\begin{align*} 534 \div 2 & = 267 \\ \\ 534 \text{ is a composite} & \text{ number since it is divisible by 2} \\ \\ \sqrt{1607} & \approx 40.087 \\ \\ \text{Since } 1607 \text{ is not divisible by } & \text{the prime numbers } 2, 3, ..., 31, 37, \text{ it is a prime number} \end{align*}

\begin{align*} 11, 17, 103, 43, 7, 127, 29, 71, 37, 53, 149, 61, 89, 5, 131, 11 \end{align*}

\begin{align*} p \times q & = \underbrace{31}_\text{Prime} \\ \\ p & = 1, q = 31 \\ \\ p + q & = 1 + 31 \\ & = 32 \end{align*}

\begin{align*} \text{Since } n(n+28) \text{ is prime, } & \text{one of its factors must be } 1. \\ \text{As } n+28 \ge 28, & n+28 \ne 1. \\ \\ \therefore n & = 1 \\ \\ n(n+28) & = 1 \times (1+28) \\ & = 29 \end{align*}

\begin{align*} 2 & | \underline{126} \\ 3 & | \underline{63} \\ 3 & | \underline{21} \\ 7 & | \underline{7} \\ & | 1 \end{align*}
$$ 126 = 2 \times 3^2 \times 7 $$

\begin{align*} 2 & | \underline{792} \\ 2 & | \underline{396} \\ 2 & | \underline{198} \\ 3 & | \underline{99} \\ 3 & | \underline{33} \\ 11 & | \underline{11} \\ & | 1 \end{align*}
$$ 792 = 2^3 \times 3^2 \times 11 $$

(i) Note: This question is a bit difficult - unlikely to be tested in exams

\begin{align*} 43 & | \underline{2021} \\ 47 & | \underline{47} \\ & | 1 \end{align*}

(ii)

\begin{align*} 2021 & = 1 \times 2021 \\ 2021 & = 43 \times 47 \\ \\ a = 1, b = 2021 & \text{ or } a = 43, b = 47 \end{align*}

\begin{align*} 195 & = 3 \times 5 \times 13 \\ \\ \text{Dimensions: } & 3 \text{ cm by 5 cm by 13 cm} \end{align*}

\begin{align*} 324 & = 2^2 \times 3^4 \\ \\ 18 \div 2 & = 9 \text{ cm} \phantom{0} [\text{This is the sum of length and breadth of top of cuboid}] \\ \\ 9 & = 1 + 8 \phantom{0} [\text{Not possible since each side is more than 2 cm}] \\ 9 & = 2 + 7 \phantom{0} [\text{Not possible}] \\ 9 & = 3 + 6 \phantom{0} [\text{Possible, since } 3 \times 6 = 18 \text{ is a factor of 324}] \\ 9 & = 4 + 5 \phantom{0} [\text{Not possible, since } 324 \div (4 \times 5) = 16.2 \text{ is not a whole number}] \\ \\ \text{Area of top} & = 3 \times 6 \\ & = 18 \text{ cm}^2 \\ \\ \text{Height} & = 324 \div 18 \\ & = 18 \text{ cm} \end{align*}

\begin{align*} 2 & | \underline{784} \\ 2 & | \underline{392} \\ 2 & | \underline{196} \\ 2 & | \underline{98} \\ 7 & | \underline{49} \\ 7 & | \underline{7} \\ & | 1 \\ \\ 784 & = 2^4 \times 7^2 \\ \\ \sqrt{784} & = \sqrt{2^4 \times 7^2} \\ & = 2^2 \times 7 \\ & = 28 \end{align*}

\begin{align*} \sqrt{7056} & = \sqrt{2^4 \times 3^2 \times 7^2} \\ & = 2^2 \times 3 \times 7 \\ & = 84 \end{align*}

\begin{align*} 2 & | \underline{2744} \\ 2 & | \underline{1372} \\ 2 & | \underline{686} \\ 7 & | \underline{343} \\ 7 & | \underline{49} \\ 7 & | \underline{7} \\ & | 1 \\ \\ 2744 & = 2^3 \times 7^3 \\ \\ \sqrt[3]{2744} & = \sqrt[3]{2^3 \times 7^3} \\ & = 2 \times 7 \\ & = 14 \end{align*}

\begin{align*} 9261 & = 3^3 \times 7^3 \\ \\ \sqrt[3]{9261} & = \sqrt[3]{3^3 \times 7^3} \\ & = 3 \times 7 \\ & = 21 \end{align*}

(i)

\begin{align*} 15 \times 135 & = (3 \times 5) \times (3^3 \times 5) \\ & = 3^4 \times 5^2 \\ \\ \text{Since the power of each prime } & \text{factor are multiples of 2, } 15 \times 135 \text{ is a perfect square} \end{align*}

(ii)

\begin{align*} 15 \times 135 & = 3^4 \times 5^2 \\ \\ 15 \times 135 \times k & = 3^4 \times 5^2 \times 3^2 \times 5 \\ & = \underbrace{3^6 \times 5^3}_\text{Powers are multiples of 3} \\ \\ k & = 3^2 \times 5 = 45 \end{align*}

(iii)

\begin{align*} 15 \times 135 & = 3^4 \times 5^2 \\ \\ 15 \times 135 \times {p \over q} & = 3^4 \times 5^2 \times {5 \over 3} \\ & = {3^4 \times 5^2 \over 1} \times {5 \over 3} \\ & = {3^4 \times 5^3 \over 3} \\ & = 3^3 \times 5^3 \\ \\ p & = 5, q = 3 \end{align*}

(i)

\begin{align*} 2016 & = 2^5 \times 3^2 \times 7 \end{align*}

(ii)

\begin{align*} 2016 & = 2^5 \times 3^2 \times 7 \\ \\ {2016 \over m} & = {2^5 \times 3^2 \times 7 \over 2 \times 7} \\ n^2 & = 2^4 \times 3^2 \\ n & = \sqrt{2^4 \times 3^2} \\ n & = 2^2 \times 3 \\ n & = 12 \\ \\ m & = 2 \times 7 = 14, n = 12 \end{align*}

(iii)

\begin{align*} 2016 & = 2^5 \times 3^2 \times 7 \\ \\ {2016 \over h} & = {2^5 \times 3^2 \times 7 \over 2^2 \times 3^2 \times 7} \\ & = \underbrace{2^3}_\text{Perfect cube} \\ \\ h & = 2^2 \times 3^2 \times 7 \\ h & = 252 \end{align*}

(a)

\begin{align*} \sqrt{123} & \approx \sqrt{121} \\ & = \sqrt{11^2} \\ & = 11 \end{align*}

(b)

\begin{align*} \sqrt[3]{123} & \approx \sqrt[3]{125} \\ & = \sqrt[3]{5^3} \\ & = 5 \end{align*}

(a)

\begin{align*} 23^2 + \sqrt{2025} - 7^3 & = 231 \end{align*}

(b)

\begin{align*} {3^2 \times \sqrt{20} \over 5^3 - \sqrt[3]{2013}} & \approx 0.35817 \\ & \approx 0.3582 \text{ (4 d.p.)} \end{align*}

\begin{align*} \text{Area of square} & = l \times l = l^2 \\ \\ l^2 & = 987 \\ l & = \sqrt{987} \\ \\ \text{Perimeter} & = 4 \times \sqrt{987} \\ & \approx 125.7 \text{ cm (to 1 d.p.)} \end{align*}

\begin{align*} \sqrt[3]{2020} & \approx 12.641 \\ \\ \text{Longest side} & = 12 \text{ cm} \\ \\ \text{Number of cubes used} & = 12 \times 12 \times 12 \\ & = 1728 \\ \\ \text{Number of leftover cubes} & = 2020 - 1728 \\ & = 292 \end{align*}

\begin{align*} 56 & = 2^3 \times 7 \\ \\ 84 & = 2^2 \times 3 \times 7 \\ \\ \text{HCF} & = 2^2 \times 7 \\ & = 28 \\ \\ \\ 2 & | \underline{56, 84} \\ 2 & | \underline{28, 42} \\ 7 & | \underline{14, 21} \\ & | 2, 3 \\ \\ \text{HCF} & = 2^2 \times 7 \\ & = 28 \end{align*}

\begin{align*} 112 & = 2^4 \times 7 \\ \\ 140 & = 2^2 \times 5 \times 7 \\ \\ \text{HCF} & = 2^2 \times 7 \\ & = 28 \end{align*}

