Power Play
Every in-text question, both “Figure it Out” exercises, and the closing game — worked step by step, with the chapter’s diagrams rebuilt and the exponent laws made plain.
Folding a Sheet of Paper
How a thickness that just “doubles” can reach the Moon.
How many times can you actually fold a sheet of paper over and over?
In practice, an ordinary sheet can be folded only about 6 to 8 times. The catch is geometric: every fold doubles the thickness while halving the width, so the stack quickly becomes too thick and too small to fold again. Thinner paper (tissue, newsprint) can manage a few more folds.
The often-quoted “7 times” limit was beaten in 2002 by Britney Gallivan, who folded a very long, thin sheet 12 times. The rest of this chapter explores what would happen if we could keep folding forever.
AnswerUsually 6–8 folds for normal paper; the record is 12. The math below assumes ideal, unlimited folding.
If the sheet is 0.001 cm thick, what is its thickness after 30 folds? After 45? Complete the doubling table.
The thickness doubles each fold, so after \(n\) folds it is
A few milestones from the completed table:
| Fold | Thickness | Real-world comparison |
|---|---|---|
| 10 | 1.024 cm | just over a fingertip |
| 17 | ≈ 131 cm | taller than a child |
| 26 | ≈ 670 m | approaching Burj Khalifa (830 m) |
| 30 | ≈ 10.7 km | cruising height of planes; Mariana Trench is 11 km deep |
| 40 | ≈ 10,995 km | nearly the Earth’s diameter |
| 45 | ≈ 3,51,844 km | almost the distance to the Moon |
| 46 | ≈ 7,03,687 km | past the Moon (3,84,400 km away)! |
AnswerAfter 30 folds ≈ 10.7 km; after 45 ≈ 3.5 lakh km; after 46 the stack reaches beyond the Moon.
By how much does the thickness increase after 2 folds? After 3 folds? After 10 folds?
Each fold multiplies by 2, so over several folds the multipliers themselves multiply:
That is why the growth table shows the thickness multiplying by 1024 over every block of 10 folds.
Answer4 times · 8 times · 1024 times.
Exponential Notation & Operations
Reading nᵃ, and the laws that make powers easy.
Which expression gives the thickness after 10 folds, if the starting thickness is the letter-number \(v\)?(i) 10v (ii) 10 + v (iii) 2×10×v (iv) 2¹⁰ (v) 2¹⁰v (vi) 10²v
Ten folds multiply the start by 2 ten times, i.e. by \(2^{10}\). Starting from \(v\), the thickness is \(v \times 2^{10} = 2^{10}v\).
Answer(v) \(2^{10}v\).
Express 32400 as a product of its prime factors in exponential form.
Dividing repeatedly by primes: \(32400 = 2\times2\times2\times2 \times 3\times3\times3\times3 \times 5\times5\).
Answer\(2^4 \times 3^4 \times 5^2\).
What is \((-1)^5\)? Is it positive or negative? What about \((-1)^{56}\)? And is \((-2)^4 = 16\)?
A negative base raised to an odd power stays negative; to an even power it turns positive (the minus signs pair off and cancel).
And \((-2)^4 = (-2)(-2)(-2)(-2) = 16\) — yes.
Answer\((-1)^5 = -1\), \((-1)^{56} = +1\), and \((-2)^4 = 16\). ✓
What is \(0^2\), \(0^5\), and \(0^n\)?
Zero multiplied by itself any number of times is still zero.
AnswerAll are 0 (for \(n \ge 1\)). Note: \(0^0\) is left undefined.
Why can \(3^7\) also be written as \(3^2 \times 3^5\)? Express \(p^4 \times p^6\) in exponential form.
Multiplying powers of the same base simply piles up the factors, so the exponents add:
The same logic gives \(p^4 \times p^6 = p^{4+6} = p^{10}\). In general, \(\;n^a \times n^b = n^{a+b}\).
AnswerBecause exponents add: \(3^2 \times 3^5 = 3^7\), and \(p^4 \times p^6 = p^{10}\).
