Key Concepts & Formulae at a Glance
- Repeated multiplication is written in exponential notation: \(n^a\) means n multiplied by itself a times, where n is the base and a is the exponent/power.
- Exponential (multiplicative) growth — where a quantity is repeatedly multiplied by a fixed factor — becomes astronomically large far faster than linear (additive) growth, where a fixed amount is repeatedly added.
- Laws of exponents (for counting numbers a, b, and later extended to any integers):
- Scientific notation (standard form) writes any number as \(x \times 10^y\), where \(1 \le x \lt 10\) and y is an integer — the exponent y tells you the scale (number of digits), and the coefficient x tells you the precision.
- For estimation problems, a reliable method is: guess first → describe the relationships between the quantities involved → make reasonable assumptions for anything unknown → compute → compare with your guess.
2.1 Experiencing the Power Play
Take a large sheet of paper and fold it repeatedly. Estu claims a sheet of paper can't be folded more than 7 times; Roxie wonders if a thinner paper (newspaper, tissue) could be folded more. Trying it with different papers is worth doing before reading on. The thickness of the paper doubles with every fold — starting from an assumed initial thickness of 0.001 cm, the thickness after n folds is \(0.001 \times 2^n\) cm.
| Fold | Thickness | Fold | Thickness | Fold | Thickness |
|---|---|---|---|---|---|
| 1 | 0.002 cm | 7 | 0.128 cm | 13 | 8.192 cm |
| 2 | 0.004 cm | 8 | 0.256 cm | 14 | 16.384 cm |
| 3 | 0.008 cm | 9 | 0.512 cm | 15 | 32.768 cm |
| 4 | 0.016 cm | 10 | 1.024 cm | 16 | 65.536 cm |
| 5 | 0.032 cm | 11 | 2.048 cm | 17 | ≈ 131 cm |
| 6 | 0.064 cm | 12 | 4.096 cm |
After 10 folds the thickness is just above 1 cm; after 17 folds it's about 131 cm (a little over 4 feet). Before reading further, it's worth guessing what the thickness might be after 30 folds, or 45 folds — most guesses turn out to be far too small, which is exactly the point of the exercise.
MTFill the table below with the paper's thickness after each fold, from fold 18 up to fold 45.
Using \(\text{thickness} = 0.001 \times 2^n\) cm and switching to metres/kilometres once the numbers get large:
| Fold | Thickness | Fold | Thickness | Fold | Thickness |
|---|---|---|---|---|---|
| 18 | ≈ 262 cm | 25 | ≈ 335.5 m | 32 | ≈ 42.95 km |
| 19 | ≈ 524 cm | 26 | ≈ 671 m | 33 | ≈ 85.9 km |
| 20 | ≈ 10.49 m | 27 | ≈ 1.34 km | 34 | ≈ 171.8 km |
| 21 | ≈ 20.97 m | 28 | ≈ 2.68 km | 35 | ≈ 343.6 km |
| 22 | ≈ 41.94 m | 29 | ≈ 5.37 km | 36 | ≈ 687.2 km |
| 23 | ≈ 83.89 m | 30 | ≈ 10.74 km | 37 | ≈ 1374.4 km |
| 24 | ≈ 167.8 m | 31 | ≈ 21.47 km | 38 | ≈ 2748.8 km |
| Fold | Thickness | Fold | Thickness |
|---|---|---|---|
| 39 | ≈ 5497.6 km | 43 | ≈ 87,961 km |
| 40 | ≈ 10,995 km | 44 | ≈ 175,922 km |
| 41 | ≈ 21,990 km | 45 | ≈ 351,844 km |
| 42 | ≈ 43,980 km |
By fold 26 the thickness (≈670 m) is already close to the height of the Burj Khalifa (830 m); by fold 30 it's about 10.7 km — the cruising altitude of planes; and at fold 46 the thickness works out to more than 7,00,000 km, far enough to reach the Moon (about 3,84,400 km away) — the doubling nature of exponential growth is what makes this possible in so few folds.
Looking at the growth in blocks of 10 folds gives a striking pattern: from fold 0 to 10 the thickness increases 1024 times; from fold 10 to 20 it again increases exactly 1024 times; and the same 1024× multiplier applies from fold 20 to 30, and 30 to 40. This is the signature of multiplicative (exponential) growth — the ratio between values stays constant, unlike additive growth, where it's the difference that stays constant.
2.2 Exponential Notation and Operations
Since the paper's thickness doubles with each fold, folding n times multiplies the initial thickness by \(2 \times 2 \times \dots \times 2\) (n times), written \(2^n\). In general, \(n^a\) denotes n multiplied by itself a times — n is the base and a is the exponent (or power). For example, \(5^4 = 5\times5\times5\times5 = 625\), read as "5 raised to the power 4."
MTWhich expression describes the thickness of a sheet of paper after it is folded 10 times, if the initial thickness is v? What is 0², 0⁵? What is 0ⁿ?
Folding n times multiplies the initial thickness v by \(2^n\), so after 10 folds the thickness is \(2^{10}v\).
