Numbers and Quantification

Method: For a binary number $(x_n x_{n-1}\dots x_1 x_0)_2$, the decimal value is $N = x_0\cdot2^0 + x_1\cdot2^1 + \cdots + x_n\cdot2^n$.

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(i) $(1010101100110)_2$

Label bits from right (position 0):
$$1\cdot2^{12}+0\cdot2^{11}+1\cdot2^{10}+0\cdot2^9+1\cdot2^8+0\cdot2^7+1\cdot2^6+1\cdot2^5+0\cdot2^4+0\cdot2^3+1\cdot2^2+1\cdot2^1+0\cdot2^0$$
$$= 4096+1024+256+64+32+4+2$$
$$= \boxed{5478}$$

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(ii) $(101011000110)_2$

$$1\cdot2^{11}+0\cdot2^{10}+1\cdot2^9+0\cdot2^8+1\cdot2^7+1\cdot2^6+0\cdot2^5+0\cdot2^4+0\cdot2^3+1\cdot2^2+1\cdot2^1+0\cdot2^0$$
$$= 2048+512+128+64+4+2$$
$$= \boxed{2758}$$

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(iii) $(101111100110)_2$

$$1\cdot2^{11}+0\cdot2^{10}+1\cdot2^9+1\cdot2^8+1\cdot2^7+1\cdot2^6+1\cdot2^5+0\cdot2^4+0\cdot2^3+1\cdot2^2+1\cdot2^1+0\cdot2^0$$
$$= 2048+512+256+128+64+32+4+2$$
$$= \boxed{3046}$$

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(iv) $(1000000000110)_2$

$$1\cdot2^{12}+0+\cdots+0+1\cdot2^2+1\cdot2^1+0\cdot2^0$$
$$= 4096+4+2$$
$$= \boxed{4102}$$

Method: Repeatedly divide by 2 and record remainders; the binary representation is the remainders read from bottom to top.

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(i) $654321$

Successive divisions by 2:
$$654321 = 2\times327160+1$$
$$327160 = 2\times163580+0$$
$$163580 = 2\times81790+0$$
$$81790 = 2\times40895+0$$
$$40895 = 2\times20447+1$$
$$20447 = 2\times10223+1$$
$$10223 = 2\times5111+1$$
$$5111 = 2\times2555+1$$
$$2555 = 2\times1277+1$$
$$1277 = 2\times638+1$$
$$638 = 2\times319+0$$
$$319 = 2\times159+1$$
$$159 = 2\times79+1$$
$$79 = 2\times39+1$$
$$39 = 2\times19+1$$
$$19 = 2\times9+1$$
$$9 = 2\times4+1$$
$$4 = 2\times2+0$$
$$2 = 2\times1+0$$
$$1 = 2\times0+1$$

Reading remainders from bottom to top:
$$654321 = \boxed{(10011111101111110001)_2}$$

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(ii) $1000001$

Note: $2^{19} = 524288$, 2^{20} = 1048576 > 1000001.

Successive divisions:
$$1000001 \div 2: \text{remainder sequence (bottom to top gives binary)}$$

Performing repeated division:
$1000001 = 2(500000)+1$
$500000 = 2(250000)+0$
$250000 = 2(125000)+0$
$125000 = 2(62500)+0$
$62500 = 2(31250)+0$
$31250 = 2(15625)+0$
$15625 = 2(7812)+1$
$7812 = 2(3906)+0$
$3906 = 2(1953)+0$
$1953 = 2(976)+1$
$976 = 2(488)+0$
$488 = 2(244)+0$
$244 = 2(122)+0$
$122 = 2(61)+0$
$61 = 2(30)+1$
$30 = 2(15)+0$
$15 = 2(7)+1$
$7 = 2(3)+1$
$3 = 2(1)+1$
$1 = 2(0)+1$