\begin{align*} 504 & = 2^3 \times 3^2 \times 7 \\ \\ 588 & = 2^2 \times 3 \times 7^2 \\ \\ \text{HCF} & = 2^2 \times 3 \times 7 \\ & = 84 \\ \\ \text{The common prime factors are } & 2, 3 \text{ and } 7. \\ \text{Taking the smallest power of each gives } & 2^2 \times 3 \times 7 = 84, \\ \text{so } 84 \text{ is the greatest whole number } & \text{that divides both numbers exactly.} \end{align*}

\begin{align*} 90 & = 2 \times 3^2 \times 5 \\ \\ 135 & = 3^3 \times 5 \\ \\ 270 & = 2 \times 3^3 \times 5 \\ \\ \text{HCF} & = 3^2 \times 5 \\ & = 45 \end{align*}

\begin{align*} 24 & = 2^3 \times 3 \\ \\ 90 & = 2 \times 3^2 \times 5 \\ \\ \text{LCM} & = 2^3 \times 3^2 \times 5 \\ & = 360 \\ \\ 2 & | \underline{24, 90} \\ 3 & | \underline{12, 45} \\ & | \underline{4, 15} \\ \\ \text{LCM} & = 2 \times 3 \times 4 \times 15 \\ & = 360 \end{align*}

\begin{align*} 120 & = 2^3 \times 3 \times 5 \\ \\ 126 & = 2 \times 3^2 \times 7 \\ \\ \text{LCM} & = 2^3 \times 3^2 \times 5 \times 7 \\ & = 2520 \end{align*}

\begin{align*} 9 & = 3^2 \\ \\ 30 & = 2 \times 3 \times 5 \\ \\ 108 & = 2^2 \times 3^3 \\ \\ \text{LCM} & = 2^2 \times 3^3 \times 5 \\ & = 540 \end{align*}

\begin{align*} 27 & = 3^3 \\ \\ 45 & = 3^2 \times 5 \\ \\ \text{LCM} & = 3^3 \times 5 \\ & = 135 \\ \\ 135 & = 1 \times 135 \\ & = 3 \times 45 \\ & = 5 \times 27 \\ & = 9 \times 15 \\ \\ \text{Factors of 135: } & 1, 3, 5, 9, 15, 27, 45, 135 \end{align*}

\begin{align*} 4 & = 2^2 \\ \\ 20 & = 2^2 \times 5 \\ \\ \text{LCM} & = 2^2 \times 5 \\ & = 20 \\ \\ 20 & = 1 \times 20 \\ & = 2 \times 10 \\ & = 4 \times 5 \phantom{000000} [\text{Only 6 factors}] \\ \\ 20 \times 2 & = 40 \\ \\ 40 & = 1 \times 40 \\ & = 2 \times 20 \\ & = 4 \times 10 \\ & = 8 \times 5 \\ \\ \text{Factors of 40: } & 1, 2, 4, 5, 8, 10, 20, 40 \end{align*}

\begin{align*} \text{LCM} & = 3^2 \times 5 \\ \\ 15 & = 3 \times 5 \\ \\ n & = 3^2 = 9 \phantom{000000} [\text{Since LCM has the factor } 3^2 \text{ but 15 does not have the factor } 3^2] \end{align*}

\begin{align*} \text{LCM} & = 2^2 \times 3^2 \times 7 \\ \\ 9 & = 3^2 \\ \\ 12 & = 2^2 \times 3 \\ \\ k & = 7 \phantom{.} \checkmark \phantom{000000} [\text{Since LCM has the factor } 7 \text{ but not present in 9 and 12}] \\ \\ k & = 7 \times 2 = 14 \text{ [NA since } k \text{ is odd}] \\ \\ k & = 7 \times 3 = 21 \phantom{.} \checkmark \\ \\ k & = 7 \times 5 = 35 \text{ [NA since LCM does not have factor } 5] \\ \\ k & = 7 \times 7 = 49 \text{ [NA since LCM does not have factor } 7^2] \\ \\ k & = 7 \times 3^2 = 63 \phantom{.} \checkmark \\ \\ \therefore k & = 7, 21, 63 \end{align*}

\begin{align*} \text{HCF} & = 245 = 5 \times 7^2 \\ \\ \text{LCM} & = 4410 = 2 \times 3^2 \times 5 \times 7^2 \\ \\ x & = 2 \times 5 \times 7^2 \\ x & = 490 \\ \\ y & = 3^2 \times 5 \times 7^2 \\ y & = 2205 \end{align*}

(i)

\begin{align*} 240 & = 2^4 \times 3 \times 5 \\ \\ 252 & = 2^2 \times 3^2 \times 7 \\ \\ \text{LCM} & = 2^4 \times 3^2 \times 5 \times 7 \\ 240n & = 5040 \\ n & = 5040 \div 240 \\ n & = 21 \end{align*}

(ii)

\begin{align*} 240 & = 2^4 \times 3 \times 5 \\ \\ 252 & = 2^2 \times 3^2 \times 7 \\ \\ \text{HCF} & = 2^2 \times 3 \\ {240 \over m} & = 2^2 \times 3 \\ {2^4 \times 3 \times 5 \over m} & = 2^2 \times 3 \\ \\ m & = 2^2 \times 5 \\ m & = 20 \end{align*}

\begin{align*} 15 & = 3 \times 5 \\ \\ 16 & = 2^4 \\ \\ 36 & = 2^2 \times 3^2 \\ \\ \text{LCM} & = 2^4 \times 3^2 \times 5 \\ & = 720 \\ \\ 720 \text{ minutes} & = 720 \div 60 \\ & = 12 \text{ hours} \\ \\ \text{They will toll tog} & \text{ether next at 2.00 am} \end{align*}

(i)

\begin{align*} 140 & = 2^2 \times 5 \times 7 \\ \\ 168 & = 2^3 \times 3 \times 7 \\ \\ 210 & = 2 \times 3 \times 5 \times 7 \\ \\ \text{HCF} & = 2 \times 7 \\ & = 14 \\ \\ \text{Length} & = 14 \text{ cm} \end{align*}

(ii)

\begin{align*} \text{Total pieces} & = (140 \div 14) + (168 \div 14) + (210 \div 14) \\ & = 37 \end{align*}

(a)

\begin{align*} 87 \div 3 & = 29 \\ \\ 87 \text{ is a composite } & \text{number since it is divisible by 3} \end{align*}

(b)

\begin{align*} \sqrt{67} & \approx 8.185 \\ \\ \text{Since } 67 \text{ is not divisible by } & \text{the prime numbers } 2, 3, 5, 7, \text{ it is a prime number} \end{align*}

(c)

\begin{align*} \sqrt{73} & \approx 8.544 \\ \\ \text{Since } 73 \text{ is not divisible by } & \text{the prime numbers } 2, 3, 5, 7, \text{ it is a prime number} \end{align*}

(d)

\begin{align*} 91 \div 7 & = 13 \\ \\ 91 \text{ is a composite } & \text{number since it is divisible by 7} \end{align*}

\begin{align*} p \times q & = \underbrace{37}_\text{Prime} \\ \\ \text{Since 37 is a prime} & \text{ number, it has only two factors: 1 and itself} \\ \\ \implies p & = 1, q = 37 \\ \\ p + q & = 1 + 37 \\ & = 38 \end{align*}

\begin{align*} \text{Since } n(n + 42) \text{ is prime,} & \text{ one factor must be } 1 \\ \\ n + 42 & \ge 42 > 1 \\ \\ \therefore n & = 1 \\ \\ n(n + 42) & = 1 \times (1 + 42) \\ & = 43 \end{align*}

(a)

\begin{align*} 2 & | \underline{72} \\ 2 & | \underline{36} \\ 2 & | \underline{18} \\ 3 & | \underline{9} \\ 3 & | \underline{3} \\ & | 1 \end{align*}
$$ 72 = 2^3 \times 3^2 $$

(b)

\begin{align*} 2 & | \underline{756} \\ 2 & | \underline{378} \\ 3 & | \underline{189} \\ 3 & | \underline{63} \\ 3 & | \underline{21} \\ 7 & | \underline{7} \\ & | 1 \end{align*}
$$ 756 = 2^2 \times 3^3 \times 7 $$

(c)

\begin{align*} 11 & | \underline{187} \\ 17 & | \underline{17} \\ & | 1 \end{align*}
$$ 187 = 11 \times 17 $$