Zero multiplied by itself any (positive) number of times is still 0: \(0^2 = 0\times0 = 0\), \(0^5=0\times0\times0\times0\times0=0\), and in general \(0^n=0\) for any counting number n.
Expressing 32400 as a product of its prime factors using the usual division-ladder method gives \(32400 = 2\times2\times2\times2\times5\times5\times3\times3\times3\times3\), which in exponential form is \(32400 = 2^4 \times 5^2 \times 3^4\).
MTWhat is (−1)⁵? Is it positive or negative? What about (−1)⁵⁶? Is (−2)⁴ = 16? Verify.
A negative number raised to an odd power stays negative (an odd number of negative signs multiply to give a negative result), while raised to an even power it becomes positive (the negative signs cancel in pairs).
\((-1)^5 = (-1)\times(-1)\times(-1)\times(-1)\times(-1) = -1\) — negative, since 5 is odd.
\((-1)^{56}\) — since 56 is even, this is positive; in fact \((-1)^{\text{even}}=1\), so \((-1)^{56}=1\).
\((-2)^4 = (-2)\times(-2)\times(-2)\times(-2) = 4\times4=16\) — yes, verified.
Figure it Out — Exponential Notation
Three questions from page 22–23 of the textbook, each with multiple parts.
1Express the following 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
(i) \(6^4\) (ii) \(y^2\) (iii) \(b^4\) (iv) \(5^2 \times 7^3\) (v) \(2^2 \times a^2\) (vi) \(a^3 \times c^4 \times d\)
2Express each of the following as a product of powers of their prime factors in exponential form: (i) 648 (ii) 405 (iii) 540 (iv) 3600
(i) \(648 = 2^3 \times 3^4\) (ii) \(405 = 3^4 \times 5\) (iii) \(540 = 2^2 \times 3^3 \times 5\) (iv) \(3600 = 2^4 \times 3^2 \times 5^2\)
3Write the numerical value of each of the following: (i) 2×10³ (ii) 7²×2³ (iii) 3×4⁴ (iv) (−3)²×(−5)² (v) 3²×10⁴ (vi) (−2)⁵×(−10)⁶
(i) \(2\times1000=2000\) (ii) \(49\times8=392\) (iii) \(3\times256=768\) (iv) \(9\times25=225\) (v) \(9\times10000=90000\) (vi) \((-32)\times1000000=-32000000\)
The Stones that Shine
A king has 3 daughters, each with 3 baskets, each basket holding 3 keys, each key opening 3 rooms, each room having 3 tables, each table holding 3 necklaces, and each necklace set with 3 diamonds — seven levels of "3" in total (daughters → baskets → keys → rooms → tables → necklaces → diamonds).
MTHow many rooms were there altogether? How many diamonds in total — can it be found with just one multiplication? Why can 3⁷ also be written as 3² × 3⁵?
The number of rooms comes from 4 levels of "3" (daughters × baskets × keys × rooms-per-key): \(3\times3\times3\times3=3^4=81\) rooms.
The number of diamonds comes from all 7 levels: \(3^7\). Since \(3^4=81\) is already known, \(3^7 = 3^4 \times 3^3 = 81 \times 27 = 2187\) — reusing the earlier product needs just one more multiplication rather than starting over from scratch.
\(3^7 = 3^2 \times 3^5\) because splitting the 7 copies of the factor 3 into a group of 2 and a group of 5 doesn't change the total product — \((3\times3)\times(3\times3\times3\times3\times3) = 3^7\) either way. This is the product rule: \(n^a \times n^b = n^{a+b}\).
MTWrite the product p⁴ × p⁶ in exponential form. Use the product rule to compute: (i) 2⁹ (ii) 5⁷ (iii) 4⁶
\(p^4 \times p^6 = (p\times p\times p\times p)\times(p\times p\times p\times p\times p\times p) = p^{10}\). In general, \(n^a \times n^b = n^{a+b}\), for counting numbers a and b.
(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) \(4^6\) can be split either as \(4^3\times4^3=64\times64=4096\), or as \(4^2\times4^2\times4^2=16\times16\times16=4096\) — both give the same value, since \(4^3\times4^3\) is the square of \(4^3\) (written \((4^3)^2\)), and \(4^2\times4^2\times4^2\) is the cube of \(4^2\) (written \((4^2)^3\)).
MTIs 2¹⁰ also equal to (2⁵)²? Write it as a product. Write 8⁶, 7¹⁵, 9¹⁴, and 5⁸ as a power of a power, in at least two different ways.
\(2^{10} = (2\times2\times2\times2\times2)\times(2\times2\times2\times2\times2) = 2^5 \times 2^5 = (2^5)^2\) — yes. In general, \((n^a)^b = (n^b)^a = n^{ab}\), the power-of-a-power rule.
\(8^6 = (8^2)^3 = (8^3)^2\), and since \(8=2^3\), also \(8^6=(2^3)^6=2^{18}\).
\(7^{15} = (7^3)^5 = (7^5)^3\).
