Power Play

Given: A sheet of paper is folded 10 times. Initial thickness = $\nu$.

Concept: Each fold doubles the thickness. So after 1 fold, thickness = $2\nu$; after 2 folds = $2^2\nu$; after $n$ folds = $2^n\nu$.

Working: After 10 folds, thickness = $2^{10} \times \nu = 2^{10}\nu$.

Answer: Option (v) $2^{10}\nu$ is correct.

The other options are incorrect because:
- (i) $10\nu$ represents additive (not multiplicative) growth.
- (ii) $10 + \nu$ is just addition.
- (iii) $2 \times 10 \times \nu = 20\nu$ is linear, not exponential.
- (iv) $2^{10}$ does not include the initial thickness $\nu$.
- (vi) $10^2\nu = 100\nu$ is not the correct base or exponent.

Given: Number = 32400.

Method: Prime factorisation by successive division.

$$32400 \div 2 = 16200$$
$$16200 \div 2 = 8100$$
$$8100 \div 2 = 4050$$
$$4050 \div 2 = 2025$$
$$2025 \div 5 = 405$$
$$405 \div 5 = 81$$
$$81 \div 3 = 27$$
$$27 \div 3 = 9$$
$$9 \div 3 = 3$$
$$3 \div 3 = 1$$

So, $32400 = 2 \times 2 \times 2 \times 2 \times 5 \times 5 \times 3 \times 3 \times 3 \times 3$.

In exponential form:
$$\boxed{32400 = 2^4 \times 5^2 \times 3^4}$$

Concept: A negative number raised to an odd power is negative; raised to an even power is positive.

Working:
$$(-1)^5 = (-1) \times (-1) \times (-1) \times (-1) \times (-1) = -1$$
Since the exponent 5 is odd, $(-1)^5$ is negative.

$$(-1)^{56} = 1$$
Since the exponent 56 is even, $(-1)^{56}$ is positive and equals $1$.

Working:
$$(-2)^4 = (-2) \times (-2) \times (-2) \times (-2)$$
$$= [(-2) \times (-2)] \times [(-2) \times (-2)]$$
$$= 4 \times 4 = 16$$

Yes, $(-2)^4 = 16$. ✓

Note: The exponent is even, so the result is positive.

Given:
- Number of daughters = 3
- Each daughter gets 3 baskets → Total baskets = $3 \times 3 = 3^2 = 9$
- Each basket has 3 keys → Total keys = $3^2 \times 3 = 3^3 = 27$
- Each key opens 3 rooms → Total rooms = $3^3 \times 3 = 3^4$

Calculation:
$$3^4 = 3 \times 3 \times 3 \times 3 = 81$$

There are $\boxed{81}$ rooms altogether.

Given (continuing from rooms):
- Total rooms = $3^4 = 81$
- Each room has 3 tables → Total tables = $3^4 \times 3 = 3^5$
- Each table has 3 necklaces → Total necklaces = $3^5 \times 3 = 3^6$
- Each necklace has 3 diamonds → Total diamonds = $3^6 \times 3 = 3^7$

Calculation:
$$3^7 = 3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3 = 2187$$

There are $\boxed{2187}$ diamonds in total.

Using products already computed: $3^4 = 81$, and $81 \times 3 = 243 = 3^5$, $243 \times 3 = 729 = 3^6$, $729 \times 3 = 2187 = 3^7$.

Concept: $n$ multiplied by itself $a$ times is written as $n^a$.

(i) $6 \times 6 \times 6 \times 6$
$$= \boxed{6^4}$$
(6 appears 4 times)

(ii) $b \times b \times b \times b$
$$= \boxed{b^4}$$
($b$ appears 4 times)

(iii) $y \times y$
$$= \boxed{y^2}$$
($y$ appears 2 times)

(iv) $2 \times 2 \times a \times a$
$$= \boxed{2^2 a^2}$$
(2 appears 2 times, $a$ appears 2 times)

(v) $5 \times 5 \times 7 \times 7 \times 7$
$$= \boxed{5^2 \times 7^3}$$
(5 appears 2 times, 7 appears 3 times)

(vi) $a \times a \times a \times c \times c \times c \times c \times d$
$$= \boxed{a^3 c^4 d}$$
($a$ appears 3 times, $c$ appears 4 times, $d$ appears 1 time)

Method: Divide repeatedly by prime numbers starting from the smallest.

