4} =$ 256 times.

Expression	Power of 7	Value
2401 × 49	7⁴ × 7² = 7⁶	1,17,649
49³	(7²)³ = 7⁶	1,17,649
343 × 2401	7³ × 7⁴ = 7⁷	8,23,543
16807 ÷ 49	7⁵ ÷ 7² = 7³	343
7 ÷ 343	7¹ ÷ 7³ = 7⁻²	1/49
16807 ÷ 823543	7⁵ ÷ 7⁷ = 7⁻²	1/49
117649 × 1/343	7⁶ × 7⁻³ = 7³	343
1/343 × 1/343	7⁻³ × 7⁻³ = 7⁻⁶	1/1,17,649

256 times; 7⁶, 7⁶, 7⁷, 7³, 7⁻², 7⁻², 7³, 7⁻⁶

2.4 Powers of 10

Write using powers of 10: (i) 172 (ii) 5642 (iii) 6374

Solution

(i) $1 \times 1 0^{2} + 7 \times 1 0^{1} + 2 \times 1 0^{0}$

(ii) $5 \times 1 0^{3} + 6 \times 1 0^{2} + 4 \times 1 0^{1} + 2 \times 1 0^{0}$

(iii) $6 \times 1 0^{3} + 3 \times 1 0^{2} + 7 \times 1 0^{1} + 4 \times 1 0^{0}$

As shown, e.g. 172 = 1 × 10² + 7 × 10¹ + 2 × 10⁰

Write the large-number facts in scientific notation: Sun's distance from the galaxy's centre, stars in our galaxy, mass of the Earth.

Solution

(i) $3 \times 1 0^{20}$ m (ii) $1 \times 1 0^{11}$ stars (iii) $5.976 \times 1 0^{24}$ kg

3 × 10²⁰ m; 1 × 10¹¹; 5.976 × 10²⁴ kg

Which is the smallest of the distances Sun–Saturn (1.4335 × 10¹² m), Saturn–Uranus (1.439 × 10¹² m), Sun–Earth (1.496 × 10¹¹ m)? Mark the Earth on the Sun–Saturn number line.

Solution

The Sun–Earth distance is the smallest: its exponent is 11, the others' is 12. It is about $\frac{1.496 \times 1 0 ^{11}}{1.4335 \times 1 0 ^{12}} \approx \frac{1}{10}$ of the Sun–Saturn distance, so mark the Earth about one-tenth of the way from the Sun to Saturn.

Sun–Earth; the Earth is about 1/10 of the way from the Sun to Saturn.

Express in standard form: (i) 59,853 (ii) 65,950 (iii) 34,30,000 (iv) 70,04,00,00,000

Solution

(i) $5.9853 \times 1 0^{4}$ (ii) $6.595 \times 1 0^{4}$ (iii) $3.43 \times 1 0^{6}$ (iv) $7.004 \times 1 0^{10}$

(i) 5.9853 × 10⁴ (ii) 6.595 × 10⁴ (iii) 3.43 × 10⁶ (iv) 7.004 × 10¹⁰

2.5 Did You Ever Wonder?

How many 1-rupee coins would equal Roxie's weight? What if 5-rupee coins or 10-rupee notes are used? (Make reasonable assumptions.)

Solution

Assume Roxie weighs 45 kg. A 1-rupee coin weighs about 3.8 g: $45000 \div 3.8 \approx 12, 000$ coins, i.e. about ₹12,000 (thousands of coins).

A 5-rupee coin weighs about 6 g: $45000 \div 6 = 7500$ coins, about ₹37,500. A 10-rupee note weighs about 1 g: about 45,000 notes, i.e. about ₹4.5 lakh.

About 12,000 one-rupee coins (₹12,000); about ₹37,500 in 5-rupee coins; about ₹4.5 lakh in 10-rupee notes (assuming 45 kg).

How many people might benefit each year from notebooks or annadāna worth one's weight? How long ago did pilgrims who walked 400 km start? How many times could a person walk around the Earth (40,000 km) in a lifetime?