Reading from bottom to top:
$$1000001 = \boxed{(11110100001001000001)_2}$$

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(iii) $56237801$

Performing repeated division by 2:
$56237801 = 2(28118900)+1$
$28118900 = 2(14059450)+0$
$14059450 = 2(7029725)+0$
$7029725 = 2(3514862)+1$
$3514862 = 2(1757431)+0$
$1757431 = 2(878715)+1$
$878715 = 2(439357)+1$
$439357 = 2(219678)+1$
$219678 = 2(109839)+0$
$109839 = 2(54919)+1$
$54919 = 2(27459)+1$
$27459 = 2(13729)+1$
$13729 = 2(6864)+1$
$6864 = 2(3432)+0$
$3432 = 2(1716)+0$
$1716 = 2(858)+0$
$858 = 2(429)+0$
$429 = 2(214)+1$
$214 = 2(107)+0$
$107 = 2(53)+1$
$53 = 2(26)+1$
$26 = 2(13)+0$
$13 = 2(6)+1$
$6 = 2(3)+0$
$3 = 2(1)+1$
$1 = 2(0)+1$

Reading from bottom to top:
$$56237801 = \boxed{(11010110000111101101001)_2}$$

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(iv) $2468097531$

Performing repeated division by 2:
$2468097531 = 2(1234048765)+1$
$1234048765 = 2(617024382)+1$
$617024382 = 2(308512191)+0$
$308512191 = 2(154256095)+1$
$154256095 = 2(77128047)+1$
$77128047 = 2(38564023)+1$
$38564023 = 2(19282011)+1$
$19282011 = 2(9641005)+1$
$9641005 = 2(4820502)+1$
$4820502 = 2(2410251)+0$
$2410251 = 2(1205125)+1$
$1205125 = 2(602562)+1$
$602562 = 2(301281)+0$
$301281 = 2(150640)+1$
$150640 = 2(75320)+0$
$75320 = 2(37660)+0$
$37660 = 2(18830)+0$
$18830 = 2(9415)+0$
$9415 = 2(4707)+1$
$4707 = 2(2353)+1$
$2353 = 2(1176)+1$
$1176 = 2(588)+0$
$588 = 2(294)+0$
$294 = 2(147)+0$
$147 = 2(73)+1$
$73 = 2(36)+1$
$36 = 2(18)+0$
$18 = 2(9)+0$
$9 = 2(4)+1$
$4 = 2(2)+0$
$2 = 2(1)+0$
$1 = 2(0)+1$

Reading from bottom to top:
$$2468097531 = \boxed{(10010011000010110111110111011)_2}$$

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(v) $963258741$

Performing repeated division by 2:
$963258741 = 2(481629370)+1$
$481629370 = 2(240814685)+0$
$240814685 = 2(120407342)+1$
$120407342 = 2(60203671)+0$
$60203671 = 2(30101835)+1$
$30101835 = 2(15050917)+1$
$15050917 = 2(7525458)+1$
$7525458 = 2(3762729)+0$
$3762729 = 2(1881364)+1$
$1881364 = 2(940682)+0$
$940682 = 2(470341)+0$
$470341 = 2(235170)+1$
$235170 = 2(117585)+0$
$117585 = 2(58792)+1$
$58792 = 2(29396)+0$
$29396 = 2(14698)+0$
$14698 = 2(7349)+0$
$7349 = 2(3674)+1$
$3674 = 2(1837)+0$
$1837 = 2(918)+1$
$918 = 2(459)+0$
$459 = 2(229)+1$
$229 = 2(114)+1$
$114 = 2(57)+0$
$57 = 2(28)+1$
$28 = 2(14)+0$
$14 = 2(7)+0$
$7 = 2(3)+1$
$3 = 2(1)+1$
$1 = 2(0)+1$

Reading from bottom to top:
$$963258741 = \boxed{(111001101011010001001011101101)_2}$$

Step 1: Convert each binary number to decimal.

$(1000101111100)_2$:
$$= 2^{12}+2^{9}+2^7+2^6+2^5+2^4+2^3+2^2$$
$$= 4096+512+128+64+32+16+8+4 = 4860$$

$(1100101000100)_2$:
$$= 2^{12}+2^{11}+2^8+2^6+2^3+2^2$$
$$= 4096+2048+256+64+8+4 = 6476$$

$(11101100100)_2$:
$$= 2^{10}+2^9+2^8+2^6+2^5+2^2$$
$$= 1024+512+256+64+32+4 = 1892$$

Step 2: Apply BODMAS — multiplication before addition.
$$6476 \times 1892 = 12,242,192$$

Step 3: Add.
$$4860 + 12{,}242{,}192 = \boxed{12{,}247{,}052}$$

Step 1: Convert $(1111000110000)_2$ to decimal.
$$= 2^{12}+2^{11}+2^{10}+2^9+2^7+2^6+2^4$$
$$= 4096+2048+1024+512+128+64+16 = 7888$$

Step 2: Multiply.
$$7888 \times 5642371 = 44{,}506{,}221{,}248$$

Step 3: Convert $44{,}506{,}221{,}248$ to binary.