(d)

\begin{align*} 2 & | \underline{630} \\ 3 & | \underline{315} \\ 3 & | \underline{105} \\ 5 & | \underline{35} \\ 7 & | \underline{7} \\ & | 1 \end{align*}
$$ 630 = 2 \times 3^2 \times 5 \times 7 $$

(i)

$$ 2026 = 2 \times 1013 $$

(ii)

\begin{align*} 2026 & = 1 \times 2026 \\ 2026 & = 2 \times 1013 \\ \\ a = 1, b = 2026 & \text{ or } a = 2, b = 1013 \end{align*}

\begin{align*} 3 & | \underline{273} \\ 7 & | \underline{91} \\ 13 & | \underline{13} \\ & | 1 \\ \\ 273 & = 3 \times 7 \times 13 \\ \\ \text{Dimensions: } & 3 \text{ cm by 7 cm by 13 cm} \end{align*}

(a)

\begin{align*} 2012 \div 2 & = 1006 \\ 2013 \div 3 & = 671 \\ 2014 \div 2 & = 1007 \\ 2015 \div 5 & = 403 \\ 2016 \div 2 & = 1008 \\ 2018 \div 2 & = 1009 \\ 2019 \div 3 & = 673 \\ 2020 \div 2 & = 1010 \\ \\ \sqrt{2011}, \sqrt{2017} & < 45 \\ \\ \text{Neither 2011 nor 2017 is divisible} & \text{ by any of the primes} \\ & 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43 \\ \\ \therefore \text{Prime numbers: } & 2011, 2017 \end{align*}

(b)

\begin{align*} 2021 \div 43 & = 47 \\ 2022 \div 2 & = 1011 \\ 2023 \div 7 & = 289 \\ 2024 \div 2 & = 1012 \\ 2025 \div 5 & = 405 \\ 2026 \div 2 & = 1013 \\ 2028 \div 2 & = 1014 \\ 2030 \div 2 & = 1015 \\ \\ \sqrt{2027}, \sqrt{2029} & < 46 \\ \\ \text{Neither 2027 nor 2029 is divisible} & \text{ by any of the primes} \\ & 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43 \\ \\ \therefore \text{Prime numbers: } & 2027, 2029 \end{align*}

\begin{align*} \text{Since 2027 is a prime number, } & 1 \text{ and } 2027 \text{ are the only factors} \\ \\ a + b & = 1 + 2027 \\ & = 2028 \end{align*}

(a)

\begin{align*} 2 & | \underline{8624} \\ 2 & | \underline{4312} \\ 2 & | \underline{2156} \\ 2 & | \underline{1078} \\ 7 & | \underline{539} \\ 7 & | \underline{77} \\ 11 & | \underline{11} \\ & | 1 \end{align*}
$$ 8624 = 2^4 \times 7^2 \times 11 $$

(b)

\begin{align*} 2 & | \underline{6804} \\ 2 & | \underline{3402} \\ 3 & | \underline{1701} \\ 3 & | \underline{567} \\ 3 & | \underline{189} \\ 3 & | \underline{63} \\ 3 & | \underline{21} \\ 7 & | \underline{7} \\ & | 1 \end{align*}
$$ 6804 = 2^2 \times 3^5 \times 7 $$

(c)

\begin{align*} 3 & | \underline{26163} \\ 3 & | \underline{8721} \\ 3 & | \underline{2907} \\ 3 & | \underline{969} \\ 17 & | \underline{323} \\ 19 & | \underline{19} \\ & | 1 \end{align*}
$$ 26163 = 3^4 \times 17 \times 19 $$

(d)

\begin{align*} 2 & | \underline{196 \phantom{.} 000} \\ 2 & | \underline{98 \phantom{.} 000} \\ 2 & | \underline{49 \phantom{.} 000} \\ 2 & | \underline{24 \phantom{.} 500} \\ 2 & | \underline{12 \phantom{.} 250} \\ 5 & | \underline{6125} \\ 5 & | \underline{1225} \\ 5 & | \underline{245} \\ 7 & | \underline{49} \\ 7 & | \underline{7} \\ & | 1 \end{align*}
$$ 196 \phantom{.} 000 = 2^5 \times 5^3 \times 7^2 $$

\begin{align*} 2 & | \underline{2022} \\ 3 & | \underline{1011} \\ 337 & | \underline{337} \\ & | 1 \\ \\ 2022 & = 2 \times 3 \times 337 \\ \\ 2022 & = 1 \times 2022 \\ 2022 & = 2 \times 1011 \\ 2022 & = 3 \times 674 \\ 2022 & = 6 \times 337 \end{align*}

\begin{align*} 2 & | \underline{210} \\ 3 & | \underline{105} \\ 5 & | \underline{35} \\ 7 & | \underline{7} \\ & | 1 \\ \\ 210 & = 2 \times 3 \times 5 \times 7 \\ \\ \text{Dimension #1: } & 2 \text{ cm by 3 cm by 35 cm} \\ \text{Dimension #2: } & 2 \text{ cm by 15 cm by 7 cm} \\ \text{Dimension #3: } & 2 \text{ cm by 5 cm by 21 cm} \\ \text{Dimension #4: } & 6 \text{ cm by 5 cm by 7 cm} \\ \text{Dimension #5: } & 10 \text{ cm by 3 cm by 7 cm} \\ \text{Dimension #6: } & 14 \text{ cm by 3 cm by 5 cm} \end{align*}

\begin{align*} 2 & | \underline{504} \\ 2 & | \underline{252} \\ 2 & | \underline{126} \\ 3 & | \underline{63} \\ 3 & | \underline{21} \\ 7 & | \underline{7} \\ & | 1 \\ \\ 504 & = 2^3 \times 3^2 \times 7 \\ \\ 20 \div 2 & = 10 \text{ cm} \phantom{000} [\text{Sum of length and breadth of top}] \\ \\ 10 & = 3 + 7 \phantom{00000} [\text{Possible answer: Length} = 7, \text{ Breadth} = 3] \\ \\ \text{Height} & = {504 \over 3 \times 7} \\ & = 24 \text{ cm} \end{align*}

(a)

\begin{align*} 1764 & = 2^2 \times 3^2 \times 7^2 \\ \\ \sqrt{1764} & = \sqrt{2^2 \times 3^2 \times 7^2} \\ & = 2 \times 3 \times 7 \\ & = 42 \end{align*}

(b)

\begin{align*} 576 & = 2^6 \times 3^2 \\ \\ \sqrt{576} & = \sqrt{2^6 \times 3^2} \\ & = 2^3 \times 3 \\ & = 24 \end{align*}

(c)

\begin{align*} 2916 & = 2^2 \times 3^6 \\ \\ \sqrt{2916} & = \sqrt{2^2 \times 3^6} \\ & = 2 \times 3^3 \\ & = 54 \end{align*}

(d)

\begin{align*} 3136 & = 2^6 \times 7^2 \\ \\ \sqrt{3136} & = \sqrt{2^6 \times 7^2} \\ & = 2^3 \times 7 \\ & = 56 \end{align*}

(a)

\begin{align*} \sqrt{3^4 \times 11^2} & = 3^2 \times 11 \\ & = 99 \end{align*}

(b)

\begin{align*} \sqrt{3^6 \times 7^2} & = 3^3 \times 7 \\ & = 189 \end{align*}

(c)

\begin{align*} \sqrt{2^4 \times 3^2 \times 13^2} & = 2^2 \times 3 \times 13 \\ & = 156 \end{align*}

(d)

\begin{align*} \sqrt{2^8 \times 3^4 \times 5^2} & = 2^4 \times 3^2 \times 5 \\ & = 720 \end{align*}

(a)

\begin{align*} \sqrt[3]{3375} & = \sqrt[3]{3^3 \times 5^3} \\ & = 3 \times 5 \\ & = 15 \end{align*}

(b)

\begin{align*} \sqrt[3]{1728} & = \sqrt[3]{2^6 \times 3^3} \\ & = 2^2 \times 3 \\ & = 12 \end{align*}

(c)

\begin{align*} \sqrt[3]{5832} & = \sqrt[3]{2^3 \times 3^6} \\ & = 2 \times 3^2 \\ & = 18 \end{align*}

(d)

\begin{align*} \sqrt[3]{8000} & = \sqrt[3]{2^6 \times 5^3} \\ & = 2^2 \times 5 \\ & = 20 \end{align*}