(i) 648:
$$648 \div 2 = 324,\quad 324 \div 2 = 162,\quad 162 \div 2 = 81$$
$$81 \div 3 = 27,\quad 27 \div 3 = 9,\quad 9 \div 3 = 3,\quad 3 \div 3 = 1$$
$$\boxed{648 = 2^3 \times 3^4}$$

(ii) 405:
$$405 \div 3 = 135,\quad 135 \div 3 = 45,\quad 45 \div 3 = 15,\quad 15 \div 3 = 5,\quad 5 \div 5 = 1$$
$$\boxed{405 = 3^4 \times 5}$$

(iii) 540:
$$540 \div 2 = 270,\quad 270 \div 2 = 135$$
$$135 \div 3 = 45,\quad 45 \div 3 = 15,\quad 15 \div 3 = 5,\quad 5 \div 5 = 1$$
$$\boxed{540 = 2^2 \times 3^3 \times 5}$$

(iv) 3600:
$$3600 \div 2 = 1800,\quad 1800 \div 2 = 900,\quad 900 \div 2 = 450,\quad 450 \div 2 = 225$$
$$225 \div 3 = 75,\quad 75 \div 3 = 25,\quad 25 \div 5 = 5,\quad 5 \div 5 = 1$$
$$\boxed{3600 = 2^4 \times 3^2 \times 5^2}$$

Concept: Evaluate each power first, then multiply.

(i) $2 \times 10^3 = 2 \times 1000 = \boxed{2000}$

(ii) $7^2 \times 2^3 = 49 \times 8 = \boxed{392}$

(iii) $3 \times 4^4 = 3 \times 256 = \boxed{768}$

(iv) $(-3)^2 \times (-5)^2 = 9 \times 25 = \boxed{225}$
(Both exponents are even, so both results are positive.)

(v) $3^2 \times 10^4 = 9 \times 10000 = \boxed{90000}$

(vi) $(-2)^5 \times (-10)^6$
$(-2)^5 = -32$ (odd power → negative)
$(-10)^6 = 1000000$ (even power → positive)
$$(-32) \times 1000000 = \boxed{-32000000}$$

Given: $2^{224} \div 4^{32}$

Step 1: Convert $4^{32}$ using $4 = 2^2$:
$$4^{32} = (2^2)^{32} = 2^{64}$$

Step 2: Apply the division law $n^a \div n^b = n^{a-b}$:
$$2^{224} \div 2^{64} = 2^{224-64} = 2^{160}$$

Step 3: Find the units digit of $2^{160}$.
The units digits of powers of 2 follow a cycle of 4: $2, 4, 8, 6, 2, 4, 8, 6, \ldots$
$$160 \div 4 = 40 \text{ (exactly divisible, remainder = 0)}$$
A remainder of 0 corresponds to the 4th position in the cycle, which gives units digit 6.

The units digit of $2^{224} \div 4^{32}$ is $\boxed{6}$.

Given: Initially 5 bottles in a container. Every day a new container is brought in.

Interpretation: Each day, the number of containers multiplies. Starting with 1 container of 5 bottles:
- After day 1: $5^1$ bottles (1 container)
- After day 2: $5^2$ bottles (each container brings another container of 5)
- After day $n$: $5^n$ bottles

After 40 days:
$$\text{Number of bottles} = 5^{40}$$

$$\boxed{5^{40} \text{ bottles}}$$

(This is an astronomically large number, illustrating exponential growth.)

Concept: Use the laws $(n^a)^b = n^{ab}$ and $n^a \times m^a = (nm)^a$, and prime factorisation.

(i) $64^3$:
Note: $64 = 2^6$, so $64^3 = (2^6)^3 = 2^{18}$.

Three ways:
- $64^3 = (2^6)^3 = 2^{18}$, i.e., $\mathbf{2^{18}}$
- $64^3 = (4^3)^3 = 4^9$, i.e., $\mathbf{4^9}$
- $64^3 = (8^2)^3 = 8^6$, i.e., $\mathbf{8^6}$

(ii) $192^8$:
Note: $192 = 2^6 \times 3$, so $192^8 = 2^{48} \times 3^8$.