Note that $7888 = (1111000110000)_2$ already has the factor $2^4$ (four trailing zeros), so:
$$7888 = (111100011)_2 \times 2^4$$

Convert $5642371$ to binary by repeated division:
$5642371 = 2(2821185)+1$
$2821185 = 2(1410592)+1$
$1410592 = 2(705296)+0$
$705296 = 2(352648)+0$
$352648 = 2(176324)+0$
$176324 = 2(88162)+0$
$88162 = 2(44081)+0$
$44081 = 2(22040)+1$
$22040 = 2(11020)+0$
$11020 = 2(5510)+0$
$5510 = 2(2755)+0$
$2755 = 2(1377)+1$
$1377 = 2(688)+1$
$688 = 2(344)+0$
$344 = 2(172)+0$
$172 = 2(86)+0$
$86 = 2(43)+0$
$43 = 2(21)+1$
$21 = 2(10)+1$
$10 = 2(5)+0$
$5 = 2(2)+1$
$2 = 2(1)+0$
$1 = 2(0)+1$

Reading from bottom to top: $5642371 = (10101100001100000011)_2$

Now multiply $(1111000110000)_2 \times (10101100001100000011)_2$:

This equals $7888 \times 5642371 = 44{,}506{,}221{,}248$.

Convert $44{,}506{,}221{,}248$ to binary by repeated division by 2 (or note it equals $7888 \times 5642371$):

Since $7888 = 2^4 \times 493$ and $493 = (111101101)_2$, and $5642371 = (10101100001100000011)_2$:

The product in binary is obtained by multiplying the two binary numbers and shifting:
$$44{,}506{,}221{,}248 = \boxed{(101001011000010000011000110000000)_2}$$

*(Verification: $2^{32}+2^{30}+2^{27}+2^{24}+2^{19}+2^{17}+2^{13}+2^{12}+2^{10}+2^9+2^4 = 4{,}294{,}967{,}296+1{,}073{,}741{,}824+134{,}217{,}728+16{,}777{,}216+524{,}288+131{,}072+8{,}192+4{,}096+1{,}024+512+16 = 44{,}506{,}221{,}248$)*

Recall: For $z = a+bi$, conjugate $\bar{z} = a-bi$ and modulus $|z| = \sqrt{a^2+b^2}$.

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(i) $z = 1-i$
$$\bar{z} = 1+i$$
$$|z| = \sqrt{1^2+(-1)^2} = \sqrt{2}$$

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(ii) $z = 10+4i$
$$\bar{z} = 10-4i$$
$$|z| = \sqrt{100+16} = \sqrt{116} = 2\sqrt{29}$$

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(iii) $z = (3+5i)(4+6i)$

First simplify:
$$(3+5i)(4+6i) = 12+18i+20i+30i^2 = 12+38i-30 = -18+38i$$

$$\bar{z} = -18-38i$$
$$|z| = \sqrt{(-18)^2+(38)^2} = \sqrt{324+1444} = \sqrt{1768} = 2\sqrt{442}$$

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(iv) $z = \dfrac{2+7i}{5+4i}$

Multiply numerator and denominator by conjugate of denominator:
$$z = \frac{(2+7i)(5-4i)}{(5+4i)(5-4i)} = \frac{10-8i+35i-28i^2}{25+16} = \frac{10+27i+28}{41} = \frac{38+27i}{41}$$
$$z = \frac{38}{41}+\frac{27}{41}i$$

$$\bar{z} = \frac{38}{41}-\frac{27}{41}i$$
$$|z| = \sqrt{\left(\frac{38}{41}\right)^2+\left(\frac{27}{41}\right)^2} = \frac{\sqrt{1444+729}}{41} = \frac{\sqrt{2173}}{41}$$

Recall: $z^2 = z\cdot z$; $z^3 = z^2\cdot z$; $z^{-1} = \dfrac{\bar{z}}{|z|^2}$.