(a)

\begin{align*} \sqrt[3]{2^6 \times 7^3} & = 2^2 \times 7 \\ & = 28 \end{align*}

(b)

\begin{align*} \sqrt[3]{2^6 \times 3^6} & = 2^2 \times 3^2 \\ & = 36 \end{align*}

(c)

\begin{align*} \sqrt[3]{2^3 \times 3^3 \times 11^3} & = 2 \times 3 \times 11 \\ & = 66 \end{align*}

(d)

\begin{align*} \sqrt[3]{2^9 \times 3^3 \times 5^3} & = 2^3 \times 3 \times 5 \\ & = 120 \end{align*}

(a)

\begin{align*} \sqrt{66} & \approx \sqrt{64} \\ & = \sqrt{8^2} \\ & = 8 \end{align*}

(b)

\begin{align*} \sqrt{80} & \approx \sqrt{81} \\ & = \sqrt{9^2} \\ & = 9 \end{align*}

(c)

\begin{align*} \sqrt[3]{218} & \approx \sqrt[3]{216} \\ & = \sqrt[3]{6^3} \\ & = 6 \end{align*}

(d)

\begin{align*} \sqrt[3]{730} & \approx \sqrt[3]{729} \\ & = \sqrt[3]{9^3} \\ & = 9 \end{align*}

(a)

\begin{align*} 7^2 - \sqrt{361} + 21^3 & = 9291 \end{align*}

(b)

\begin{align*} { \sqrt{555} + 5^2 \over 2^3 \times \sqrt[3]{222} } & \approx 1.002 \phantom{.} 437 \\ & \approx 1.0024 \text{ (to 4 d.p.)} \end{align*}

(c)

\begin{align*} \sqrt{ 4^3 + \sqrt[3]{4913} } & = 9 \end{align*}

(i)

\begin{align*} 112 & = 2^4 \times 7 \\ \\ 2^4 \times 7 \times 7 & = \underbrace{2^4 \times 7^2}_\text{Perfect square} \\ \\ \text{Smallest whole number} & = 7 \end{align*}

(ii)

\begin{align*} 112 & = 2^4 \times 7 \\ \\ 2^4 \times 7 \times 2^2 \times 7^2 & = \underbrace{2^6 \times 7^3}_\text{Perfect cube} \\ \\ \text{Smallest whole number} & = 2^2 \times 7^2 \\ & = 196 \end{align*}

\begin{align*} \sqrt{420} & \approx \sqrt{400} \\ & = \sqrt{20^2} \\ & = 20 \\ \\ 20 \times 21 & = 420 \\ \\ \text{Pages } & 20 \text{ and } 21 \end{align*}

\begin{align*} \text{Length of photo frame} & = \sqrt{250} \text{ cm} \\ & \approx 15.811 \text{ cm} \\ \\ \text{Perimeter of photo frame} & = 4 \times \sqrt{250} \\ & \approx 63.2 \text{ cm (to 1 d.p.)} \end{align*}

\begin{align*} \text{Length of box} & = \sqrt[3]{2197} \\ & = 13 \text{ cm} \\ \\ \text{Area of one side} & = 13 \times 13 \\ & = 169 \text{ cm}^2 \end{align*}

\begin{align*} 44^2 & < 2020 < 45^2 \\ \\ \text{Number of tiles used} & = 44 \times 44 \\ & = 1936 \\ \\ \text{Number of left over tiles} & = 2020 - 1936 \\ & = 84 \end{align*}

(i)

\begin{align*} 6 & = 2 \times 3 \\ \\ 54 & = 2 \times 3^3 \\ \\ 6 \times 54 & = 2^2 \times 3^4 \\ \\ \text{Since the powers of each prime factor} & \text{ are multiples of 2, } 6 \times 54 \text{ is a perfect square} \end{align*}

(ii)

\begin{align*} 6 \times 54 & = 2^2 \times 3^4 \\ \\ 6 \times 54 \times k & = 2^2 \times 3^4 \times 2 \times 3^2 \\ & = 2^3 \times 3^6 \\ \\ k & = 2 \times 3^2 \\ k & = 18 \end{align*}

(iii)

\begin{align*} 6 \times 54 & = 2^2 \times 3^4 \\ \\ 6 \times 54 \times {p \over q} & = {2^2 \times 3^4 \over 1} \times {2 \over 3} \\ & = {2^3 \times 3^4 \over 3} \\ & = 2^3 \times 3^3 \\ \\ p & = 2, q = 3 \end{align*}

(i)

\begin{align*} 4704 & = 2^5 \times 3 \times 7^2 \end{align*}

(ii)

\begin{align*} {4704 \over m} & = {2^5 \times 3 \times 7^2 \over m} \\ & = {2^5 \times 3 \times 7^2 \over 2 \times 3} \\ & = \underbrace{2^4 \times 7^2}_\text{Perfect square} \\ \\ m & = 2 \times 3 \\ m & = 6 \\ \\ n^2 & = 2^4 \times 7^2 \\ n & = \sqrt{2^4 \times 7^2} \\ n & = 2^2 \times 7 \\ n & = 28 \end{align*}

(iii)

\begin{align*} {4704 \over h} & = {2^5 \times 3 \times 7^2 \over 2^2 \times 3 \times 7^2} \\ & = \underbrace{2^3}_\text{Perfect cube} \\ \\ h & = 2^2 \times 3 \times 7^2 \\ & = 588 \end{align*}

(a)

\begin{align*} 12 & = 2^2 \times 3 \\ \\ 30 & = 2 \times 3 \times 5 \\ \\ \text{HCF} & = 2 \times 3 \\ & = 6 \end{align*}

(b)

\begin{align*} 13 & = 13 \\ \\ 91 & = 7 \times 13 \\ \\ \text{HCF} & = 13 \end{align*}

(c)

\begin{align*} 126 & = 2 \times 3^2 \times 7 \\ \\ 240 & = 2^4 \times 3 \times 5 \\ \\ \text{HCF} & = 2 \times 3 \\ & = 6 \end{align*}

(d)

\begin{align*} 180 & = 2^2 \times 3^2 \times 5 \\ \\ 450 & = 2 \times 3^2 \times 5^2 \\ \\ \text{HCF} & = 2 \times 3^2 \times 5 \\ & = 90 \end{align*}

(e)

\begin{align*} 11 & = 11 \\ \\ 31 & = 31 \\ \\ \text{Only com} & \text{mon factor is } 1 \end{align*}

(f)

\begin{align*} 64 & = 2^6 \\ \\ 81 & = 3^4 \\ \\ \text{Only com} & \text{mon factor is } 1 \end{align*}

\begin{align*} 156 & = 2^2 \times 3 \times 13 \\ \\ 168 & = 2^3 \times 3 \times 7 \\ \\ \text{HCF} & = 2^2 \times 3 \\ & = 12 \end{align*}

(a)

\begin{align*} 15 & = 3 \times 5 \\ \\ 60 & = 2^2 \times 3 \times 5 \\ \\ 75 & = 3 \times 5^2 \\ \\ \text{HCF} & = 3 \times 5 \\ & = 15 \end{align*}

(b)

\begin{align*} 360 & = 2^3 \times 3^2 \times 5 \\ \\ 540 & = 2^2 \times 3^3 \times 5 \\ \\ 882 & = 2 \times 3^2 \times 7^2 \\ \\ \text{HCF} & = 2 \times 3^2 \\ & = 18 \end{align*}

(c)

\begin{align*} 336 & = 2^4 \times 3 \times 7 \\ \\ 392 & = 2^3 \times 7^2 \\ \\ 504 & = 2^3 \times 3^2 \times 7 \\ \\ \text{HCF} & = 2^3 \times 7 \\ & = 56 \end{align*}

(d)

\begin{align*} 77 & = 7 \times 11 \\ \\ 91 & = 7 \times 13 \\ \\ 143 & = 11 \times 13 \\ \\ \text{Only com} & \text{mon factor is } 1 \end{align*}

(e)

\begin{align*} 7 & = 7 \\ \\ 17 & = 17 \\ \\ 23 & = 23 \\ \\ \text{Only com} & \text{mon factor is } 1 \end{align*}