Use \(n^a \times n^b = n^{a+b}\) to compute \(2^9\), \(5^7\) and \(4^6\). Then write \(8^6, 7^{15}, 9^{14}, 5^8\) as a power of a power in two ways. Is \(2^{10} = (2^5)^2\)?
Splitting to compute:
Power of a power uses \((n^a)^b = n^{a\times b}\), so an exponent can be split into factors:
And yes, \(2^{10} = (2^5)\times(2^5) = (2^5)^2\).
Answer\(2^9=512,\ 5^7=78125,\ 4^6=4096\); the power-of-power forms above; and \(2^{10}=(2^5)^2\). ✓
Figure it Out — Set A
Pages 22–23 · writing powers and finding values.
Express the following in exponential form.
| Expression | Exponential form |
|---|---|
| (i) 6×6×6×6 | 6⁴ |
| (ii) y×y | y² |
| (iii) b×b×b×b | b⁴ |
| (iv) 5×5×7×7×7 | 5² × 7³ |
| (v) 2×2×a×a | 2² × a² |
| (vi) a×a×a×c×c×c×c×d | a³ × c⁴ × d |
AnswerGroup equal factors and count them — see the table.
Express each as a product of powers of its prime factors.(i) 648 (ii) 405 (iii) 540 (iv) 3600
Answer648 = 2³×3⁴ · 405 = 3⁴×5 · 540 = 2²×3³×5 · 3600 = 2⁴×3²×5².
Write the numerical value of each.(i) 2×10³ (ii) 7²×2³ (iii) 3×4⁴ (iv) (−3)²×(−5)² (v) 3²×10⁴ (vi) (−2)⁵×(−10)⁶
In (vi), \((-2)^5 = -32\) (odd power, negative) and \((-10)^6 = +10^6\) (even power, positive), so the product is negative: \(-32000000\).
Answer2000 · 392 · 768 · 225 · 90000 · −32000000.
Stones, Ponds & Counting
Multiplication that branches: \(3^7\) diamonds, doubling lotuses, password locks.
3 daughters, each with 3 baskets, each holding 3 keys, each opening 3 rooms, each with 3 tables, each with 3 necklaces, each with 3 diamonds. How many rooms in all? How many diamonds?
At every stage the count is multiplied by 3, so the totals are powers of 3.
Diamonds need three more factors of 3 (tables, necklaces, diamonds): \(3^4 \times 3^3 = 3^7\). We can reuse work: \(3^7 = 3^4 \times 3^3 = 81 \times 27 = 2187\).
Answer81 rooms (= 3⁴) and 2187 diamonds (= 3⁷).
Lotuses double daily and cover the pond on day 30. On which day was it half covered? Write the lotus count (exponential) when (i) fully and (ii) half covered.
If the pond doubles each day and is full on day 30, then one day earlier it was exactly half as covered.
AnswerHalf full on day 29; fully covered = \(2^{30}\), half covered = \(2^{29}\).
A lotus grows 4 days in a doubling pond, then moves to a tripling pond for 4 more days. How many lotuses at the end? Does the order matter? Compute \(2^5 \times 5^5\) and \(\dfrac{10^4}{5^4}\).
After 4 days doubling: \(1\times2^4 = 2^4\). Then 4 days tripling multiplies by \(3^4\):
Order does not matter — multiplication can be regrouped: \(2^4\times3^4 = (3\times2)^4\). This is the law \(\;m^a \times n^a = (mn)^a\). Using it:
Answer\(6^4 = 1296\) either way · \(2^5\times5^5 = 10^5\) · \(10^4/5^4 = 16\).
Estu has 4 dresses and 3 caps. Roxie has 7 dresses, 2 hats and 3 pairs of shoes. How many different outfits can each make?
For each independent choice we multiply the number of options.
AnswerEstu: 12 outfits · Roxie: 42 ways.
How many passwords does a 5-digit lock have (digits 0–9)? What about a lock with 6 slots using the letters A–Z?
Each slot is an independent choice, so we multiply the options slot by slot. Building up: a 2-digit lock has \(10\times10 = 100\), a 3-digit lock \(100\times10 = 1000\), and so on.