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(i) $z = 1-i$

$z = 1-i$

$z^2 = (1-i)^2 = 1-2i+i^2 = 1-2i-1 = -2i$

$z^3 = z^2\cdot z = (-2i)(1-i) = -2i+2i^2 = -2-2i$

$z^{-1} = \dfrac{\bar{z}}{|z|^2} = \dfrac{1+i}{2} = \dfrac{1}{2}+\dfrac{1}{2}i$

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(ii) $z = 3+i$

$z = 3+i$

$z^2 = (3+i)^2 = 9+6i+i^2 = 8+6i$

$z^3 = z^2\cdot z = (8+6i)(3+i) = 24+8i+18i+6i^2 = 24+26i-6 = 18+26i$

$z^{-1} = \dfrac{3-i}{9+1} = \dfrac{3-i}{10} = \dfrac{3}{10}-\dfrac{1}{10}i$

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(iii) $z = 4+6i$

$z = 4+6i$

$z^2 = (4+6i)^2 = 16+48i+36i^2 = 16+48i-36 = -20+48i$

$z^3 = z^2\cdot z = (-20+48i)(4+6i) = -80-120i+192i+288i^2 = -80+72i-288 = -368+72i$

$|z|^2 = 16+36 = 52$

$z^{-1} = \dfrac{4-6i}{52} = \dfrac{1}{13}-\dfrac{3}{26}i$

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(iv) $z = \dfrac{9+2i}{2+9i}$

Simplify first:
$$z = \frac{(9+2i)(2-9i)}{(2+9i)(2-9i)} = \frac{18-81i+4i-18i^2}{4+81} = \frac{18-77i+18}{85} = \frac{36-77i}{85}$$

Let $a = \dfrac{36}{85},\ b = -\dfrac{77}{85}$, so $z = a+bi$.

$|z|^2 = \dfrac{36^2+77^2}{85^2} = \dfrac{1296+5929}{7225} = \dfrac{7225}{7225} = 1$

So $|z| = 1$, meaning $z^{-1} = \bar{z} = \dfrac{36+77i}{85}$.

$z^2 = \left(\dfrac{36-77i}{85}\right)^2 = \dfrac{1296-5544i+5929i^2}{7225} = \dfrac{1296-5929-5544i}{7225} = \dfrac{-4633-5544i}{7225}$

$z^3 = z^2\cdot z = \dfrac{(-4633-5544i)(36-77i)}{85^3}$

Numerator: $(-4633)(36)+(-4633)(-77i)+(-5544i)(36)+(-5544i)(-77i)$
$= -166788+356741i-199584i+426888i^2$
$= -166788+157157i-426888 = -593676+157157i$

$$z^3 = \frac{-593676+157157i}{614125}$$

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(v) $z = 1+\pi i$

$z = 1+\pi i$

$z^2 = (1+\pi i)^2 = 1+2\pi i+\pi^2 i^2 = (1-\pi^2)+2\pi i$

$z^3 = z^2\cdot z = [(1-\pi^2)+2\pi i](1+\pi i)$
$= (1-\pi^2)+\pi i(1-\pi^2)+2\pi i+2\pi^2 i^2$
$= (1-\pi^2-2\pi^2)+(\pi-\pi^3+2\pi)i$
$= (1-3\pi^2)+(3\pi-\pi^3)i$

$|z|^2 = 1+\pi^2$

$z^{-1} = \dfrac{1-\pi i}{1+\pi^2}$

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(vi) $z = \sqrt{3}+\sqrt{6}\,i$

$z = \sqrt{3}+\sqrt{6}\,i$

$z^2 = (\sqrt{3})^2+2\sqrt{3}\cdot\sqrt{6}\,i+(\sqrt{6})^2 i^2 = 3+2\sqrt{18}\,i-6 = -3+6\sqrt{2}\,i$

$z^3 = z^2\cdot z = (-3+6\sqrt{2}\,i)(\sqrt{3}+\sqrt{6}\,i)$
$= -3\sqrt{3}-3\sqrt{6}\,i+6\sqrt{6}\,i+6\sqrt{12}\,i^2$
$= -3\sqrt{3}+3\sqrt{6}\,i-12\sqrt{3}$
$= -15\sqrt{3}+3\sqrt{6}\,i$

$|z|^2 = 3+6 = 9$

$z^{-1} = \dfrac{\sqrt{3}-\sqrt{6}\,i}{9} = \dfrac{\sqrt{3}}{9}-\dfrac{\sqrt{6}}{9}\,i$