\(9^{14} = (9^2)^7\), and since \(9=3^2\), also \(9^{14}=(3^2)^{14}=3^{28}\).
\(5^8 = (5^2)^4 = (5^4)^2\).
Magical Pond
A lotus pond doubles the number of lotuses every day, and is completely covered on the 30th day.
MTOn which day was the pond half full? Write the number of lotuses (in exponential form) when the pond was (i) fully covered, and (ii) half covered.
Since the lotus count doubles every day, whatever covers the pond on any day is exactly half of what covers it the next day. So the pond must have been half covered the day before it was fully covered — the 29th day.
If day 1 has \(2^0=1\) lotus's worth of coverage as a starting reference (or more directly, tracking powers of 2 relative to day 30 being "full"), the fully covered state corresponds to \(2^{30}\) and the half-covered state to \(2^{29}\), since \(2^{30}\div2=2^{29}\).
MTA doubling pond and a tripling pond both start empty. A lotus is placed in the doubling pond; after 4 days its lotuses are moved to the tripling pond, which then grows for 4 more days. How many lotuses result — and would it change if the ponds were used in the other order? Can the product be written as a single power mⁿ? Use this to compute 2⁵ × 5⁵, and simplify 10⁴ ÷ 5⁴.
After 4 days in the doubling pond: \(1\times2\times2\times2\times2=2^4\) lotuses. Moving these into the tripling pond for 4 more days multiplies this count by \(3^4\): total = \(2^4 \times 3^4\).
Swapping the order (tripling first, then doubling) gives \(3^4 \times 2^4\) — since multiplication doesn't care about order, this is the exact same value, so the final count would not change.
Regrouping the factors, \(2^4 \times 3^4 = (2\times3)\times(2\times3)\times(2\times3)\times(2\times3) = (2\times3)^4 = 6^4\) — so yes, it can be written as a single power. In general, \(m^a \times n^a = (mn)^a\).
Using this rule, \(2^5 \times 5^5 = (2\times5)^5 = 10^5 = 100000\).
For division, the analogous rule is \(\frac{m^a}{n^a}=\left(\frac{m}{n}\right)^a\), so \(\frac{10^4}{5^4} = \left(\frac{10}{5}\right)^4 = 2^4 = 16\).
How Many Combinations
MTEstu has 4 dresses and 3 caps — how many different combinations are possible? Roxie has 7 dresses, 2 hats, and 3 pairs of shoes — how many different ways can she dress up?
For Estu, each of the 3 caps can be paired with any of the 4 dresses, giving \(4\times3=12\) combinations (equivalently, each of the 4 dresses can be paired with any of the 3 caps: \(3\times4=12\)).
For Roxie, the same reasoning extends across all three categories: \(7\times2\times3=42\) different outfit combinations.
To count 5-digit lock passwords, it helps to start smaller: a 2-digit lock has 10 choices for each digit, giving \(10\times10=100\) passwords; a 3-digit lock extends this to \(100\times10=1000\) passwords. Following the same logic, a 5-digit lock (digits 0–9) has \(10\times10\times10\times10\times10=10^5=1,00,000\) possible passwords — Estu and Roxie would have had to check all of them in the worst case.
MTHow many passwords are possible with a 6-slot lock using the letters A to Z?
Each of the 6 slots has 26 possible letters, so the total is \(26\times26\times26\times26\times26\times26=26^6\).
\(26^6 = 308{,}915{,}776\) — over 30 crore possible codes, far more than the 1,00,000 of a 5-digit numeric lock.
Try This: The same combinatorial idea shows up in Indian PIN codes (6 digits), mobile numbers (10 digits), and vehicle registration numbers (a mix of letters and digits) — each extra digit or letter slot multiplies the number of possible codes, which is exactly why these systems can comfortably assign a unique code to every address, phone connection, or vehicle in the country without running out.
2.3 The Other Side of Powers
Halving a line of length 16 units (\(=2^4\)) once gives \(2^4\div2=2^3=8\) units; halving it again gives \(2^4\div2^2=2^2=4\) units; and halving it a third time gives \(2^4\div2^3=2^1=2\) units. This gives the quotient rule: \(n^a \div n^b = n^{a-b}\), for \(n\ne0\) and \(a \gt b\).
MTWhat is 2¹⁰⁰ ÷ 2²⁵ in powers of 2? Why can't n be 0 in the quotient rule? We had required a and b to be counting numbers — can they be any integers, and will the rule still hold?
\(2^{100} \div 2^{25} = 2^{100-25} = 2^{75}\).
n cannot be 0 because the rule comes from cancelling common factors in a fraction like \(\frac{n^a}{n^b}\) — if \(n=0\), this becomes \(\frac{0}{0}\), which is undefined.
Yes — the rule \(n^a\div n^b=n^{a-b}\) continues to hold even when a and b are any integers (not just counting numbers), as long as \(n\ne0\); this is exactly what allows negative and zero exponents to be defined consistently.
To make sense of a zero exponent, the quotient rule is applied to \(2^4 \div 2^4\): this must equal \(2^{4-4}=2^0\), but it's also just a number divided by itself, which is 1. So \(2^0=1\), and in general \(x^0=1\) for any \(x\ne0\).