For 6 slots each chosen from 26 letters:
Answer5-digit lock: \(10^5 = 1{,}00{,}000\) passwords · 6-letter lock: \(26^6\) passwords.
The Other Side of Powers
Dividing powers, the meaning of \(n^0\), and negative exponents.
What is \(2^{100} \div 2^{25}\) in powers of 2?
Dividing powers of the same base subtracts exponents (common factors cancel): \(n^a \div n^b = n^{a-b}\).
Answer\(2^{75}\).
In the rule \(n^a \div n^b = n^{a-b}\), why must \(n \neq 0\)? And why is \(n^0 = 1\)?
Division by \(n\) is hidden inside the rule, and we can never divide by zero — so \(n=0\) is not allowed. To find \(n^0\), set \(a=b\):
(For \(n=0\) this would read \(0/0\), which is undefined — another reason to exclude it.)
AnswerBecause the rule divides by \(n\); and any non-zero number to the power 0 equals 1.
Can the exponents \(a, b\) be any integers (not just counting numbers)? Write equivalent forms of \(2^{-4}, 10^{-5}, (-7)^{-2}, (-5)^{-3}, 10^{-100}\).
Yes — the laws keep working for all integers. A negative exponent simply means the reciprocal: \(n^{-a} = \dfrac{1}{n^{a}}\).
AnswerYes; each is the reciprocal of the positive power (see above).
Simplify and write in exponential form.(i) 2⁻⁴×2⁷ (ii) 3²×3⁻⁵×3⁶ (iii) p³×p⁻¹⁰ (iv) 2⁴×(−4)⁻² (v) 8^p × 8^q
- \(2^{-4}\times2^{7} = 2^{-4+7} = 2^{3}\).
- \(3^{2}\times3^{-5}\times3^{6} = 3^{2-5+6} = 3^{3}\).
- \(p^{3}\times p^{-10} = p^{3-10} = p^{-7}\).
- \(2^{4}\times(-4)^{-2} = 2^{4}\times\dfrac{1}{16} = 2^{4}\times 2^{-4} = 2^{0} = 1\).
- \(8^{p}\times8^{q} = 8^{p+q}\).
Answer(i) 2³ · (ii) 3³ · (iii) p⁻⁷ · (iv) 1 · (v) 8^(p+q).
On the power line of 4, how many times larger than \(4^{-2}\) is \(4^{2}\)? Then use the power line of 7 to evaluate the listed products and quotients.
“How many times larger” means divide: \(4^{2} \div 4^{-2} = 4^{2-(-2)} = 4^{4} = 256\) times.
Reading values off the line and adding/subtracting exponents:
| Expression | As a power of 7 |
|---|---|
| 2,401 × 49 | 7⁴ × 7² = 7⁶ |
| 49³ | (7²)³ = 7⁶ |
| 343 × 2,401 | 7³ × 7⁴ = 7⁷ |
| 16,807 ÷ 49 | 7⁵ ÷ 7² = 7³ |
| 7 ÷ 343 | 7¹ ÷ 7³ = 7⁻² |
| 16,807 ÷ 8,23,543 | 7⁵ ÷ 7⁷ = 7⁻² |
| 1,17,649 × (1 ÷ 343) | 7⁶ × 7⁻³ = 7³ |
| (1 ÷ 343) × (1 ÷ 343) | 7⁻³ × 7⁻³ = 7⁻⁶ |
Answer\(4^2\) is \(4^4 = 256\) times larger than \(4^{-2}\); the powers of 7 are listed above.
Powers of 10 & Scientific Notation
Writing huge numbers as \(x \times 10^{y}\) so the exponent does the talking.
Write 172, 5642 and 6374 in expanded form using powers of 10.
AnswerEach digit is multiplied by its place value, written as a power of 10.
Write these facts in scientific form: the Sun is 30,00,00,00,00,00,00,00,00,000 m from the galaxy centre; the galaxy has 1,00,00,00,00,000 stars; the Earth’s mass is 59,76,00,00,00,00,00,00,00,00,000 kg.