(f)

\begin{align*} 252 & = 2^2 \times 3^2 \times 7 \\ \\ 588 & = 2^2 \times 3 \times 7^2 \\ \\ 840 & = 2^3 \times 3 \times 5 \times 7 \\ \\ \text{HCF} & = 2^2 \times 3 \times 7 \\ & = 84 \end{align*}

\begin{align*} 288 & = 2^5 \times 3^2 \\ \\ 336 & = 2^4 \times 3 \times 7 \\ \\ 480 & = 2^5 \times 3 \times 5 \\ \\ \text{HCF} & = 2^4 \times 3 \\ & = 48 \\ \\ \text{The common prime factors are } & 2 \text{ and } 3 \\ \text{Their smallest powers are } & 2^4 \text{ and } 3 \\ \therefore 48 & \text{ is the largest common factor} \end{align*}

(a)

\begin{align*} 3 & | \underline{45} \\ 3 & | \underline{15} \\ 5 & | \underline{5} \\ & | \underline{1} \\ \\ 2 & | \underline{60} \\ 2 & | \underline{30} \\ 3 & | \underline{15} \\ 5 & | \underline{5} \\ & | \underline{1} \\ \\ 45 & = 3^2 \times 5 \\ \\ 60 & = 2^2 \times 3 \times 5 \\ \\ \text{LCM} & = 2^2 \times 3^2 \times 5 \\ & = 180 \end{align*}

(b)

\begin{align*} 2 & | \underline{42} \\ 3 & | \underline{21} \\ 7 & | \underline{7} \\ & | \underline{1} \\ \\ 2 & | \underline{462} \\ 3 & | \underline{231} \\ 7 & | \underline{77} \\ 11 & | \underline{11} \\ & | \underline{1} \\ \\ 42 & = 2 \times 3 \times 7 \\ \\ 462 & = 2 \times 3 \times 7 \times 11 \\ \\ \text{LCM} & = 2 \times 3 \times 7 \times 11 \\ & = 462 \end{align*}

(c)

\begin{align*} 2 & | \underline{54} \\ 3 & | \underline{27} \\ 3 & | \underline{9} \\ 3 & | \underline{3} \\ & | \underline{1} \\ \\ 2 & | \underline{240} \\ 2 & | \underline{120} \\ 2 & | \underline{60} \\ 2 & | \underline{30} \\ 3 & | \underline{15} \\ 5 & | \underline{5} \\ & | \underline{1} \\ \\ 54 & = 2 \times 3^3 \\ \\ 240 & = 2^4 \times 3 \times 5 \\ \\ \text{LCM} & = 2^4 \times 3^3 \times 5 \\ & = 2160 \end{align*}

(d)

\begin{align*} 11 & | \underline{11} \\ & | \underline{1} \\ \\ 19 & | \underline{19} \\ & | \underline{1} \\ \\ 11 & = 11 \\ \\ 19 & = 19 \\ \\ \text{LCM} & = 11 \times 19 \\ & = 209 \end{align*}

(e)

\begin{align*} 3 & | \underline{27} \\ 3 & | \underline{9} \\ 3 & | \underline{3} \\ & | \underline{1} \\ \\ 2 & | \underline{32} \\ 2 & | \underline{16} \\ 2 & | \underline{8} \\ 2 & | \underline{4} \\ 2 & | \underline{2} \\ & | \underline{1} \\ \\ 27 & = 3^3 \\ \\ 32 & = 2^5 \\ \\ \text{LCM} & = 2^5 \times 3^3 \\ & = 864 \end{align*}

(f)

\begin{align*} 2 & | \underline{78} \\ 3 & | \underline{39} \\ 13 & | \underline{13} \\ & | \underline{1} \\ \\ 2 & | \underline{352} \\ 2 & | \underline{176} \\ 2 & | \underline{88} \\ 2 & | \underline{44} \\ 2 & | \underline{22} \\ 11 & | \underline{11} \\ & | \underline{1} \\ \\ 78 & = 2 \times 3 \times 13 \\ \\ 352 & = 2^5 \times 11 \\ \\ \text{LCM} & = 2^5 \times 3 \times 11 \times 13 \\ & = 13728 \end{align*}

\begin{align*} 2 & | \underline{80} \\ 2 & | \underline{40} \\ 2 & | \underline{20} \\ 2 & | \underline{10} \\ 5 & | \underline{5} \\ & | \underline{1} \\ \\ 2 & | \underline{104} \\ 2 & | \underline{52} \\ 2 & | \underline{26} \\ 13 & | \underline{13} \\ & | \underline{1} \\ \\ 80 & = 2^4 \times 5 \\ \\ 104 & = 2^3 \times 13 \\ \\ \text{LCM} & = 2^4 \times 5 \times 13 \\ & = 1040 \\ \\ \text{The prime factors needed are } & 2, 5 \text{ and } 13 \\ \text{Their highest powers are } & 2^4, 5 \text{ and } 13 \\ \text{Any common multiple must be divisible by } & 2^4 \times 5 \times 13 \\ \therefore 1040 & \text{ is the smallest non-zero common multiple} \end{align*}

(a)

\begin{align*} 2 & | \underline{12} \\ 2 & | \underline{6} \\ 3 & | \underline{3} \\ & | \underline{1} \\ \\ 2 & | \underline{18} \\ 3 & | \underline{9} \\ 3 & | \underline{3} \\ & | \underline{1} \\ \\ 3 & | \underline{81} \\ 3 & | \underline{27} \\ 3 & | \underline{9} \\ 3 & | \underline{3} \\ & | \underline{1} \\ \\ 12 & = 2^2 \times 3 \\ \\ 18 & = 2 \times 3^2 \\ \\ 81 & = 3^4 \\ \\ \text{LCM} & = 2^2 \times 3^4 \\ & = 324 \end{align*}

(b)

\begin{align*} 3 & | \underline{63} \\ 3 & | \underline{21} \\ 7 & | \underline{7} \\ & | \underline{1} \\ \\ 2 & | \underline{80} \\ 2 & | \underline{40} \\ 2 & | \underline{20} \\ 2 & | \underline{10} \\ 5 & | \underline{5} \\ & | \underline{1} \\ \\ 2 & | \underline{102} \\ 3 & | \underline{51} \\ 17 & | \underline{17} \\ & | \underline{1} \\ \\ 63 & = 3^2 \times 7 \\ \\ 80 & = 2^4 \times 5 \\ \\ 102 & = 2 \times 3 \times 17 \\ \\ \text{LCM} & = 2^4 \times 3^2 \times 5 \times 7 \times 17 \\ & = 85680 \end{align*}

(c)

\begin{align*} 3 & | \underline{3} \\ & | \underline{1} \\ \\ 7 & | \underline{7} \\ & | \underline{1} \\ \\ 11 & | \underline{11} \\ & | \underline{1} \\ \\ 3 & = 3 \\ \\ 7 & = 7 \\ \\ 11 & = 11 \\ \\ \text{LCM} & = 3 \times 7 \times 11 \\ & = 231 \end{align*}

(d)

\begin{align*} 5 & | \underline{25} \\ 5 & | \underline{5} \\ & | \underline{1} \\ \\ 3 & | \underline{27} \\ 3 & | \underline{9} \\ 3 & | \underline{3} \\ & | \underline{1} \\ \\ 2 & | \underline{32} \\ 2 & | \underline{16} \\ 2 & | \underline{8} \\ 2 & | \underline{4} \\ 2 & | \underline{2} \\ & | \underline{1} \\ \\ 25 & = 5^2 \\ \\ 27 & = 3^3 \\ \\ 32 & = 2^5 \\ \\ \text{LCM} & = 2^5 \times 3^3 \times 5^2 \\ & = 21600 \end{align*}

(e)

\begin{align*} 2 & | \underline{26} \\ 13 & | \underline{13} \\ & | \underline{1} \\ \\ 2 & | \underline{52} \\ 2 & | \underline{26} \\ 13 & | \underline{13} \\ & | \underline{1} \\ \\ 2 & | \underline{104} \\ 2 & | \underline{52} \\ 2 & | \underline{26} \\ 13 & | \underline{13} \\ & | \underline{1} \\ \\ 26 & = 2 \times 13 \\ \\ 52 & = 2^2 \times 13 \\ \\ 104 & = 2^3 \times 13 \\ \\ \text{LCM} & = 2^3 \times 13 \\ & = 104 \end{align*}