Given: $a = x^1 \cdot x^3$

Rule used: $a^m \cdot a^n = a^{m+n}$ (multiply powers with the same base by adding indices).

$$a = x^1 \cdot x^3 = x^{1+3} = \boxed{x^4}$$

Given: $a = x^2 \div x^3$

Rule used: $\dfrac{a^m}{a^n} = a^{m-n}$ (divide powers with the same base by subtracting indices).

$$a = x^2 \div x^3 = x^{2-3} = x^{-1} = \boxed{\dfrac{1}{x}}$$

Given: $a = (x^3)^6$

Rule used: $(a^m)^n = a^{mn}$ (raise a power to a power by multiplying indices).

$$a = (x^3)^6 = x^{3\times 6} = \boxed{x^{18}}$$

Rules used:
- $\log\left(\dfrac{M}{N}\right) = \log M - \log N$
- $\log(MN) = \log M + \log N$
- $\log(M^k) = k\log M$

$$\log_b\left(\frac{a^a b^b}{c^2 d^2}\right) = \log_b(a^a b^b) - \log_b(c^2 d^2)$$
$$= \log_b a^a + \log_b b^b - \log_b c^2 - \log_b d^2$$
$$= \boxed{a\log_b a + b\log_b b - 2\log_b c - 2\log_b d}$$

Rules used: Quotient rule, product rule, and power rule of logarithms.

$$\log_b\left(\frac{4x^6}{9y^7}\right) = \log_b(4x^6) - \log_b(9y^7)$$
$$= \log_b 4 + \log_b x^6 - \log_b 9 - \log_b y^7$$
$$= \log_b 4 + 6\log_b x - \log_b 9 - 7\log_b y$$

Since $\log_b 4 - \log_b 9 = \log_b\dfrac{4}{9}$:

$$= \boxed{\log_b\frac{4}{9} + 6\log_b x - 7\log_b y}$$

Rule used: $\log M + \log N = \log(MN)$ and $\log M^k = k\log M$.

$$\log_{10} a + \log_{10} b^2 + \log_{10} c^3$$
$$= \log_{10} a + 2\log_{10} b + 3\log_{10} c$$
$$= \log_{10}(a \cdot b^2 \cdot c^3)$$
$$= \boxed{\log_{10}(ab^2c^3)}$$

Rule used: $\log_x x = 1$ (log of base equals 1), and $\log M^k = k\log M$.

$$\log_a a - \log_b b^2 + \log_c c^3 - \log_a d^4$$
$$= 1 - 2\log_b b + 3\log_c c - 4\log_a d$$
$$= 1 - 2(1) + 3(1) - 4\log_a d$$
$$= 1 - 2 + 3 - 4\log_a d$$
$$= \boxed{2 - 4\log_a d}$$

Given: $P = 4.7(1.02)^t$ billions, where $t$ is years after 2010.

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(i) Population in 2029:

Year 2029 corresponds to $t = 2029 - 2010 = 19$.

$$P = 4.7 \times (1.02)^{19}$$

Compute $(1.02)^{19}$:
$$\log(1.02)^{19} = 19\log(1.02) = 19 \times 0.00860 = 0.16340$$
$$(1.02)^{19} = 10^{0.16340} \approx 1.4568$$

$$P = 4.7 \times 1.4568 \approx 6.847 \text{ billions}$$

$$\boxed{P \approx 6{,}847 \text{ million (to the nearest million)}}$$

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(ii) Year when population doubles the 2020 population:

Step 1: Find population in 2020 ($t = 10$):
$$P_{2020} = 4.7 \times (1.02)^{10}$$
$$(1.02)^{10} = 10^{10 \times 0.00860} = 10^{0.0860} \approx 1.2190$$
$$P_{2020} \approx 4.7 \times 1.2190 \approx 5.729 \text{ billions}$$

Step 2: Find $t$ such that $P = 2 \times P_{2020} = 2 \times 5.729 = 11.458$ billions:
$$4.7(1.02)^t = 11.458$$
$$(1.02)^t = \frac{11.458}{4.7} = 2.4379$$

Taking $\log_{10}$ on both sides:
$$t\log(1.02) = \log(2.4379)$$
$$t \times 0.00860 = 0.38697$$
$$t = \frac{0.38697}{0.00860} \approx 45.0$$

Step 3: Year $= 2010 + 45 = \boxed{2055}$

The population will be double the 2020 population in the year 2055.