Scientific form writes a number as \(x \times 10^{y}\) with \(1 \le x < 10\): keep the leading digits, count the remaining places for the exponent.
Answer≈ 3×10²⁰ m · 1×10¹¹ stars · ≈ 5.976×10²⁴ kg.
Of the Sun–Saturn (1.4335×10¹² m), Saturn–Uranus (1.439×10¹² m) and Sun–Earth (1.496×10¹¹ m) distances, which is smallest? Mark the Earth on the Sun–Saturn line.
Compare exponents first: two distances have \(10^{12}\), but Sun–Earth has \(10^{11}\) — one power of 10 smaller — so it is the smallest, regardless of the leading digits.
AnswerThe Sun–Earth distance (1.496×10¹¹ m) is the smallest.
Express in standard form.(i) 59,853 (ii) 65,950 (iii) 34,30,000 (iv) 70,04,00,00,000
Answer5.9853×10⁴ · 6.595×10⁴ · 3.43×10⁶ · 7.004×10¹⁰.
Getting a Sense for Large Numbers
Estimation, linear vs exponential growth, and the names of giants.
What is the worth of jaggery equal to Roxie’s weight, and wheat equal to Estu’s weight? Roughly how many 1-rupee coins equal Roxie’s weight?
Set up the relationship, then put in reasonable values.
Assuming Roxie ≈ 45 kg, jaggery ≈ ₹70/kg → \(45\times70 = ₹3150\). Assuming Estu ≈ 50 kg, wheat ≈ ₹50/kg → \(50\times50 = ₹2500\).
Coins: a 1-rupee coin weighs about 4 g. Roxie’s 45 kg = 45000 g, so the number of coins ≈ \(45000 \div 4 \approx 11{,}250 \approx 1.1\times10^{4}\) coins (about eleven thousand — in the thousands, not lakhs).
Estimate≈ ₹3150 jaggery · ≈ ₹2500 wheat · ≈ 1.1×10⁴ coins (assumptions stated).
How many times could a person walk all the way around the Earth (≈ 40,000 km) in a lifetime if they walked non-stop?
- Walking speed ≈ 5 km/h, non-stop for a lifetime ≈ 70 years.
- Distance \(= 5 \times 24 \times 365 \times 70 \approx 30{,}66{,}000\) km.
- Number of trips \(= 30{,}66{,}000 \div 40{,}000 \approx 77\).
EstimateAbout 75–80 times around the Earth (with these assumptions).
A ladder to the Moon (3,84,400 km) has steps 20 cm apart. How many steps? How does this compare with folding paper?
How many 20 cm gaps fit into the distance? Convert to the same unit first: 3,84,400 km = 38,440,000,000 cm.
A ladder grows the same amount each step — linear (additive) growth. Paper folding doubles each time — exponential (multiplicative) growth, which is why a mere 46 folds beats nearly two billion steps.
AnswerAbout \(1.922\times10^{9}\) steps; linear growth is additive, exponential is multiplicative.
Use powers of 10 to compare quantities. Fill the blanks: starlings ≈ 1.3 billion; mosquitoes ≈ 110 trillion. Are there about 20,000 people per African elephant?
Filling the blanks in scientific form:
People per elephant: humans ≈ \(8\times10^{9}\), African elephants ≈ \(4\times10^{5}\):
Answer1.3×10⁹ starlings · 1.1×10¹⁴ mosquitoes · yes, ≈ 20,000 people per African elephant.
Find, in scientific notation: ants per human; flocks of starlings (10,000 each); leaves on all trees (10⁴ each); sheets of paper to reach the Moon; time to count all stars (1/sec); time to drink all of Earth’s water (200 ml per 10 s).
(The Moon is \(3.844\times10^{8}\) m \(= 3.844\times10^{10}\) cm away, and each sheet is 0.001 cm thick. For the water: \(2\times10^{25}\) drops at 16 drops/ml is \(1.25\times10^{24}\) ml, drunk 200 ml every 10 s.)
