ch2 POWER PLAY figure it out CLASS 8

Q1

Find the units digit in the value of          

$2^{224}$   

4324^{32}

$2^{224}\div 4^{32}$. (Hint: $4=2^2$.)

Solution and explanation

Write everything as powers of 2:

$$4^{32}=(2^2)^{32}=2^{64}.$$

So

$$2^{224}\div 4^{32}=2^{224}\div 2^{64}=2^{224-64}=2^{160}.$$

Units digits of powers of 2 cycle every 4: $2,4,8,6,$ then repeat. We find the position in the cycle:

$$160 \bmod 4 = 0,$$

so $2^{160}$ is at the 4th position of the cycle → units digit = 6.

Answer: 6

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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?

Solution and explanation

Each container has 5 bottles. After 40 days there are 40 containers (one per day), so total bottles:

$$40 \times 5 = 200.$$

Answer: 200 bottles

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Q3

Write the given number as the product of two or more powers in three different ways.
(i) $64^3$ \qquad (ii) $192^8$ \qquad (iii) $32^{-5}$

Solution and explanation

We use prime-factor forms and power rules $(a^b)^c=a^{bc}$.

(i) $64^3$
$64=2^6$. So

$$64^3=(2^6)^3=2^{18}.$$

Other equivalent ways:

$$64^3=(8^2)^3=8^6,\qquad 64^3=(4^3)^3=4^9.$$

(All three equal $2^{18}$.)

Three forms: $2^{18},\;8^{6},\;4^{9}.$

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(ii) $192^8$
Prime factorise $192$:

$$192=64\times 3 = 2^6\times 3.$$

So

$$192^8=(2^6\cdot3)^8 =2^{48}\cdot 3^8.$$

Other ways (same exponents grouped differently):

$$192^8=(2^{24})^2\cdot 3^8,\qquad 192^8=(2^{16})^3\cdot 3^8.$$

Three forms: $2^{48}\cdot3^8,\; (2^{24})^2\cdot3^8,\; (2^{16})^3\cdot3^8.$

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(iii) $32^{-5}$
$32=2^5$. So

$$32^{-5}=(2^5)^{-5} = 2^{-25}.$$

Other forms:

$$32^{-5}=(2^{-5})^{5},\qquad 32^{-5}=\frac{1}{32^{5}}.$$

Three forms: $2^{-25},\; (2^{-5})^{5},\; \dfrac{1}{32^{5}}.$

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Q4

Examine each statement and say if it is ‘Always True’, ‘Only Sometimes True’, or ‘Never True’. Explain.

(i) Cube numbers are also square numbers.

• A number that is both a cube and a square must be both $n^2$ and $m^3$. Such numbers are 6th powers ($k^6$). Example: $64=2^6$ is both. But $8=2^3$ is not a square.
  Answer: Only Sometimes True (exactly those numbers that are 6th powers).

(ii) Fourth powers are also square numbers.

• $x^4=(x^2)^2$, so every 4th power is a square (square of $x^2$).
  Answer: Always True.

(iii) The fifth power of a number is divisible by the cube of that number.

• $n^5 = n^3\cdot n^2$, so yes $n^5$ is divisible by $n^3$ for any integer $n$.
  Answer: Always True.

(iv) The product of two cube numbers is a cube number.

• If numbers are $a^3$ and $b^3$, product $a^3b^3=(ab)^3$, which is a cube.
  Answer: Always True.

(v) $q^{46}$ is both a 4th power and a 6th power (q is a prime number).

• For a power $q^{46}$ to be a 4th power, exponent 46 must be divisible by 4; to be a 6th power, 46 must be divisible by 6. 46 is divisible by neither 4 nor 6 (46 ÷ 4 = 11.5; 46 ÷ 6 ≈ 7.67). So it is neither.
  Answer: Never True.

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Q5

Simplify and write these in exponential form.

(i) $10^{-2}\times 10^{-5}$
(ii) $5^7\div5^4$
(iii) $9^{-7}\div9^4$
(iv) $(13^{-2})^{-3}$
(v) $m^5 n^{12}(mn)^9$

Solution and explanation (use exponent laws)

• Law used: $a^m a^n = a^{m+n}$, $a^m\div a^n = a^{m-n}$, $(a^m)^n=a^{mn}$, $(ab)^n=a^n b^n$.