Solution

Notebooks: a notebook weighs about 200 g, so 50 kg ≈ 250 notebooks: about 50 children getting 5 notebooks each.
Annadāna: 50 kg of rice at about 150 g per meal ≈ 330 meals.
Pādayātra: walking about 25 km a day, 400 km takes about 16 days, so they started about 2 weeks earlier.
Around the Earth: walking non-stop at 5 km/h gives $5 \times 24 \times 365 \approx 43, 800$ km a year; in 70 years about $3 \times 1 0^{6}$ km, i.e. about 75 times around the Earth.

About 250 notebooks or 330 meals a year; the 400 km walk took about 16 days; non-stop walking for 70 years ≈ 75 rounds of the Earth.

Give examples of linear growth and of exponential growth.

Solution

Linear: saving ₹100 every week; a candle burning down 1 cm an hour; the steps of a ladder; a taxi meter adding a fixed amount per km.
Exponential: bacteria doubling every 20 minutes; a rumour where each person tells two more; money with compound interest; folding paper; a chess board with 1, 2, 4, 8, … grains of rice.

Linear: fixed amount added each time (weekly savings). Exponential: multiplied each time (doubling bacteria, compound interest).

Write the blanks: 1.3 billion starlings; 110 trillion mosquitoes. With 8 × 10⁹ people and 4 × 10⁵ African elephants, are there nearly 20,000 people per elephant?

Solution

Starlings $\approx 1.3 \times 1 0^{9}$ ; mosquitoes $\approx 1.1 \times 1 0^{14}$ .

$\frac{8 \times 1 0 ^{9}}{4 \times 1 0 ^{5}} = 2 \times 1 0^{4} = 20, 000$ : yes.

1.3 × 10⁹; 1.1 × 10¹⁴; yes, 2 × 10⁴ people per elephant.

Using scientific notation: (i) ants per human (ii) flocks of 10,000 starlings (iii) leaves if each tree has 10⁴ leaves (iv) sheets of paper (0.001 cm) to reach the Moon.

Solution

(i) $\frac{2 \times 1 0 ^{16}}{8.2 \times 1 0 ^{9}} \approx 2.4 \times 1 0^{6}$ ants per person

(ii) $\frac{1.3 \times 1 0 ^{9}}{1 0 ^{4}} = 1.3 \times 1 0^{5}$ flocks

(iii) $3 \times 1 0^{12} \times 1 0^{4} = 3 \times 1 0^{16}$ leaves

(iv) $\frac{3.844 \times 1 0 ^{10} cm}{1 0 ^{- 3} cm} = 3.844 \times 1 0^{13}$ sheets

(i) ≈ 2.4 × 10⁶ (ii) 1.3 × 10⁵ (iii) 3 × 10¹⁶ (iv) 3.844 × 10¹³

Roxie is 4840 days old: how many hours? Estu is 4070 days old: what is his date of birth? How old would you be after a million seconds?

Solution

$4840 \times 24 =$ 1,16,160 hours (and $6, 969, 600$ minutes).

4070 days before 5 October 2026 is 14 August 2015 (about 11 years 2 months).

$1 0^{6}$ seconds $= 1 0^{6} \div 86400 \approx$ 11.6 days.

1,16,160 hours; born on 14 August 2015 (counting from 5 Oct 2026); about 11.6 days.

Think of events of the order of 10⁵ seconds and 10⁶ seconds. Fill in the blanks for the terror bird (15 million years) and land plants (470 million years).

Solution

$1 0^{5}$ s (about a day): one rotation of the Earth, $8.64 \times 1 0^{4}$ s; a weekend, $1.7 \times 1 0^{5}$ s.
$1 0^{6}$ s (about 11 days): the Moon's orbit around the Earth (27.3 days), $2.4 \times 1 0^{6}$ s; a two-week holiday, $1.2 \times 1 0^{6}$ s.
Terror bird: $1.5 \times 1 0^{7} years \times 3.15 \times 1 0^{7} s \approx 4.7 \times 1 0^{14}$ s.
Land plants: $4.7 \times 1 0^{8} \times 3.15 \times 1 0^{7} \approx 1.5 \times 1 0^{16}$ s.

E.g. a day ≈ 8.64 × 10⁴ s, a lunar month ≈ 2.4 × 10⁶ s; terror bird ≈ 4.7 × 10¹⁴ s; land plants ≈ 1.5 × 10¹⁶ s.