To make sense of negative exponents, halving a line of length \(2^4\) units 5 times gives \(2^4\div2^5=2^{-1}\) units, but also equals \(\frac{1}{2}\) unit directly — so \(2^{-1}=\frac{1}{2}\). Halving it 10 times gives \(2^4\div2^{10}=2^{-6}\), which also equals \(\frac{1}{2^6}=\frac{1}{64}\) directly — so \(2^{-6}=\frac{1}{64}\). In general, \(n^{-a}=\frac{1}{n^a}\) and \(n^a=\frac{1}{n^{-a}}\), for \(n\ne0\).
MTWrite equivalent forms of: (i) 2⁻⁴ (ii) 10⁻⁵ (iii) (−7)⁻² (iv) (−5)⁻³ (v) 10⁻¹⁰⁰
Each follows the rule \(n^{-a}=\frac{1}{n^a}\):
(i) \(2^{-4}=\dfrac{1}{2^4}\) (ii) \(10^{-5}=\dfrac{1}{10^5}\) (iii) \((-7)^{-2}=\dfrac{1}{(-7)^2}\) (iv) \((-5)^{-3}=\dfrac{1}{(-5)^3}\) (v) \(10^{-100}=\dfrac{1}{10^{100}}\)
MTSimplify and write in exponential form: (i) 2⁻⁴ × 2⁷ (ii) 3² × 3⁻⁵ × 3⁶ (iii) p³ × p⁻¹⁰ (iv) 2⁴ × (−4)⁻² (v) 8ᵖ × 8ᑫ
Applying the product rule \(n^a\times n^b=n^{a+b}\) throughout (adding exponents, keeping the same base):
(i) \(2^{-4}\times2^7=2^{-4+7}=2^3\)
(ii) \(3^2\times3^{-5}\times3^6=3^{2-5+6}=3^3\)
(iii) \(p^3\times p^{-10}=p^{3-10}=p^{-7}\)
(iv) \(2^4\times(-4)^{-2}=16\times\dfrac{1}{16}=1\) (since \((-4)^2=16\))
(v) \(8^p\times8^q=8^{p+q}\)
Power Lines
Arranging the powers of 4 along a vertical number line (from \(4^{-2}=\frac{1}{16}\) up to \(4^8=65536\)) makes the relationships between powers visually clear: moving up one step multiplies by 4, moving down one step divides by 4.
MTCan we say that 16384 (4⁷) is 16 (4²) times larger than 1024 (4⁵)? How many times larger than 4⁻² is 4²?
Yes — since \(4^7\div4^5=4^{7-5}=4^2=16\), 16384 is indeed 16 times 1024.
\(4^2\div4^{-2}=4^{2-(-2)}=4^4=256\) times larger.
MTUse the power line for 7 (7⁻⁴ = 1/2401 up to 7⁷ = 8,23,543) to answer: 2401×49, 49³, 343×2401, 16807÷49, 7÷343, 16807÷8,23,543, 1,17,649×(1/343), and (1/343)×(1/343) — each as a power of 7.
Since \(2401=7^4\) and \(49=7^2\): \(2401\times49=7^4\times7^2=7^6\).
\(49^3=(7^2)^3=7^6\).
Since \(343=7^3\): \(343\times2401=7^3\times7^4=7^7\).
Since \(16807=7^5\): \(16807\div49=7^5\div7^2=7^3\).
\(7\div343=7^1\div7^3=7^{-2}\).
Since \(823543=7^7\): \(16807\div823543=7^5\div7^7=7^{-2}\).
Since \(117649=7^6\): \(117649\times\frac{1}{343}=7^6\times7^{-3}=7^3\).
\(\frac{1}{343}\times\frac{1}{343}=7^{-3}\times7^{-3}=7^{-6}\).
2.4 Powers of 10 & Scientific Notation
Numbers can be written in expanded form using powers of 10 — e.g. \(47561 = (4\times10^4)+(7\times10^3)+(5\times10^2)+(6\times10^1)+(1\times10^0)\), and this extends naturally to decimals: \(561.903 = (5\times10^2)+(6\times10^1)+(1\times10^0)+(9\times10^{-1})+(0\times10^{-2})+(3\times10^{-3})\).
MTWrite these numbers in expanded form using powers of 10: (i) 172 (ii) 5642 (iii) 6374
(i) \(172 = (1\times10^2)+(7\times10^1)+(2\times10^0)\)
(ii) \(5642 = (5\times10^3)+(6\times10^2)+(4\times10^1)+(2\times10^0)\)
(iii) \(6374 = (6\times10^3)+(3\times10^2)+(7\times10^1)+(4\times10^0)\)
Very large numbers are hard to read correctly digit by digit — miscounting a zero can turn ₹5,000 into ₹50,000. Writing a number as the product of a value between 1 and 10 and a power of 10 (its scientific notation, e.g. \(5900=5.9\times10^3\)) fixes this, because the exponent alone tells you the scale.