(i) $10^{-2}\times 10^{-5} = 10^{-2 + (-5)} = 10^{-7}.$

(ii) $5^7\div5^4 = 5^{7-4} = 5^{3} = 125.$

(iii) $9^{-7}\div9^4 = 9^{-7-4} = 9^{-11} = \dfrac{1}{9^{11}}.$

(iv) $(13^{-2})^{-3} = 13^{(-2)\cdot(-3)} = 13^{6}.$

(v) $m^5 n^{12}(mn)^9 = m^5 n^{12} \cdot m^9 n^9 = m^{5+9} n^{12+9} = m^{14} n^{21}.$

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Q6

If $12^2=144$, find:

(i) $(1.2)^2$
(ii) $(0.12)^2$
(iii) $(0.012)^2$
(iv) $120^2$

Solution and explanation (scale by powers of 10)

$(1.2) = 12/10$. So

$$(1.2)^2=\left(\frac{12}{10}\right)^2=\frac{12^2}{10^2}=\frac{144}{100}=1.44.$$

$(0.12)=12/100$. So

$$(0.12)^2=\frac{12^2}{100^2}=\frac{144}{10,000}=0.0144.$$

$(0.012)=12/1000$. So

$$(0.012)^2=\frac{12^2}{1000^2}=\frac{144}{1,000,000}=0.000144.$$

$120=12\times10$. So

$$120^2=(12\times10)^2=12^2\times10^2=144\times100=14,400.$$

Answers: (i) 1.44, (ii) 0.0144, (iii) 0.000144, (iv) 14,400

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Q7

Circle the numbers that are the same —
$2^{4}\times3^{6},\;6^{4}\times3^{2},\;6^{10},\;18^{2}\times6^{2},\;6^{24}.$

Solution and explanation

Compute prime factor form:

• $2^{4}\cdot3^{6}$ is already $2^4 3^6$.

• $6^{4}\cdot3^{2} = (2\cdot3)^{4}\cdot3^2 = 2^{4}3^{4}\cdot3^{2} = 2^{4}3^{6}$. → same as first.

• $6^{10} = 2^{10}3^{10}$ → different.

• $18^{2}\cdot6^{2} = (2\cdot3^2)^2 \cdot (2\cdot3)^2 = (2^2 3^4)\cdot(2^2 3^2) = 2^{4}3^{6}$. → same as first.

• $6^{24}=2^{24}3^{24}$ → different.

So the equal ones are:
$\;2^{4}\cdot3^{6},\;6^{4}\cdot3^{2},\;18^{2}\cdot6^{2}.$

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Q8

Identify the greater number in each of the following—

(i) $4^3$ or $3^4$
(ii) $2^8$ or $8^2$
(iii) $100^2$ or $2^{100}$

Solution and explanation

(i) $4^3=64$. $3^4=81$. So $3^4$ is greater.
Answer: $3^4$.

(ii) $2^8=256$. $8^2=64$. So $2^8$ is greater.
Answer: $2^8$.

(iii) $100^2=10{,}000$. $2^{100}$ is extremely large (approx $1.27\times10^{30}$). So $2^{100}$ is far greater.
Answer: $2^{100}$.

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Q9

A dairy plans to produce 8.5 billion packets of milk in a year. They want a unique ID for each packet using digits 0–9. How many digits should the code consist of?

Solution and explanation

Number of packets = $8.5$ billion $= 8.5\times10^9$.

A code of length $n$ (digits 0–9) has $10^n$ possible codes. We need the smallest $n$ with $10^n\ge 8.5\times10^9$.

Check:

• $10^9 = 1\times10^9$ (too small).

• $10^{10} = 1\times10^{10} = 10{,}000{,}000{,}000$ (which is $>8.5\times10^9$).

Therefore $n=10$ digits are needed.

Answer: 10 digits

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Q10

64 is a square number ($8^2$) and a cube number ($4^3$). Are there other numbers that are both squares and cubes? Describe them in general.

Solution and explanation

A number that is both a perfect square and a perfect cube must be a power that is simultaneously of the form $a^2$ and $b^3$. This happens exactly when the exponent is a common multiple of 2 and 3, i.e. a multiple of $\mathrm{lcm}(2,3)=6$.

Thus numbers of the form $k^6$ (6th powers) are both squares and cubes:

$$k^6 = (k^3)^2 = (k^2)^3.$$

Examples: $1^6=1,\;2^6=64,\;3^6=729,\;4^6=4096$, etc.

Answer: Yes. Exactly the sixth powers $k^6$.