(i) Counting one star a second, how long to count all the stars in the universe? (ii) Drinking a 200 mL glass every 10 seconds, how long to finish all the water on Earth?

Solution

(i) $2 \times 1 0^{23}$ stars → $2 \times 1 0^{23}$ seconds (about $6 \times 1 0^{15}$ years).

(ii) $2 \times 1 0^{25}$ drops ÷ 16 drops per mL $= 1.25 \times 1 0^{24}$ mL $= 6.25 \times 1 0^{21}$ glasses; × 10 s $= 6.25 \times 1 0^{22}$ seconds.

(i) 2 × 10²³ s (ii) 6.25 × 10²² s

What does the first part of the names million, billion, trillion, … denote?

Solution

The Latin prefix counts how many times 1000 is multiplied after the first thousand: mi(llion) = 1, bi = 2, tri = 3, quadri = 4, …, so the n-th name is $1 0^{3 n + 3}$ (billion $= 1 0^{3 \times 2 + 3} = 1 0^{9}$ ).

A Latin number n: million n = 1, billion 2, trillion 3, …, giving 10^(3n + 3).

Figure it Out (page 44)

Find the units digit of 2²²⁴ ÷ 4³².

Solution

$4^{32} = 2^{64}$ , so the value is $2^{224 - 64} = 2^{160}$ . Units digits of powers of 2 repeat 2, 4, 8, 6; 160 is a multiple of 4, so the units digit is 6.

There are 5 bottles in a container, and a new container is brought in every day. How many bottles after 40 days?

Solution

40 new containers of 5 bottles: $40 \times 5 =$ 200 (this is linear growth). (If there was already one container at the start, it is $41 \times 5 = 205$ .)

200 = 2 × 10² bottles (205 if one container was there at the start).

Write as a product of two or more powers in three ways: (i) 64³ (ii) 192⁸ (iii) 32⁻⁵

Solution

(i) $6 4^{3} = 2^{18} = 2^{10} \times 2^{8} = 4^{5} \times 4^{4} = 8^{3} \times 8^{3}$

(ii) $19 2^{8} = (2^{6} \times 3)^{8} = 2^{48} \times 3^{8} = 2^{40} \times 6^{8} = 6 4^{8} \times 3^{8}$

(iii) $3 2^{- 5} = 2^{- 25} = 2^{- 10} \times 2^{- 15} = 4^{- 12} \times 2^{- 1} = 8^{- 8} \times 2^{- 1}$

E.g. 64³ = 2¹⁰ × 2⁸; 192⁸ = 2⁴⁸ × 3⁸; 32⁻⁵ = 2⁻¹⁰ × 2⁻¹⁵

Always, sometimes or never true? (i) Cube numbers are also square numbers. (ii) Fourth powers are square numbers. (iii) The fifth power of a number is divisible by its cube. (iv) The product of two cubes is a cube. (v) q⁴⁶ is both a 4th power and a 6th power (q prime).

Solution

(i) Sometimes: 64 and 729 are both, but 8 and 27 are not squares. (It happens for sixth powers.)

(ii) Always: $n^{4} = (n^{2})^{2}$ .

(iii) Always (for n ≠ 0): $n^{5} = n^{3} \times n^{2}$ .

(iv) Always: $a^{3} \times b^{3} = (ab)^{3}$ .

(v) Never: $q^{46}$ is a 4th power only if 4 divides 46, and a 6th power only if 6 divides 46; neither does.

(i) sometimes (ii) always (iii) always (iv) always (v) never

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

Solution

(i) $1 0^{- 7}$ (ii) $5^{3}$ (iii) $9^{- 11}$ (iv) $1 3^{6}$ (v) $m^{14} n^{21}$

(i) 10⁻⁷ (ii) 5³ (iii) 9⁻¹¹ (iv) 13⁶ (v) m¹⁴n²¹

If 12² = 144, find (i) 1.2² (ii) 0.12² (iii) 0.012² (iv) 120²

Solution

(i) 1.44 (ii) 0.0144 (iii) 0.000144 (iv) 14400

1.44, 0.0144, 0.000144, 14400

Circle the numbers that are equal: 2⁴ × 3⁶, 6⁴ × 3², 6¹⁰, 18² × 6², 6²⁴

Solution

$2^{4} \times 3^{6} = 11664$ ; $6^{4} \times 3^{2} = 2^{4} \times 3^{6} = 11664$ ; $1 8^{2} \times 6^{2} = 10 8^{2} = 11664$ .