MTWrite these large-number facts in scientific notation: (i) the Sun is 30,00,00,00,00,00,00,00,00,000 m from the centre of the Milky Way (ii) the number of stars in our galaxy is 1,00,00,00,00,000 (iii) the mass of the Earth is 59,76,00,00,00,00,00,00,00,00,00,000 kg
(i) The distance has 21 digits total, leading with "3" — this is \(3\times10^{20}\) m.
(ii) The star count has 12 digits, leading with "1" — this is \(1\times10^{11}\), i.e. about 100 billion stars, matching the commonly cited estimate for the Milky Way.
(iii) The mass has 25 digits, leading with "5976" — this is \(5.976\times10^{24}\) kg, the standard scientific value for Earth's mass.
In scientific notation \(x\times10^y\), the exponent y matters far more than the coefficient x — changing Mumbai's population from \(2\times10^7\) to \(3\times10^7\) is a 50% change, but changing the exponent from \(10^7\) to \(10^8\) is a 10× change. The number of digits kept in x also signals precision: writing Kohima's population as \(1.42\times10^5\) claims more precision than \(1.4\times10^5\).
MTThe Sun–Saturn distance is 1.4335 × 10¹² m, the Saturn–Uranus distance is 1.439 × 10¹² m, and the Sun–Earth distance is 1.496 × 10¹¹ m. Which of the three is smallest? Mark Earth's position on a number line running from the Sun to Saturn.
Comparing the exponents first: the Sun–Earth distance has exponent 11, while the other two both have exponent 12 — since a smaller exponent means a smaller order of magnitude (all coefficients being between 1 and 10), the Sun–Earth distance is the smallest of the three by a wide margin (roughly one-tenth the size).
Earth sits at about 10.4% of the way from Sun to Saturn on this line, since (1.496×10¹¹) ÷ (1.4335×10¹²) ≈ 0.104.
MTExpress the following numbers in standard form: (i) 59,853 (ii) 65,950 (iii) 34,30,000 (iv) 70,04,00,00,000
(i) \(59853=5.9853\times10^4\) (ii) \(65950=6.595\times10^4\) (iii) \(3430000=3.43\times10^6\) (iv) \(700400000000=7.004\times10^{10}\)
2.5 Did You Ever Wonder?
This section works through a series of estimation problems using a consistent method: guess first, describe the relationships between the quantities involved, make reasonable assumptions where exact figures aren't known, then compute and compare against the guess. As the textbook itself notes, different reasonable assumptions can lead to somewhat different final numbers — that's expected, and the modelling process matters more than matching one "correct" figure exactly.
MTNanjundappa wants to donate jaggery equal to Roxie's weight and wheat equal to Estu's weight. What would each donation be worth in rupees?
Worth of jaggery = Roxie's weight (kg) × cost per kg of jaggery. Worth of wheat = Estu's weight (kg) × cost per kg of wheat.
Assuming Roxie (age 13) weighs about 45 kg and jaggery costs about ₹70/kg: worth = \(45\times70=3150\), i.e. ₹3150.
Assuming Estu (age 11) weighs about 50 kg and wheat costs about ₹50/kg: worth = \(50\times50=2500\), i.e. ₹2500.
MTInstead of jaggery, how many 1-rupee coins would it take to equal Roxie's weight? Would the count be in hundreds, thousands, lakhs, or crores — guess first, then estimate.
A real 1-rupee coin weighs about 6 grams. Assuming Roxie weighs 45 kg = 45,000 g:
Number of coins \(= \dfrac{45000}{6} = 7500\) coins.
This falls in the thousands — a useful check on whether an initial "guess" of lakhs or crores was reasonable.
MTSomeone did a padayatra of about 400 km, arriving early one morning — how long ago would they have started? How many times could a person circumnavigate the Earth in a lifetime if they walked non-stop (Earth's circumference ≈ 40,000 km)?
Padayatra: assuming a walking pace of about 20 km per day (a common sustainable rate for multi-day pilgrimage walking), covering 400 km would take \(400\div20=20\) days — so the journey would have started roughly 20 days before they arrived.
Circumnavigating the Earth: assuming a person can walk about 25 km per day and does so for, say, 50 years out of a lifetime (about 18,250 days), the total distance covered would be \(25\times18250=456{,}250\) km. Dividing by Earth's circumference: \(456250\div40000\approx11.4\) times.
For the "ladder to the Moon" thought experiment, assuming a rung spacing of 20 cm and the Earth–Moon distance of 3,84,400 km, the number of rungs needed is \(3{,}84{,}400 \text{ km} \div 20\text{ cm} = 1{,}92{,}20{,}00{,}000\) rungs — about 192 crore, or 1.92 billion. This is linear growth (each rung adds a fixed 20 cm), and it takes nearly 2 billion rungs to cover the same distance that just 46 doublings of a folded paper (exponential growth) can reach — a vivid illustration of how much faster multiplicative growth outpaces additive growth.
MTWith a global human population of about 8×10⁹ and about 4×10⁵ African elephants, can we say there are nearly 20,000 people for every African elephant?