Answer2.5×10⁶ ants · 1.3×10⁵ flocks · 3×10¹⁶ leaves · 3.844×10¹³ sheets · 2×10²³ s · 6.25×10²² s.
Roxie is 4840 days old. How old is she in hours? If you have lived a million seconds, how old are you?
(In minutes that is \(4840\times1440 \approx 69{,}70{,}000\) — the number in Roxie’s speech bubble!)
A million seconds is \(10^{6} \div 86400 \approx 11.6\) days — so you would be just under 12 days old.
Answer≈ 1,16,160 hours (≈ 1.16×10⁵) · a million seconds ≈ 12 days.
In the names million, billion, trillion, quadrillion, …, what does the first part of each name denote?
Each name stacks more factors of 1000 on top of a thousand. The Latin prefix counts how many extra thousands are multiplied:
So bi-, tri-, quad-, quint- … (2, 3, 4, 5, …) tell you how many times a thousand is multiplied with another thousand.
AnswerIt tells how many times 1000 is multiplied with 1000 (the number of thousand-factors).
Figure it Out — Set B
Pages 44–45 · the full set, worked out.
Find the units digit of \(2^{224} \div 4^{32}\). [Hint: \(4 = 2^2\)]
- Rewrite the base: \(4^{32} = (2^2)^{32} = 2^{64}\).
- Divide: \(2^{224} \div 2^{64} = 2^{224-64} = 2^{160}\).
- Units digits of \(2^1,2^2,2^3,2^4,\dots\) repeat in a cycle of 4: 2, 4, 8, 6. Since 160 is a multiple of 4, the units digit matches \(2^4\), which is 6.
AnswerThe units digit is 6.
A container has 5 bottles. A new container arrives every day. How many bottles after 40 days?
One container (5 bottles) is added each day — this is linear, not exponential, growth.
Answer200 bottles (= 2×10²).
Write each number as a product of two or more powers in three different ways (exponents can be any integers).(i) 64³ (ii) 192⁸ (iii) 32⁻⁵
First find the prime form, then split the exponent however you like.
AnswerThree sample splittings for each are shown above (many others work too).
Is each statement Always True, Only Sometimes True, or Never True? Explain.
- Sometimes(i) Cube numbers are also square numbers. Only when the number is a 6th power, since it must be both a square and a cube. Example: \(64 = 2^6 = (2^3)^2 = (2^2)^3\).
- Always(ii) Fourth powers are also square numbers. \(n^4 = (n^2)^2\) — always a perfect square.
- Always(iii) The fifth power of a number is divisible by its cube. \(n^5 \div n^3 = n^2\), a whole number — always divisible.
- Always(iv) The product of two cube numbers is a cube number. \(a^3 \times b^3 = (ab)^3\) — always a cube.
- Never(v) \(q^{46}\) is both a 4th power and a 6th power (\(q\) prime). A 4th power needs the exponent divisible by 4, a 6th power by 6. But 46 is divisible by neither — never true.
Answer(i) Sometimes · (ii) Always · (iii) Always · (iv) Always · (v) Never.
Simplify 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
- \(10^{-2}\times10^{-5} = 10^{-7} = \dfrac{1}{10^{7}}\).
- \(5^{7}\div5^{4} = 5^{3}\).
- \(9^{-7}\div9^{4} = 9^{-7-4} = 9^{-11} = \dfrac{1}{9^{11}}\).
- \((13^{-2})^{-3} = 13^{(-2)\times(-3)} = 13^{6}\).
- \(m^{5}n^{12}(mn)^{9} = m^{5}n^{12}\cdot m^{9}n^{9} = m^{14}n^{21}\).
Answer(i) 10⁻⁷ · (ii) 5³ · (iii) 9⁻¹¹ · (iv) 13⁶ · (v) m¹⁴n²¹.
If \(12^2 = 144\), what is (i) (1.2)² (ii) (0.12)² (iii) (0.012)² (iv) 120²?
Each number is \(12\) shifted by powers of 10, and squaring shifts the decimal twice as far.