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Q11

A digital locker has an alphanumeric passcode of length 5 (digits 0–9 and letters A–Z). How many such codes are possible?

Solution and explanation

Assume uppercase letters only → 26 letters + 10 digits = 36 possible characters per slot. For 5 positions, total possibilities:

$$36^5.$$

Compute value (optional): $36^2=1296,\;36^3=46,656,\;36^4=1,679,616,\;36^5=60,466,176.$

Answer: $36^5 = 60,466,176$ possible codes.

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Q12

The worldwide population of sheep (2024) is about $10^9$, and goats are about the same. What is the total population of sheep and goats?

Solution and explanation

Sheep ≈ $10^9$ and goats ≈ $10^9$. Sum:

$$10^9 + 10^9 = 2\times10^9.$$

Answer: $2\times10^9$

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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 in all humans in the world (use world pop ≈ $8.2\times10^9$).
(iv) Total time spent eating in a lifetime in seconds. (Make and state reasonable assumptions.)

Solution and explanation

We show method + result in scientific notation.

(i) Let world population ≈ $8.2\times10^9$ people. Pieces per person = 30.
Total $=30\times 8.2\times10^9 =246\times10^9 =2.46\times10^{11}.$

Answer (i): $\boxed{2.46\times10^{11}}$ pieces of clothing (using 8.2 billion people).

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(ii) Colonies = $100$ million = $1.0\times10^8$. Bees per colony = 50,000 = $5.0\times10^4$.
Total bees $=(1.0\times10^8)\times(5.0\times10^4)=5.0\times10^{12}.$

Answer (ii): $\boxed{5.0\times10^{12}}$ honeybees.

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(iii) Bacteria per person ≈ 38 trillion = $3.8\times10^{13}$. World pop ≈ $8.2\times10^9$.
Total bacteria ≈ $(3.8\times10^{13})\times(8.2\times10^9)$.

Multiply coefficients: $3.8\times8.2=31.16$. Add exponents: $13+9=22$. So

$$31.16\times10^{22}=3.116\times10^{23}.$$

Answer (iii): $\boxed{3.116\times10^{23}}$ (approx).

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(iv) Total time spent eating in a lifetime (example model & result).
We must assume average daily time spent eating and an expected lifetime. One reasonable model:

• Eating time ≈ 1.5 hours/day = $1.5\times60\times60 = 5400$ seconds/day.

• Lifetime ≈ 70 years. Use 1 year ≈ 365 days (ignore leap days for a rough estimate).

Total seconds $=5400$ s/day $\times 70\times365$ days
First compute days: $70\times365=25,550$ days.
Now multiply: $25,550\times5,400$.

We compute carefully:

• $25,550\times5,400 = 25,550\times(54\times100) = (25,550\times54)\times100.$

• $25,550\times54 = 25,550\times(50+4) = 25,550\times50 + 25,550\times4.$

• $25,550\times50 = 1,277,500$.

• $25,550\times4 = 102,200$.

• Sum = $1,277,500 + 102,200 = 1,379,700.$
  Multiply by 100 → $137,970,000$ seconds.

So about $1.3797\times10^8$ seconds.

Answer (iv): ≈ $\boxed{1.38\times10^{8}\ \text{seconds}}$ (using 1.5 h/day and 70 years).
(If you pick different assumptions — e.g. 1 hour/day or 80 years — change numbers accordingly; show assumptions when giving to students.)

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Q14

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

Solution and explanation

1 arab = 1 billion seconds = $10^9$ seconds.

Convert $10^9$ seconds to days:

$$1\,\text{day}=86400\ \text{seconds}. \quad \text{Compute } \frac{10^9}{86400}.$$

We compute exact integer days and remainder (division):

• $86400\times 11{,}574 = 86400\times 11{,}574 = 999{,}993{,}600$ seconds. (Check by multiplication.)

• Remainder: $10^9 - 999{,}993{,}600 = 4{,}400$ seconds.

So $10^9$ seconds = 11,574 days + 4,400 seconds.

Convert 4,400 seconds to hours/minutes:

• $4,400 \div 3{,}600 = 1$ hour remainder $800$ seconds.

• $800$ seconds = $13$ minutes $20$ seconds.

So $10^9$ seconds = 11,574 days, 1 hour, 13 minutes, 20 seconds.

Now convert days to years roughly:

$$11,574\ \text{days}\div365\approx31.7\ \text{years}.$$

So about 31.7 years ago.