People per elephant \(= \dfrac{8\times10^9}{4\times10^5} = 2\times10^4 = 20{,}000\).
MTCalculate and write in scientific notation: (i) how many ants are there for every human? (ii) how many flocks of 10,000 starlings could there be worldwide? (iii) if each tree has about 10⁴ leaves, how many leaves are there on all trees globally? (iv) how many sheets of 0.001 cm paper stacked up would reach the Moon?
Using the estimates given earlier in the chapter — ants ≈ \(2\times10^{16}\), humans ≈ \(8\times10^9\), starlings ≈ \(1.3\times10^9\), trees ≈ \(3\times10^{12}\), and the Earth–Moon distance ≈ \(3.844\times10^{10}\) cm:
(i) Ants per human \(=\dfrac{2\times10^{16}}{8\times10^9}=2.5\times10^6\).
(ii) Flocks of starlings \(=\dfrac{1.3\times10^9}{10^4}=1.3\times10^5\).
(iii) Total leaves \(=3\times10^{12}\times10^4=3\times10^{16}\).
(iv) Sheets needed \(=\dfrac{3.844\times10^{10}\text{ cm}}{0.001\text{ cm}}=3.844\times10^{13}\) sheets.
MTRoxie says "I completed 13 years a few weeks ago," then "I'm 4840 days old today," then "I'm ___ hours old." Estu says he is 4070 days old — what is his date of birth? If you have lived for a million seconds, how old would you be?
Converting Roxie's 4840 days into hours: \(4840\times24=1{,}16{,}160\) hours.
Converting Estu's 4070 days into years: \(4070\div365.25\approx11.14\) years — consistent with him being around 11 years old. Counting back 4070 days from today's date gives his date of birth (the exact date depends on today's date when you solve it — for example, counting back from 19 July 2026 lands on 28 May 2015).
A million seconds, converted to days: \(1{,}000{,}000\div86400\approx11.57\) days — so a "million-second-old" person is not even two weeks old!
Working through a table of "orders of magnitude" of time in seconds reveals the pace of exponential growth strikingly: \(10^6\) seconds is less than a fortnight (about 11.6 days), while \(10^9\) seconds is a whopping 31.7 years — roughly half a human lifespan — and \(10^{18}\) seconds ago, according to modern physics, the universe itself did not yet exist.
MTThink of events or phenomena whose duration is of the order of (i) 10⁵ seconds (≈1.16 days) and (ii) 10⁶ seconds (≈11.57 days). A fossil of the terror bird Kelenken guillermoi is dated to 15 million years ago — write this in seconds. Land plants started 470 million years ago — write this in seconds too.
(i) An example around \(10^5\) seconds: roughly how long the International Space Station takes to complete about 16 orbits of the Earth (each orbit takes around 90 minutes, so 16 orbits ≈ 24 hours ≈ \(8.6\times10^4\) s, of order \(10^5\)).
(ii) An example around \(10^6\) seconds: roughly the length of a long school study break of about 12 days.
15 million years in seconds: \(15\times10^6 \times 3.156\times10^7 \approx 4.73\times10^{14}\) seconds (using 1 year ≈ \(3.156\times10^7\) s).
470 million years in seconds: \(470\times10^6\times3.156\times10^7\approx1.48\times10^{16}\) seconds.
Try This
If one star is counted every second, how long would it take to count all the stars in the universe (≈2×10²³)? Time = \(2\times10^{23}\) seconds directly (since one star is counted per second).
If a glass of water (200 mL) is drunk every 10 seconds, how long to finish all the water on Earth (≈2×10²⁵ drops, 16 drops/mL)? A 200 mL glass holds \(200\times16=3200\) drops. There are \(\dfrac{2\times10^{25}}{3200}\approx6.25\times10^{21}\) glasses' worth of water; drinking one glass every 10 seconds takes \(6.25\times10^{21}\times10=6.25\times10^{22}\) seconds.
A Pinch of History
The Lalitavistara (a Buddhist treatise, 1st century BCE) names powers of ten up to \(10^{53}\) — a hundred koṭis is an ayuta (\(10^9\)), a hundred ayutas is a niyuta (\(10^{11}\)), and so on up to a tallakshana (\(10^{53}\)). Mahāvīrācārya's Gaṇita-sāra-saṅgraha lists 24 terms (up to \(10^{23}\)); the Jaina Amalasiddhi goes up to \(10^{96}\); and a Pāli grammar text of Kāccāyana lists names up to \(10^{140}\).
The modern Indian naming system builds each new unit as a hundred times the previous one (lakh → crore → arab → kharab → neel → padma → shankh → maha shankh), while the international system builds each new unit as a thousand times the previous one (million → billion → trillion → quadrillion → …).
MTLooking at the international names — million, billion, trillion, quadrillion, quintillion, sextillion, septillion, octillion, nonillion, decillion — what does the first part of each name denote?