Three ways:
- $192^8 = 2^{48} \times 3^8$
- $192^8 = (64 \times 3)^8 = 64^8 \times 3^8$
- $192^8 = (192^4)^2$, i.e., $(192^4)^2$

(iii) $32^{-5}$:
Note: $32 = 2^5$, so $32^{-5} = (2^5)^{-5} = 2^{-25}$.

Three ways:
- $32^{-5} = 2^{-25}$
- $32^{-5} = (2^5)^{-5} = 2^{-25}$, equivalently $(4^{-5})^{\frac{5}{2}}$ — or more simply: $32^{-5} = \left(\frac{1}{32}\right)^5$
- $32^{-5} = \left(\frac{1}{2}\right)^{25}$
- $32^{-5} = 4^{-5} \times 8^{-5} \div 1$ — using $32 = 4 \times 8$: $(4 \times 8)^{-5} = 4^{-5} \times 8^{-5}$

(i) Cube numbers are also square numbers.

Only Sometimes True.

A cube number is $n^3$. For it to also be a square, we need $n^3 = m^2$ for some integer $m$. This happens when $n$ itself is a perfect square. For example:
- $n = 1$: $1^3 = 1 = 1^2$ ✓ (both square and cube)
- $n = 4$: $4^3 = 64 = 8^2$ ✓
- $n = 9$: $9^3 = 729 = 27^2$ ✓
- $n = 2$: $2^3 = 8$ — not a perfect square ✗

So it is only sometimes true.

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(ii) Fourth powers are also square numbers.

Always True.

If a number is a fourth power, it can be written as $n^4 = (n^2)^2$, which is always a perfect square.

For example: $2^4 = 16 = 4^2$; $3^4 = 81 = 9^2$. ✓

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

Always True.

$$\frac{n^5}{n^3} = n^{5-3} = n^2$$

Since $n^2$ is always an integer (for integer $n \neq 0$), $n^5$ is always divisible by $n^3$.

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

Always True.

Let the two cube numbers be $a^3$ and $b^3$.
$$a^3 \times b^3 = (ab)^3$$
This is the cube of $ab$, so the product is always a cube number.

Example: $8 \times 27 = 216 = 6^3$ ✓

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

Always True.

- As a 4th power: $q^{46} = (q^{46/4})$? — $46/4$ is not an integer, so this approach needs care.

Actually, $q^{46}$ is a 4th power if $46$ is divisible by 4. $46 = 4 \times 11 + 2$ — not divisible by 4.

Let us re-examine: $q^{46}$ is a 4th power means $q^{46} = (q^k)^4 = q^{4k}$, so we need $4k = 46$, i.e., $k = 11.5$ — not an integer.

So $q^{46}$ is not always a 4th power.

However, $q^{46}$ is a 6th power: $q^{46} = (q^{46/6})$? $46/6$ is not an integer either.

Wait — let us check: is $46$ divisible by 6? $46 = 6 \times 7 + 4$ — No.

Correction: $q^{46}$ is a 4th power if $46$ is a multiple of 4 — it is not. It is a 6th power if $46$ is a multiple of 6 — it is not.

But note: $q^{46} = (q^{23})^2$ — it IS a perfect square (2nd power). Also $q^{46} = (q^2)^{23}$.

For it to be a 4th power: need $q^{46} = x^4$ for some integer $x$. Since $q$ is prime, $x = q^{46/4}$ — not an integer. So not a 4th power.

For it to be a 6th power: need $q^{46} = x^6$, so $x = q^{46/6}$ — not an integer. So not a 6th power.

Conclusion: Never True (for prime q > 1, $q^{46}$ is neither a 4th power nor a 6th power, since 46 is divisible by neither 4 nor 6).