• If you want a calendar date, subtract ~31 years and 259 days (approx) from today. For example, if "today" were mid-2025, one billion seconds earlier would be around 1993 (roughly 31 years and 8–9 months earlier). Always state the reference date when giving an exact calendar date.

Answer: Approximately 11,574 days ≈ 31.7 years ago (i.e. ~31 years, 8 months, 1 hour, 13 min, 20 s earlier).

Q: Think about how many combinations are possible in different contexts

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(i) PIN codes of places in India

• A PIN code has 6 digits.

• Each digit has 10 choices (0–9).

• So theoretically:

  $$10^6 = 1,000,000 \text{ possible PINs.}$$

But in practice:

• The first digit is 1–9 (denoting zones). “0” is not used in the first place.

• So usable combinations:

  $$9 \times 10^5 = 900,000.$$

How allotted?

• 1st digit → postal zone (8 zones + 9 = Army Postal Service).

• 2nd digit → sub-zone/region.

• 3rd digit → sorting district (major hub).

• Last 3 digits → delivery post office.

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(ii) Mobile numbers

• Indian mobile numbers have 10 digits.

• The first digit must be 6, 7, 8, or 9 (4 options).

• The remaining 9 digits each have 10 options.

• Total possible mobile numbers:

  $$4 \times 10^9 = 4,000,000,000.$$

How allotted?

• The Department of Telecom allocates blocks of numbers (based on the first few digits) to different telecom operators.

• Operators then issue them to subscribers.

• Even if a person changes the operator (via MNP), the original number block remains the same.

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(iii) Vehicle registration numbers

Format (typical):
SS NN LL DDDD

• SS = State/UT code (e.g., KA for Karnataka).

• NN = RTO code (00–99).

• LL = Series letters (like A, B … AA, AB …).

• DDDD = numbers 0001–9999.

Combinations per RTO:

• For 1-letter series: $26 \times 9999 = 259,974$.

• For 2-letter series: $26^2 \times 9999 = 6,759,324$.

So millions of numbers are possible per RTO.

How allotted?

• Registration is done in the RTO jurisdiction.

• Numbers are issued in order: KA-01-A-0001, 0002, … 9999.

• Then next series (KA-01-B-0001 …).

• Some states also use the new BH series for vehicles moving across states.

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✅ Final Answers (with explanation)

1. PIN codes → up to 1,000,000 possible, practically 900,000 (hierarchically allotted by zones, sub-zones, districts, post offices).

2. Mobile numbers → about 4 billion possible (first digit 6/7/8/9; blocks allotted to operators).

3. Vehicle registration numbers → Millions per RTO (State code + RTO code + series letters + 4-digit numbers; allotted sequentially).
   

   (i) PIN codes

   • 6 digits → $10^6 = 1,000,000$ possible.

   • First digit only 1–9 → usable ≈ 900,000.

   • Allotment:

     • 1st digit → Zone (8 zones + 9 for APS).

     • 2nd digit → Sub-zone/region.

     • 3rd digit → Sorting district.

     • Last 3 digits → Post office.

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   (ii) Mobile numbers

   • 10 digits.

   • First digit = 6, 7, 8, or 9 → 4 options.

   • Remaining 9 digits → 10 choices each.

   • Total = $4 \times 10^9 = 4,000,000,000$.

   • Allotment: Blocks of numbers given to operators by DoT.

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   (iii) Vehicle registration numbers

   • Format: SS NN LL DDDD

     • SS = State code (KA, DL, MH …)

     • NN = RTO code (00–99)

     • LL = Series letters (A, B … AA, AB …)

     • DDDD = Numbers 0001–9999

   • Combinations:

     • 1-letter series: $26 \times 9999 ≈ 2.6 \times 10^5$

     • 2-letter series: $26^2 \times 9999 ≈ 6.7 \times 10^6$

   • Allotment: Issued sequentially within each RTO; new series starts after 9999. Special BH series for vehicles moving between states.