(f)

\begin{align*} 7 & | \underline{91} \\ 13 & | \underline{13} \\ & | \underline{1} \\ \\ 11 & | \underline{143} \\ 13 & | \underline{13} \\ & | \underline{1} \\ \\ 13 & | \underline{169} \\ 13 & | \underline{13} \\ & | \underline{1} \\ \\ 91 & = 7 \times 13 \\ \\ 143 & = 11 \times 13 \\ \\ 169 & = 13^2 \\ \\ \text{LCM} & = 7 \times 11 \times 13^2 \\ & = 13013 \end{align*}

\begin{align*} 3 & | \underline{315} \\ 3 & | \underline{105} \\ 5 & | \underline{35} \\ 7 & | \underline{7} \\ & | \underline{1} \\ \\ 2 & | \underline{378} \\ 3 & | \underline{189} \\ 3 & | \underline{63} \\ 3 & | \underline{21} \\ 7 & | \underline{7} \\ & | \underline{1} \\ \\ 2 & | \underline{392} \\ 2 & | \underline{196} \\ 2 & | \underline{98} \\ 7 & | \underline{49} \\ 7 & | \underline{7} \\ & | \underline{1} \\ \\ 315 & = 3^2 \times 5 \times 7 \\ \\ 378 & = 2 \times 3^3 \times 7 \\ \\ 392 & = 2^3 \times 7^2 \\ \\ \text{LCM} & = 2^3 \times 3^3 \times 5 \times 7^2 \\ & = 52\,920 \end{align*}

\begin{align*} 2 & | \underline{10} \\ 5 & | \underline{5} \\ & | \underline{1} \\ \\ 5 & | \underline{125} \\ 5 & | \underline{25} \\ 5 & | \underline{5} \\ & | \underline{1} \\ \\ 10 & = 2 \times 5 \\ \\ 125 & = 5^3 \\ \\ \text{LCM} & = 2 \times 5^3 \\ & = 250 \\ \\ 250 & = 1 \times 250 \\ 250 & = 2 \times 125 \\ 250 & = 5 \times 50 \\ 250 & = 10 \times 25 \\ \\ \text{Factors of 250} & : 1, 2, 5, 10, 25, 50, 125, 250 \end{align*}

\begin{align*} 2 & | \underline{40} \\ 2 & | \underline{20} \\ 2 & | \underline{10} \\ 5 & | \underline{5} \\ & | \underline{1} \\ \\ 2 & | \underline{100} \\ 2 & | \underline{50} \\ 5 & | \underline{25} \\ 5 & | \underline{5} \\ & | \underline{1} \\ \\ 40 & = 2^3 \times 5 \\ \\ 100 & = 2^2 \times 5^2 \\ \\ \text{LCM} & = 2^3 \times 5^2 \\ & = 200 \\ \\ 200 & = 1 \times 200 \\ 200 & = 2 \times 100 \\ 200 & = 4 \times 50 \\ 200 & = 5 \times 40 \\ 200 & = 8 \times 25 \\ 200 & = 10 \times 20 \\ \\ \text{Factors of 200} & : 1, 2, 4, 5, 8, 10, 20, 25, 40, 50, 100, 200 \end{align*}

\begin{align*} 2 & | \underline{6} \\ 3 & | \underline{3} \\ & | \underline{1} \\ \\ 2 & | \underline{60} \\ 2 & | \underline{30} \\ 3 & | \underline{15} \\ 5 & | \underline{5} \\ & | \underline{1} \\ \\ \text{LCM} = 60 & = 2^2 \times 3 \times 5 \\ \\ 6 & = 2 \times 3 \\ \\ \text{Comparing } 6 \text{ and LCM, } & k \text{ must minimally contain the factors } 2^2 \text{ and } 5 \\ \text{The factor } 3 & \text{ is already supplied by } 6 \\ \\ \text{Smallest value of } k & = 2^2 \times 5 \\ & = 20 \end{align*}

\begin{align*} 792 & = 2^3 \times 3^2 \times 11 \\ \\ 990 & = 2 \times 3^2 \times 5 \times 11 \\ \\ \text{HCF} & = 2 \times 3^2 \times 11 \\ & = 198 \\ \\ \text{The common} & \text{ prime factors are 2, 3 and 11} \\ \text{Their small} & \text{est powers are } 2, 3^2 \text{ and } 11 \\ \therefore 198 \text{ is the } & \text{largest common factor} \end{align*}

\begin{align*} 1080 & = 2^3 \times 3^3 \times 5 \\ \\ 1188 & = 2^2 \times 3^3 \times 11 \\ \\ 1815 & = 3 \times 5 \times 11^2 \\ \\ \text{HCF} & = 3 \end{align*}

\begin{align*} 176 & = 2^4 \times 11 \\ \\ 342 & = 2 \times 3^2 \times 19 \\ \\ \text{LCM} & = 2^4 \times 3^2 \times 11 \times 19 \\ & = 30 \, 096 \end{align*}

\begin{align*} 1120 & = 2^5 \times 5 \times 7 \\ \\ 1276 & = 2^2 \times 11 \times 29 \\ \\ 1450 & = 2 \times 5^2 \times 29 \\ \\ \text{LCM} & = 2^5 \times 5^2 \times 7 \times 11 \times 29 \\ & = 1 \, 786 \, 400 \end{align*}

\begin{align*} 2 & | \underline{4} \\ 2 & | \underline{2} \\ & | \underline{1} \\ \\ 2 & | \underline{26} \\ 13 & | \underline{13} \\ & | \underline{1} \\ \\ 4 & = 2^2 \\ \\ 26 & = 2 \times 13 \\ \\ \text{LCM} & = 2^2 \times 13 \\ & = 52 \\ \\ 52 & = 1 \times 52 \\ 52 & = 2 \times 26 \\ 52 & = 4 \times 13 \\ \\ \text{Factors of 52} & : 1, 2, 4, 13, 26, 52 \\ \\ \text{52 has only 6 factors, so try } & 2 \times 52 = 104 \\ \\ 104 & = 1 \times 104 \\ 104 & = 2 \times 52 \\ 104 & = 4 \times 26 \\ 104 & = 8 \times 13 \\ \\ \text{Factors of 104} & : 1, 2, 4, 8, 13, 26, 52, 104 \end{align*}

\begin{align*} 2 & | \underline{28} \\ 2 & | \underline{14} \\ 7 & | \underline{7} \\ & | \underline{1} \\ \\ 7 & | \underline{49} \\ 7 & | \underline{7} \\ & | \underline{1} \\ \\ 2 & | \underline{588} \\ 2 & | \underline{294} \\ 3 & | \underline{147} \\ 7 & | \underline{49} \\ 7 & | \underline{7} \\ & | \underline{1} \\ \\ \text{LCM} = 588 & = 2^2 \times 3 \times 7^2 \\ \\ 28 & = 2^2 \times 7 \\ \\ 49 & = 7^2 \\ \\ \text{The factors } 2^2 \text{ and } 7^2 & \text{ in 588 are provided by 28 and 49 respectively} \\ \therefore n & \text{ must have the factor } 3 \\ \\ n & = 3 \\ \\ n & = 2 \times 3 \\ n & = 6 \text{ (Reject since } n \text{ is odd}) \\ \\ n & = 2^2 \times 3 \\ n & = 12 \text{ (Reject since } n \text{ is odd}) \\ \\ n & = 3 \times 7 \\ n & = 21 \\ \\ n & = 3 \times 7^2 \\ n & = 147 \\ \\ \text{Possible values of } n : \, & 3, 21, 147 \end{align*}

\begin{align*} 5 & | \underline{175} \\ 5 & | \underline{35} \\ 7 & | \underline{7} \\ & | \underline{1} \\ \\ 2 & | \underline{12 \, 600} \\ 2 & | \underline{6 \, 300} \\ 2 & | \underline{3 \, 150} \\ 3 & | \underline{1 \, 575} \\ 3 & | \underline{525} \\ 5 & | \underline{175} \\ 5 & | \underline{35} \\ 7 & | \underline{7} \\ & | \underline{1} \\ \\ \text{HCF} = 175 & = 5^2 \times 7 \\ \\ \text{LCM} = 12 \, 600 & = 2^3 \times 3^2 \times 5^2 \times 7 \\ \\ \text{Since HCF and LCM contains the } & \text{ } 5^2 \text{ and } 7, \text{ both numbers must have the factors } 5^2 \text{ and } 7 \\ \\ \text{The factors } 2^3 \text{ and } 3^2 & \text{ must each occur entirely in one number} \\ \text{Neither } 2 \text{ nor } 3 & \text{ can divide both numbers,} \\ & \text{otherwise the HCF would exceed } 175 \\ \text{Putting both powers in one number gives } & 175 \text{ and } 12\,600 \\ \text{This is rejected because } & \text{both numbers must be greater than } 175 \\ \therefore 2^3 \text{ and } 3^2 & \text{ must go to different numbers} \\ \\ \text{First number} & = 2^3 \times 5^2 \times 7 \\ & = 1400 \\ \\ \text{Second number} & = 3^2 \times 5^2 \times 7 \\ & = 1575 \end{align*}