Answer1.44 · 0.0144 · 0.000144 · 14400.
Circle the numbers that are the same.2⁴×3⁶ 6⁴×3² 6¹⁰ 18²×6² 6²⁴
Reduce each to powers of the primes 2 and 3.
The first, second and fourth all equal \(2^4\times3^6 = 11664\).
Answer2⁴×3⁶, 6⁴×3² and 18²×6² are the same (= 11664).
Identify the greater number.(i) 4³ or 3⁴ (ii) 2⁸ or 8² (iii) 100² or 2¹⁰⁰
Answer(i) 3⁴ · (ii) 2⁸ · (iii) 2¹⁰⁰.
A dairy makes 8.5 billion packets a year and wants a unique code per packet using digits 0–9. How many digits must the code have?
An \(n\)-digit code (digits 0–9) gives \(10^n\) possibilities. We need at least 8.5 billion = \(8.5\times10^9\):
So \(10^9\) is too few and \(10^{10}\) is enough — the code needs 10 digits.
Answer10 digits.
64 is both a square (8²) and a cube (4³). Are there other numbers that are both? Describe them in general.
A number that is both a square and a cube must have every prime exponent divisible by both 2 and 3, hence by 6. So such numbers are exactly the 6th powers, \(n^6\).
There are infinitely many, and each \(n^6 = (n^3)^2 = (n^2)^3\).
AnswerYes — infinitely many; they are the 6th powers \(n^6\) (1, 64, 729, 4096, …).
A digital locker has an alphanumeric passcode of length 5 (digits and letters). How many codes are possible?
Each slot can be any of 26 letters or 10 digits = 36 characters, and there are 5 independent slots.
Answer\(36^5\) codes (about 6 crore).
Sheep ≈ 10⁹ and goats ≈ 10⁹. What is their total?(i) 20⁹ (ii) 10¹¹ (iii) 10¹⁰ (iv) 10¹⁸ (v) 2×10⁹ (vi) 10⁹ + 10⁹
Adding equal powers is not the same as multiplying them — the exponents do not add. We simply add the counts:
So the correct expression is (vi) \(10^9 + 10^9\), which equals \(2\times10^9\) (option (v) is the same value).
Answer(vi) \(10^9 + 10^9 = 2\times10^9\).
Calculate in scientific notation: (i) total clothing if each of the world’s people has 30 pieces; (ii) honeybees if 100 million colonies have 50,000 bees each; (iii) bacteria in all humans if each body has 38 trillion; (iv) total time spent eating in a lifetime (seconds).
(For (iv): assume ~70 years and ~1 hour = 3600 s of eating per day.)
Answer2.46×10¹¹ pieces · 5×10¹² bees · 3.116×10²³ bacteria · ≈ 9.198×10⁷ s.
What was the date 1 arab / 1 billion seconds ago?
Convert seconds to years: \(1\times10^9\) s \(\div\) (number of seconds in a year).
So count back about 31 years 8 months from today’s date to get the answer (the exact date depends on when you solve it).
AnswerAbout 31.7 years (≈ 11,574 days) before today.
Tremendous in Ten!
Who can write the bigger number? Exponents settle it instantly.
Roxie writes 10000000000000 and Estu writes 999999 × 999999. Who is bigger?
Compare by powers of 10. Roxie’s number is \(10^{13}\). Estu’s is a product of two numbers each below \(10^6\), so it is less than \((10^6)^2 = 10^{12}\).
AnswerRoxie wins — \(10^{13}\) beats anything below \(10^{12}\).
Roxie writes \(10^{1000} + 10^{1000} + 10^{1000} + 10^{1000}\) and Estu writes \(10^{1000000} \times 9000\). Which is greater?
Simplify each. Adding four equal powers only multiplies by 4 — the exponent does not grow:
Estu’s exponent (about a million) towers over Roxie’s (just 1000). The exponent decides it.
AnswerEstu wins by a colossal margin (\(\approx 9\times10^{1000003}\) vs \(4\times10^{1000}\)).
End of Chapter 2 · Power Play · solutions by EduGrown