So 2⁴ × 3⁶, 6⁴ × 3² and 18² × 6² are equal. (6¹⁰ and 6²⁴ are much larger.)

2⁴ × 3⁶ = 6⁴ × 3² = 18² × 6² (= 11664)

Which is greater? (i) 4³ or 3⁴ (ii) 2⁸ or 8² (iii) 100² or 2¹⁰⁰

Solution

(i) $64 < 81$ : 3⁴ (ii) $256 > 64$ : 2⁸ (iii) $1 0^{4} < 2^{100}$ (which is more than $1 0^{30}$ ): 2¹⁰⁰

(i) 3⁴ (ii) 2⁸ (iii) 2¹⁰⁰

A dairy needs a unique ID for 8.5 billion packets using digits 0–9. How many digits should the code have?

Solution

n digits give $1 0^{n}$ codes. $1 0^{9} < 8.5 \times 1 0^{9} < 1 0^{10}$ , so at least 10 digits are needed.

10 digits

64 is a square and a cube. Are there other such numbers? Describe them in general.

Solution

Yes, infinitely many: exactly the sixth powers $n^{6} = (n^{3})^{2} = (n^{2})^{3}$ : 1, 64, 729, 4096, 15625, …

Yes: the sixth powers n⁶ (1, 64, 729, 4096, …).

An alphanumeric passcode of length 5 (digits and letters A–Z): how many codes?

Solution

Each place has $10 + 26 = 36$ choices: $3 6^{5} =$ 6,04,66,176 codes.

36⁵ = 6,04,66,176

Sheep and goats are each about 10⁹. What is the total? (i) 20⁹ (ii) 10¹¹ (iii) 10¹⁰ (iv) 10¹⁸ (v) 2 × 10⁹ (vi) 10⁹ + 10⁹

Solution

$1 0^{9} + 1 0^{9} = 2 \times 1 0^{9}$ , so both (v) and (vi) are correct. (Adding numbers does not add exponents: $1 0^{18}$ would be their product.)

(v) and (vi): 2 × 10⁹

In scientific notation: (i) clothing if each person has 30 pieces (ii) honeybees in 100 million colonies of 50,000 (iii) bacteria in all humans (38 trillion each) (iv) time spent eating in a lifetime, in seconds

Solution

(i) $8.2 \times 1 0^{9} \times 30 = 2.46 \times 1 0^{11}$

(ii) $1 0^{8} \times 5 \times 1 0^{4} = 5 \times 1 0^{12}$

(iii) $3.8 \times 1 0^{13} \times 8.2 \times 1 0^{9} \approx 3.1 \times 1 0^{23}$

(iv) Assuming 1.5 hours a day for 70 years: $5400 \times 365 \times 70 \approx 1.4 \times 1 0^{8}$ seconds.

(i) 2.46 × 10¹¹ (ii) 5 × 10¹² (iii) ≈ 3.1 × 10²³ (iv) ≈ 1.4 × 10⁸ s (1.5 h a day for 70 years)

What was the date 1 billion seconds ago?

Solution

$1 0^{9}$ s $\approx 11, 574$ days $\approx 31.7$ years. Counting back from 5 October 2026 gives about 27 January 1995. (Count back from your own date in the same way.)

About 31.7 years ago: around 27 January 1995 (from 5 October 2026).

Game: Tremendous in Ten!

Round 2: Roxie wrote 10¹⁰⁰⁰ + 10¹⁰⁰⁰ + 10¹⁰⁰⁰ + 10¹⁰⁰⁰ and Estu wrote 10¹⁰⁰⁰⁰⁰⁰ × 9000. Which is greater?

Solution

Roxie's number is $4 \times 1 0^{1000}$ (a 1001-digit number). Estu's is $9 \times 1 0^{1000003}$ (over a million digits). Estu's number is far greater: the exponent matters much more than the coefficient.

Estu's (9 × 10¹⁰⁰⁰⁰⁰³ against 4 × 10¹⁰⁰⁰).

← Chapter 1: A Square and a Cube Chapter 3: A Story of Numbers →

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