(i)

\begin{align*} 2 & | \underline{1050} \\ 3 & | \underline{525} \\ 5 & | \underline{175} \\ 5 & | \underline{35} \\ 7 & | \underline{7} \\ & | \underline{1} \\ \\ 1050 & = 2 \times 3 \times 5^2 \times 7 \end{align*}

(ii)

\begin{align*} \text{HCF} = 21 & = 3 \times 7 \\ \\ \text{LCM} = 1050 & = 2 \times 3 \times 5^2 \times 7 \\ \\ \text{Since HCF and LCM contains the } & \text{ } 3 \text{ and } 7, \text{ both numbers must have the factors } 3 \text{ and } 7 \\ \\ \text{The factors } 2 \text{ and } 5^2 & \text{ must each occur entirely in one number} \\ \text{Neither } 2 \text{ nor } 5 & \text{ can divide both numbers,} \\ & \text{otherwise the HCF would exceed } 21 \\ \text{Putting both factors in one number gives } & 21 \text{ and } 1050 \\ \text{This is rejected because } & \text{both numbers must be greater than } 40 \\ \therefore 2 \text{ and } 5^2 & \text{ must go to different numbers} \\ \\ \text{First number} & = 2 \times 3 \times 7 \\ & = 42 \\ \\ \text{Second number} & = 3 \times 5^2 \times 7 \\ & = 525 \end{align*}

(i)

\begin{align*} 3 & | \underline{171} \\ 3 & | \underline{57} \\ 19 & | \underline{19} \\ & | \underline{1} \\ \\ 3 & | \underline{63} \\ 3 & | \underline{21} \\ 7 & | \underline{7} \\ & | \underline{1} \\ \\ 3 & | \underline{27} \\ 3 & | \underline{9} \\ 3 & | \underline{3} \\ & | \underline{1} \\ \\ 171 & = 3^2 \times 19 \\ \\ 63 & = 3^2 \times 7 \\ \\ 27 & = 3^3 \\ \\ \text{Largest number of gift bags} & = \text{HCF of } 171, 63, 27 \\ & = 3^2 \\ & = 9 \end{align*}

(ii)

\begin{align*} \text{No. of pens} & = 171 \div 9 \\ & = 19 \\ \\ \text{No. of pencils} & = 63 \div 9 \\ & = 7 \\ \\ \text{No. of erasers} & = 27 \div 9 \\ & = 3 \end{align*}

\begin{align*} 2 & | \underline{120} \\ 2 & | \underline{60} \\ 2 & | \underline{30} \\ 3 & | \underline{15} \\ 5 & | \underline{5} \\ & | \underline{1} \\ \\ 2 & | \underline{18} \\ 3 & | \underline{9} \\ 3 & | \underline{3} \\ & | \underline{1} \\ \\ 120 & = 2^3 \times 3 \times 5 \\ \\ 18 & = 2 \times 3^2 \\ \\ \text{Least no. of sweets} & = \text{LCM of } 120, 18 \\ & = 2^3 \times 3^2 \times 5 \\ & = 360 \\ \\ \text{Least no. of packs of sweets} & = 360 \div 120 \\ & = 3 \end{align*}

(i)

\begin{align*} 2 & | \underline{60} \\ 2 & | \underline{30} \\ 3 & | \underline{15} \\ 5 & | \underline{5} \\ & | \underline{1} \\ \\ 2 & | \underline{80} \\ 2 & | \underline{40} \\ 2 & | \underline{20} \\ 2 & | \underline{10} \\ 5 & | \underline{5} \\ & | \underline{1} \\ \\ 60 & = 2^2 \times 3 \times 5 \\ \\ 80 & = 2^4 \times 5 \\ \\ \text{LCM} & = 2^4 \times 3 \times 5 \\ & = 240 \\ \\ \text{Time taken} & = 240 \text{ seconds} \end{align*}

(ii)

\begin{align*} \text{Laps completed by car X in 240 seconds} & = 240 \div 60 \\ & = 4 \\ \\ \text{Laps completed by car Y in 240 seconds} & = 240 \div 80 \\ & = 3 \\ \\ [ \text{This implies that for every 240 seconds, } & \text{car X completes one more lap than car Y}] \\ \\ \text{Time taken for car X to be 5 laps ahead} & = 240 \times 5 \\ & = 1200 \text{ seconds} \\ & = (1200 \div 60) \text{ minutes} \\ & = 20 \text{ minutes} \end{align*}

(i)

\begin{align*} 5 & | \underline{65} \\ 13 & | \underline{13} \\ & | \underline{1} \\ \\ 2 & | \underline{50} \\ 5 & | \underline{25} \\ 5 & | \underline{5} \\ & | \underline{1} \\ \\ 65 & = 5 \times 13 \\ \\ 50 & = 2 \times 5^2 \\ \\ \text{HCF} & = 5 \\ \\ \text{Length} & = 5 \text{ cm} \end{align*}

(ii)

\begin{align*} \text{No. of squares along the length} & = 65 \div 5 \\ & = 13 \\ \\ \text{No. of squares along the breadth} & = 50 \div 5 \\ & = 10 \\ \\ \text{No. of squares cut} & = 13 \times 10 \\ & = 130 \end{align*}

\begin{align*} 2 & | \underline{8} \\ 2 & | \underline{4} \\ 2 & | \underline{2} \\ & | \underline{1} \\ \\ 2 & | \underline{12} \\ 2 & | \underline{6} \\ 3 & | \underline{3} \\ & | \underline{1} \\ \\ 8 & = 2^3 \\ \\ 12 & = 2^2 \times 3 \\ \\ \text{LCM} & = 2^3 \times 3 \\ & = 24 \\ \\ \text{The required numbers must be } & \text{multiples of } 24 \\ \\ 24 & = 1 \times 24 \\ 24 & = 2 \times 12 \\ 24 & = 3 \times 8 \\ 24 & = 4 \times 6 \\ \\ \text{Factors of 24} & : 1, 2, 3, 4, 6, 8, 12, 24 \\ \\ \text{24 has only 8 factors, so try } & 5 \times 24 = 120 \\ \\ 120 & = 1 \times 120 \\ 120 & = 2 \times 60 \\ 120 & = 3 \times 40 \\ 120 & = 4 \times 30 \\ 120 & = 5 \times 24 \\ 120 & = 6 \times 20 \\ 120 & = 8 \times 15 \\ 120 & = 10 \times 12 \\ \\ \text{Factors of 120} & : 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120 \\ \\ \text{For another number, try } & 7 \times 24 = 168 \\ \\ 168 & = 1 \times 168 \\ 168 & = 2 \times 84 \\ 168 & = 3 \times 56 \\ 168 & = 4 \times 42 \\ 168 & = 6 \times 28 \\ 168 & = 7 \times 24 \\ 168 & = 8 \times 21 \\ 168 & = 12 \times 14 \\ \\ \text{Factors of 168} & : 1, 2, 3, 4, 6, 7, 8, 12, 14, 21, 24, 28, 42, 56, 84, 168 \\ \\ \text{Both numbers have exactly } & 16 \text{ factors, including } 8 \text{ and } 12 \\ \therefore \text{Two possible numbers are } & 120 \text{ and } 168 \end{align*}

\begin{align*} 7 & | \underline{49} \\ 7 & | \underline{7} \\ & | \underline{1} \\ \\ 3 & | \underline{63} \\ 3 & | \underline{21} \\ 7 & | \underline{7} \\ & | \underline{1} \\ \\ 2 & | \underline{882} \\ 3 & | \underline{441} \\ 3 & | \underline{147} \\ 7 & | \underline{49} \\ 7 & | \underline{7} \\ & | \underline{1} \\ \\ \text{LCM} = 882 & = 2 \times 3^2 \times 7^2 \\ \\ 49 & = 7^2 \\ \\ 63 & = 3^2 \times 7 \\ \\ \text{The factors } 3^2 \text{ and } 7^2 & \text{ in 882 are provided by 63 and 49 respectively} \\ \therefore m & \text{ must have the factor } 2 \\ \text{Also, } m & \text{ must be a factor of } 882 \\ \\ m & = 2 \\ \\ m & = 2 \times 3 \\ m & = 6 \\ \\ m & = 2 \times 7 \\ m & = 14 \\ \\ m & = 2 \times 3^2 \\ m & = 18 \\ \\ m & = 2 \times 3 \times 7 \\ m & = 42 \\ \\ m & = 2 \times 7^2 \\ m & = 98 \\ \\ \text{Six possible values of } m : \, & 2, 6, 14, 18, 42, 98 \end{align*}