The Latin-derived prefixes (bi-, tri-, quadri-, quinti-, sexti-, septi-, octi-, noni-, deci-) directly count how many groups of three zeros (i.e. how many extra factors of 1000, or equivalently \(10^3\)) have been multiplied onto the base unit of a million (\(10^6\)). So "bi" (2) in billion means \(10^6\times10^3\times... \), more precisely each prefix number k corresponds to \(10^{3(k+1)}\): billion = \(10^{3\times2+3}=10^9\)? — more simply, each name's number tells you how many thousands have been multiplied together beyond the first (a million is "thousand thousand"; a billion is "thousand million," i.e. one more factor of a thousand; and so on).
The number \(10^{100}\) is called a googol, and \(10^{\text{googol}}\) is a googolplex — numbers far beyond anything physically countable (the estimated number of atoms in the entire observable universe is only around \(10^{78}\) to \(10^{82}\)).
Figure it Out — Chapter Review
Fourteen questions from page 44–45 of the textbook.
1Find the units digit in the value of 2²²⁴ ÷ 4³². [Hint: 4 = 2²]
Since \(4=2^2\), \(4^{32}=(2^2)^{32}=2^{64}\), so \(2^{224}\div4^{32}=2^{224}\div2^{64}=2^{224-64}=2^{160}\).
The units digit of powers of 2 cycles every 4 exponents: 2, 4, 8, 6, 2, 4, 8, 6, … (for exponents 1, 2, 3, 4, 5, …). Since \(160\div4\) leaves remainder 0, it lines up with exponent 4 in the cycle (units digit 6).
2There are 5 bottles in a container. Every day, a new container is brought in. How many bottles would there be after 40 days?
Each day adds a fresh container of 5 bottles (this is linear, additive growth — not multiplicative), so after 40 days: \(5\times40=200\) bottles.
3Write the given number as the product of two or more powers, in three different ways (powers can be any integers): (i) 64³ (ii) 192⁸ (iii) 32⁻⁵
(i) \(64^3\): since \(64=2^6\), this is \(2^{18}\), which splits as \(2^{10}\times2^8\), or \(4^5\times4^4\), or \(8^3\times8^3\).
(ii) \(192^8\): since \(192=2^6\times3\), this is \(2^{48}\times3^8\), which can also be written as \(2^{40}\times2^8\times3^8\), or as \(2^{40}\times6^8\) (grouping one factor of 2 with each 3).
(iii) \(32^{-5}\): since \(32=2^5\), this is \(2^{-25}\), which splits as \(2^{-10}\times2^{-15}\), or \(2^{-5}\times2^{-20}\), or \(4^{-12}\times2^{-1}\).
4Examine each statement and decide if it's 'Always True', 'Only Sometimes True', or 'Never True': (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 prime).
(i) Only Sometimes True. A number is both a cube and a square exactly when it's a 6th power (\(n^6=(n^2)^3=(n^3)^2\)) — e.g. 64 works, but 27 (a cube, \(3^3\)) is not a square.
(ii) Always True. Any fourth power can be written as \(n^4=(n^2)^2\), which is a square by definition.
(iii) Always True. \(\dfrac{n^5}{n^3}=n^2\), a whole number, so \(n^5\) is always divisible by \(n^3\).
(iv) Always True. \((n_1)^3\times(n_2)^3=(n_1\times n_2)^3\), which is itself a perfect cube.
(v) Never True. For \(q^{46}\) to be a 4th power, 46 would need to be divisible by 4; for it to be a 6th power, 46 would need to be divisible by 6. Since 46 is divisible by neither 4 nor 6, it's never true.
5Simplify and write in exponential form: (i) 10⁻² × 10⁻⁵ (ii) 5⁷ ÷ 5⁴ (iii) 9⁻⁷ ÷ 9⁴ (iv) (13⁻²)⁻³ (v) m⁵n¹²(mn)⁹
(i) \(10^{-2}\times10^{-5}=10^{-7}\)
(ii) \(5^7\div5^4=5^3\)
(iii) \(9^{-7}\div9^4=9^{-11}\)
(iv) \((13^{-2})^{-3}=13^{(-2)\times(-3)}=13^6\)
(v) \(m^5n^{12}(mn)^9=m^5n^{12}\times m^9n^9=m^{14}n^{21}\)
6If 12² = 144, what is (i) (1.2)² (ii) (0.12)² (iii) (0.012)² (iv) 120²?
Each shift of the decimal point in the base shifts the decimal point in the square by twice as many places:
(i) \((1.2)^2=1.44\) (ii) \((0.12)^2=0.0144\) (iii) \((0.012)^2=0.000144\) (iv) \(120^2=14400\)
7Circle the numbers that are the same: 2⁴×3⁶, 6⁴×3², 6¹⁰, 18²×6², 6²⁴
\(2^4\times3^6=16\times729=11664\).
\(6^4\times3^2=1296\times9=11664\).
\(18^2\times6^2=324\times36=11664\).
These three all equal 11664. \(6^{10}\) and \(6^{24}\) are vastly larger and different from each other and from the first three.
8Identify the greater number in each: (i) 4³ or 3⁴ (ii) 2⁸ or 8² (iii) 100² or 2¹⁰⁰
(i) \(4^3=64\), \(3^4=81\) — \(3^4\) is greater.