Laws used:
- $n^a \times n^b = n^{a+b}$
- $n^a \div n^b = n^{a-b}$
- $(n^a)^b = n^{ab}$
- $(mn)^a = m^a n^a$

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

(ii) $5^7 \div 5^4 = 5^{7-4} = \boxed{5^3}$

(iii) $9^{-7} \div 9^4 = 9^{-7-4} = \boxed{9^{-11}}$

(iv) $(13^{-2})^{-3} = 13^{(-2)\times(-3)} = \boxed{13^{6}}$

(v) $m^5 n^{12}(mn)^9$

First expand $(mn)^9 = m^9 n^9$:
$$m^5 n^{12} \times m^9 n^9 = m^{5+9} \times n^{12+9} = \boxed{m^{14} n^{21}}$$

Given: $12^2 = 144$

Concept: Express each number in terms of 12 and use the power law.

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

(ii) $(0.12)^2$
$$0.12 = \frac{12}{100}$$
$$(0.12)^2 = \left(\frac{12}{100}\right)^2 = \frac{144}{10000} = \boxed{0.0144}$$

(iii) $(0.012)^2$
$$0.012 = \frac{12}{1000}$$
$$(0.012)^2 = \frac{144}{1000000} = \boxed{0.000144}$$

(iv) $120^2$
$$120 = 12 \times 10$$
$$(120)^2 = (12 \times 10)^2 = 12^2 \times 10^2 = 144 \times 100 = \boxed{14400}$$

Strategy: Evaluate or simplify each expression.

Expression 1: $2^4 \times 3^6$
$$= 16 \times 729 = 11664$$

Expression 2: $6^4 \times 3^2$
$$= 1296 \times 9 = 11664$$

Verification: $6^4 \times 3^2 = (2 \times 3)^4 \times 3^2 = 2^4 \times 3^4 \times 3^2 = 2^4 \times 3^6$ ✓

Expression 3: $6^{10}$
$$= 60466176 \neq 11664$$

Expression 4: $18^2 \times 6^2$
$$= (18 \times 6)^2 = 108^2 = 11664$$

Verification: $18^2 \times 6^2 = (2 \times 3^2)^2 \times (2 \times 3)^2 = 2^2 \times 3^4 \times 2^2 \times 3^2 = 2^4 \times 3^6$ ✓

Expression 5: $6^{24}$ — clearly much larger.

Conclusion: $\boxed{2^4 \times 3^6, \quad 6^4 \times 3^2, \quad 18^2 \times 6^2}$ are all equal to $11664$.

(i) $4^3$ vs $3^4$:
$$4^3 = 64$$
$$3^4 = 81$$
81 > 64
$$\boxed{3^4 \text{ is greater.}}$$

(ii) $2^8$ vs $8^2$:
$$2^8 = 256$$
$$8^2 = 64$$
256 > 64
$$\boxed{2^8 \text{ is greater.}}$$

(iii) $100^2$ vs $2^{100}$:
$$100^2 = 10000 = 10^4$$
$$2^{100} = (2^{10})^{10} = 1024^{10}$$
Since 1024 > 10, we have $1024^{10} \gg 10^4$.

More precisely: $2^{100} \approx 1.27 \times 10^{30}$, while $100^2 = 10^4$.
$$\boxed{2^{100} \text{ is far greater.}}$$

Given: Total packets = 8.5 billion = $8.5 \times 10^9$.

Available digits: 0–9 (10 choices per position).

Concept: A code of $n$ digits can represent $10^n$ unique codes.

We need: $10^n \geq 8.5 \times 10^9$

Check $n = 10$: $10^{10} = 10,000,000,000 = 10$ billion > 8.5 billion ✓

Check $n = 9$: $10^9 = 1$ billion < 8.5 billion ✗

Therefore, the code should consist of $\boxed{10}$ digits.

With a 10-digit code, up to 10 billion unique IDs are possible, which is sufficient for 8.5 billion packets.

Given: $64 = 8^2 = 4^3$.

Finding other such numbers:
A number that is both a perfect square and a perfect cube must be a perfect sixth power.

Reason: If $N = a^2 = b^3$, then the prime factorisation of $N$ must have all exponents divisible by both 2 and 3, i.e., divisible by $\text{lcm}(2,3) = 6$.

So $N = k^6$ for some positive integer $k$.

Examples:
- $k = 1$: $1^6 = 1$ (both $1^2$ and $1^3$)
- $k = 2$: $2^6 = 64$ (both $8^2$ and $4^3$)
- $k = 3$: $3^6 = 729$ (both $27^2$ and $9^3$)
- $k = 4$: $4^6 = 4096$ (both $64^2$ and $16^3$)

General description: Numbers that are both perfect squares and perfect cubes are exactly the perfect sixth powers: $1, 64, 729, 4096, 15625, \ldots$, i.e., numbers of the form $k^6$ where $k$ is a positive integer.