Source / System	Progression Rule	Examples (with Power of 10)	Special Notes / Names
Ancient Buddhist Text (Lalitavistara, 1st c. BCE)	Each higher name = 100 × previous	100 koti = 1 ayuta (10⁹) 100 ayuta = 1 niyuta (10¹¹) 100 niyuta = 1 kankara (10¹³) … continues till 100 vibhutangama = 1 tallakshana (10⁵³)	Covers odd powers of 10 up to 10⁵³
Jaina Texts	Sequential naming of powers of 10	Mahaviracharya (Ganita-sara-sangraha): names up to 10²³ (24 terms) Amalasiddhi: names each power up to 10⁹⁶ (dasha-ananta)	Extremely detailed naming system
Pali Grammar (Kāccāyana)	Similar to Buddhist/Jaina expansions	Names numbers up to 10¹⁴⁰ (asaṅkhyeya)	Shows cultural interest in very large numbers
Indian System (Modern)	Each higher name = 100 × previous	1 lakh = 10⁵ 1 crore = 10⁷ 1 arab = 10⁹ 1 kharab = 10¹¹ 1 neel = 10¹³ 1 padma = 10¹⁵ 1 shankh = 10¹⁷ 1 maha shankh = 10¹⁹	Still in use today in India
International / American System	Each higher name = 1000 × previous (step of 10³)	1 million = 10⁶ 1 billion = 10⁹ 1 trillion = 10¹² 1 quadrillion = 10¹⁵ 1 quintillion = 10¹⁸ 1 sextillion = 10²¹ 1 septillion = 10²⁴ … continues	Based on Latin prefixes: bi = 2, tri = 3, quadri = 4, quint = 5, etc.
Special Numbers	—	1 googol = 10¹⁰⁰ 1 googolplex = 10^(10¹⁰⁰)	Estimated atoms in universe = 10⁷⁸ – 10⁸²
Currency Context	—	India: Highest note = ₹2000 Zimbabwe (2009): 100 trillion note = 10¹⁴ Hungary (1946): Printed but not issued = 1 sextillion pengő = 10²¹	Shows real-world use of very large numbers

Powers and Exponents – Questions & Step-by-Step Solutions

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1. Simplify: \( 2^3 \times 2^4 \)

View Solution

Using the law \( a^m \times a^n = a^{m+n} \):
\( 2^3 \times 2^4 = 2^{3+4} = 2^7 = 128 \)

2. Simplify: \( 5^6 \div 5^2 \)

View Solution

Using \( a^m \div a^n = a^{m-n} \):
\( 5^6 \div 5^2 = 5^{6-2} = 5^4 = 625 \)

3. Simplify: \( (3^2)^4 \)

View Solution

Using \( (a^m)^n = a^{m \times n} \):
\( (3^2)^4 = 3^{2 \times 4} = 3^8 = 6561 \)

4. Write 81 as a power of 3.

View Solution

\( 3^4 = 3 \times 3 \times 3 \times 3 = 81 \)

5. Simplify: \( 2^{-3} \)

View Solution

Negative exponent rule: \( a^{-n} = \frac{1}{a^n} \)
\( 2^{-3} = \frac{1}{2^3} = \frac{1}{8} \)

6. Simplify: \( 10^0 \)

View Solution

Any non-zero number raised to the power 0 is 1.
\( 10^0 = 1 \)

7. Express \( 1/125 \) as a power of 5.

View Solution

\( 125 = 5^3 \), so \( 1/125 = 5^{-3} \)

8. Simplify: \( (2^3 \times 3^2)^2 \)

View Solution

First simplify inside the bracket:
\( 2^3 = 8, \; 3^2 = 9 \) → \( 8 \times 9 = 72 \)
Then square: \( 72^2 = 5184 \)

9. Simplify: \( (4^2 \div 4)^3 \)

View Solution

Inside: \( 4^2 \div 4 = 4^{2-1} = 4^1 = 4 \)
Then: \( 4^3 = 64 \)

10. Evaluate: \( (5^{-1})^{-3} \)

View Solution

Multiply powers: \( (5^{-1})^{-3} = 5^{(-1)\times(-3)} = 5^3 = 125 \)

11. Simplify: \( 9^{3/2} \)

View Solution

Fractional exponent: \( a^{m/n} = \sqrt[n]{a^m} \)
\( 9^{3/2} = (\sqrt{9})^3 = 3^3 = 27 \)

12. Simplify: \( 27^{2/3} \)

View Solution

\( 27^{2/3} = (\sqrt[3]{27})^2 = 3^2 = 9 \)

13. Simplify: \( (16^{1/2})^3 \)

View Solution

\( 16^{1/2} = 4 \), so \( 4^3 = 64 \)

14. If \( 2^x = 32 \), find \( x \).

View Solution

\( 32 = 2^5 \) → Therefore, \( x = 5 \)