(ii) \(2^8=256\), \(8^2=64\) — \(2^8\) is greater.
(iii) \(100^2=10000\), while \(2^{100}\) is an astronomically larger number (over 30 digits) — \(2^{100}\) is far greater.
9A dairy plans to produce 8.5 billion packets of milk in a year and wants a unique ID code for each, using digits 0–9. How many digits should the code have?
Total packets = 8,500,000,000. With n-digit codes using 10 possible digits each, there are \(10^n\) possible codes, so we need \(10^n \ge 8{,}500{,}000{,}000\).
Since \(10^9=1{,}000{,}000{,}000 \lt 8{,}500{,}000{,}000\) but \(10^{10}=10{,}000{,}000{,}000 \gt 8{,}500{,}000{,}000\), the code needs 10 digits.
1064 is both a square (8²) and a cube (4³). Are there other numbers that are both squares and cubes? Is there a general way to describe them?
Yes — infinitely many. A number that is simultaneously a perfect square and a perfect cube must be a perfect 6th power (since \(n^6=(n^2)^3=(n^3)^2\)), because the LCM of 2 and 3 is 6.
Examples: \(1^6=1\), \(2^6=64\), \(3^6=729\), \(4^6=4096\), and so on.
11A digital locker has an alphanumeric passcode of length 5 (digits and letters both allowed). How many such codes are possible?
There are 26 letters + 10 digits = 36 possible characters for each of the 5 slots, so the total number of codes is \(36^5\).
\(36^5 = 60{,}466{,}176\).
12The worldwide sheep population (2024) is about 10⁹, and the goat population is also about 10⁹. What is the total population of sheep and goats? (i) 20⁹ (ii) 10¹¹ (iii) 10¹⁰ (iv) 10¹⁸ (v) 2×10⁹ (vi) 10⁹+10⁹
Adding the two populations directly: \(10^9+10^9\), which is option (vi) exactly as stated, and simplifies to \(2\times10^9\) — option (v). These two options are mathematically identical and both correct; the other options are wrong by orders of magnitude (20⁹ is astronomically too large, and \(10^{10}\), \(10^{11}\), \(10^{18}\) are each the result of mistakenly multiplying rather than adding).
13Calculate and write in scientific notation: (i) if each person in the world had 30 pieces of clothing, the total pieces of clothing (ii) with about 100 million bee colonies and 50,000 bees each, the total number of honeybees (iii) with about 38 trillion bacterial cells per human, the total bacterial population residing in all humans (iv) total time spent eating in a lifetime, in seconds
(i) World population ≈ \(8.2\times10^9\); total clothing \(=8.2\times10^9\times30=2.46\times10^{11}\) pieces.
(ii) Bee colonies = \(10^8\); honeybees \(=50{,}000\times10^8=5.0\times10^{12}\).
(iii) Bacterial cells per human \(=38\times10^{12}\); total \(=38\times10^{12}\times8.2\times10^9=3.116\times10^{23}\).
(iv) Assuming a 70-year lifetime and about 1 hour (3600 s) of eating per day: total \(=3600\times365\times70=9.198\times10^7\) seconds (roughly 3 years' worth of continuous eating time, spread across a lifetime).
14What was the date 1 arab / 1 billion seconds ago?
1 billion seconds \(= 1{,}000{,}000{,}000 \div 86400 \approx 11{,}574\) days \(\approx 31.7\) years.
Counting back roughly 31.7 years from today's date gives the answer — for instance, counting back from 19 July 2026 lands on approximately 10 November 1994.
Puzzle Time — Tremendous in Ten!
The Challenge
Find a partner. In 10 seconds, each person writes a number or expression (using only digits 0–9 and arithmetic operations) — whoever's expression gives the larger value wins the round.
Round 1: Roxie wrote 10000000000000; Estu wrote 999999 × 999999. Roxie's number is \(10^{13}\), while Estu's is less than \((10^6)^2=10^{12}\) (since 999999 is just under a million) — so Roxie wins, because her exponent (13) beats Estu's (12), even before the coefficients matter.
Round 2: Roxie wrote \(10^{1000}+10^{1000}+10^{1000}+10^{1000}\); Estu wrote \(10^{1000000}\times9000\). Roxie's expression is just \(4\times10^{1000}\) — a modest multiple of \(10^{1000}\). Estu's expression involves \(10^{1000000}\), an exponent a thousand times larger; no ordinary multiplier like 9000 can make up for such a vast gap in the exponent itself. So Estu's expression is overwhelmingly larger.
This captures the core intuition of the whole chapter: once two numbers are both written as powers of 10 (or any common base), the exponent almost always decides who wins — the coefficient in front only matters when the exponents are equal or very close.
Variants worth trying: (i) exponents banned, only addition allowed; (ii) exponents banned, addition and multiplication allowed; (iii) exponents allowed, only addition allowed; (iv) exponents allowed, any operation allowed. Each variant changes which strategies actually win.
Frequently Asked Questions
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