Given:
- Code length = 5
- Characters available: digits 0–9 (10 options) + letters A–Z (26 options) = 36 options per position.

Concept: Each position in the code is independent, with 36 choices.

Total number of codes:
$$36 \times 36 \times 36 \times 36 \times 36 = 36^5$$

Calculation:
$$36^2 = 1296$$
$$36^3 = 1296 \times 36 = 46656$$
$$36^4 = 46656 \times 36 = 1679616$$
$$36^5 = 1679616 \times 36 = 60466176$$

$$\boxed{36^5 = 60{,}466{,}176 \text{ codes are possible.}}$$

Given: Sheep population $\approx 10^9$, Goat population $\approx 10^9$.

Total:
$$10^9 + 10^9 = 2 \times 10^9$$

Correct options: (v) $2 \times 10^9$ and (vi) $10^9 + 10^9$ — both represent the same value.

Justification for incorrect options:
- (i) $20^9 \neq 2 \times 10^9$ ($20^9$ is a much larger number)
- (ii) $10^{11} = 100 \times 10^9 \neq 2 \times 10^9$
- (iii) $10^{10} = 10 \times 10^9 \neq 2 \times 10^9$
- (iv) $10^{18}$ is the product, not the sum

$$\boxed{\text{Total} = 2 \times 10^9 \approx 2{,}000{,}000{,}000}$$

Assumptions used:
- World population $\approx 8 \times 10^9$ people
- 1 million = $10^6$; 1 billion = $10^9$; 1 trillion = $10^{12}$
- Average human lifespan $\approx 70$ years; time spent eating per day $\approx 1$ hour

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(i) Total pieces of clothing:

$$\text{Total} = 8 \times 10^9 \times 30 = 240 \times 10^9 = 2.4 \times 10^{11}$$

$$\boxed{2.4 \times 10^{11} \text{ pieces of clothing}}$$

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(ii) Number of honeybees:

$$\text{Bee colonies} = 100 \text{ million} = 10^8$$
$$\text{Bees per colony} = 50{,}000 = 5 \times 10^4$$
$$\text{Total bees} = 10^8 \times 5 \times 10^4 = 5 \times 10^{12}$$

$$\boxed{5 \times 10^{12} \text{ honeybees}}$$

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(iii) Total bacterial cells in all humans:

$$\text{Bacteria per human} = 38 \text{ trillion} = 3.8 \times 10^{13}$$
$$\text{World population} = 8 \times 10^9$$
$$\text{Total bacteria} = 3.8 \times 10^{13} \times 8 \times 10^9 = 30.4 \times 10^{22} = 3.04 \times 10^{23}$$

$$\boxed{3.04 \times 10^{23} \text{ bacterial cells}}$$

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(iv) Total time spent eating in a lifetime (in seconds):

Assumptions: Lifespan = 70 years; eating time = 1 hour/day.

$$\text{Eating time} = 70 \times 365 \times 1 \text{ hour} = 25550 \text{ hours}$$
$$= 25550 \times 3600 \text{ seconds} = 91{,}980{,}000 \text{ s} \approx 9.2 \times 10^7 \text{ seconds}$$

$$\boxed{\approx 9.2 \times 10^7 \text{ seconds}}$$

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

Convert $10^9$ seconds to years:

$$1 \text{ year} = 365 \times 24 \times 60 \times 60 = 31{,}536{,}000 \approx 3.15 \times 10^7 \text{ seconds}$$

$$\text{Number of years} = \frac{10^9}{3.15 \times 10^7} \approx \frac{10^9}{3.15 \times 10^7} \approx 31.7 \text{ years}$$

Conclusion: 1 billion seconds is approximately 31 years and 8 months.

If the current year is 2025:
$$2025 - 31.7 \approx 1993$$

1 billion seconds ago was approximately in the year 1993 (around April–May 1993, depending on the exact current date).

$$\boxed{\text{Approximately 31 years and 8 months ago} \approx \text{year } 1993}$$