(i)

\begin{align*} \text{LCM of 528 and 540} & = 2^4 \times 3^3 \times 5 \times 11 \\ & = 23 \, 760 \\ \\ 528h & = 23 \, 760 \\ h & = 23 \, 760 \div 528 \\ h & = 45 \end{align*}

(ii)

\begin{align*} \text{HCF of 528 and 540} & = 2^2 \times 3 \\ & = 12 \\ \\ \frac{528}{k} & = 12 \\ \\ k & = 528 \div 12 \\ k & = 44 \end{align*}

(i)

\begin{align*} \text{LCM of 630 and 1248} & = 2^5 \times 3^2 \times 5 \times 7 \times 13 \\ & = 131 \, 040 \\ \\ 630n & = 131 \, 040 \\ n & = 131 \, 040 \div 630 \\ n & = 208 \end{align*}

(ii)

\begin{align*} \text{HCF of 630 and 1248} & = 2 \times 3 \\ & = 6 \\ \\ \frac{1248}{m} & = 6 \\ \\ m & = 1248 \div 6 \\ m & = 208 \end{align*}

(i)

\begin{align*} 3 & | \underline{15} \\ 5 & | \underline{5} \\ & | \underline{1} \\ \\ 15 & = 3 \times 5 \\ \\ \text{Total number of girls and form teacher} & = 21 \\ \\ 3 & | \underline{21} \\ 7 & | \underline{7} \\ & | \underline{1} \\ \\ 21 & = 3 \times 7 \\ \\ \text{LCM of 15 and 21} & = 3 \times 5 \times 7 \\ & = 105 \\ \\ \text{No. of chocolate bars} & = 105 \\ \\ \text{No. of boys} & = 105 \div 15 \\ & = 7 \\ \\ \text{Total no. of students} & = 20 + 7 \\ & = 27 \text{ (Reject)} \\ \\ \\ \text{No. of chocolate bars} & = 105 \times 2 \\ & = 210 \\ \\ \text{No. of boys} & = 210 \div 15 \\ & = 14 \\ \\ \text{Total no. of students} & = 20 + 14 \\ & = 34 \end{align*}

(ii)

\begin{align*} \text{No. of chocolate bars form teacher received} & = 210 \div 21 \\ & = 10 \end{align*}

(a)

\begin{align*} 753 \div 3 & = 251 \\ \\ 753 \text{ is a composite} & \text{ number since it is divisible by 3} \end{align*}

(b)

\begin{align*} 27^2 & < 757 < 28^2 \\ \\ \text{Only prime divisors less than } 28 & \text{ need to be checked.} \\ \text{These primes are } & 2, 3, 5, 7, 11, 13, 17, 19, 23. \\ \\ 757 & \text{ is not divisible by any of these primes.} \\ \therefore 757 & \text{ is prime.} \end{align*}

(a)

\begin{align*} & \begin{array}{r|r} 5 & \underline{1225} \\ 5 & \underline{245} \\ 7 & \underline{49} \\ 7 & \underline{7} \\ & 1 \end{array} \\ \\ \sqrt{1225} & = \sqrt{ 5^2 \times 7^2 } \\ & = 5 \times 7 \\ & = 35 \end{align*}

(b)

\begin{align*} 13 \, 824 & = 2^9 \times 3^3 \\ \\ \sqrt[3]{13 \, 824} & = \sqrt[3]{2^9 \times 3^3} \\ & = 2^3 \times 3 \\ & = 24 \end{align*}

(a)

\begin{align*} \sqrt{63} & \approx \sqrt{64} \\ & = 8 \end{align*}

(b)

\begin{align*} \sqrt[3]{345} & \approx \sqrt[3]{343} \\ & = 7 \end{align*}

(i)

\begin{align*} \text{HCF of 840 and 8316} & = 2^2 \times 3 \times 7 \\ & = 84 \end{align*}

(ii)

\begin{align*} \text{LCM of 840 and 8316} & = 2^3 \times 3^3 \times 5 \times 7 \times 11 \\ & = 83 \, 160 \end{align*}

(i)

\begin{align*} & \begin{array}{r|r} 2 & \underline{108} \\ 2 & \underline{54} \\ 3 & \underline{27} \\ 3 & \underline{9} \\ 3 & \underline{3} \\ & 1 \end{array} \\ \\ 108 & = 2^2 \times 3^3 \\ \\ & \begin{array}{r|r} 3 & \underline{81} \\ 3 & \underline{27} \\ 3 & \underline{9} \\ 3 & \underline{3} \\ & 1 \end{array} \\ \\ 81 & = 3^4 \\ \\ & \begin{array}{r|r} 2 & \underline{54} \\ 3 & \underline{27} \\ 3 & \underline{9} \\ 3 & \underline{3} \\ & 1 \end{array} \\ \\ 54 & = 2 \times 3^3 \\ \\ \text{HCF} & = 3^3 \\ & = 27 \\ \\ \text{Largest number of baskets} & = 27 \end{align*}

(ii)

\begin{align*} \text{Roses per basket} & = 108 \div 27 \\ & = 4 \text{ stalks} \\ \\ \text{Lilies per basket} & = 81 \div 27 \\ & = 3 \text{ stalks} \\ \\ \text{Orchids per basket} & = 54 \div 27 \\ & = 2 \text{ stalks} \end{align*}

\begin{align*} 15 \text{ minutes} & = (15 \times 60) \text{ seconds} \\ & = 900 \text{ seconds} \\ \\ & \begin{array}{r|r} 2 & \underline{900} \\ 2 & \underline{450} \\ 3 & \underline{225} \\ 3 & \underline{75} \\ 5 & \underline{25} \\ 5 & \underline{5} \\ & 1 \end{array} \\ \\ 900 & = 2^2 \times 3^2 \times 5^2 \\ \\ & \begin{array}{r|r} 2 & \underline{360} \\ 2 & \underline{180} \\ 2 & \underline{90} \\ 3 & \underline{45} \\ 3 & \underline{15} \\ 5 & \underline{5} \\ & 1 \end{array} \\ \\ 360 & = 2^3 \times 3^2 \times 5 \\ \\ 6\div 2 \text{ minutes} & = 3 \text{ minutes} \\ 3 \text{ minutes} & = (3 \times 60) \text{ seconds} \\ & = 180 \text{ seconds} \\ \\ & \begin{array}{r|r} 2 & \underline{180} \\ 2 & \underline{90} \\ 3 & \underline{45} \\ 3 & \underline{15} \\ 5 & \underline{5} \\ & 1 \end{array} \\ \\ 180 & = 2^2 \times 3^2 \times 5 \\ \\ \text{LCM of 900, 360 and 180} & = 2^3 \times 3^2 \times 5^2 \\ & = 1800 \\ \\ 1800 \text{ seconds} & = \frac{1800}{60} \text{ minutes} \\ & = 30 \text{ minutes} \\ \\ \text{Required time} & = 5.45 \text{ pm} + 30 \text{ minutes} \\ & = 6.15 \text{ pm} \end{align*}

(i)

\begin{align*} \text{Vani's days off (every 4 days): } & 29 \text{ April}, 3 \text{ May}, 7 \text{ May}, 11 \text{ May}, \ldots \\ \\ \text{Li Ting's days off (every 6 days): } & 1 \text{ May}, 7 \text{ May}, 13 \text{ May}, 19 \text{ May}, \ldots \\ \\ \text{Their first shared day off after 1 May} & = 7 \text{ May} \end{align*}

(ii)

\begin{align*} 4 & = 2^2 \\ \\ 6 & = 2 \times 3 \\ \\ \text{LCM of 4 and 6} & = 2^2 \times 3 \\ & = 12 \\ \\ \text{After 7 May, they will have} & \text{ the same day off every 12 days} \end{align*}

(i)

\begin{align*} \text{LCM of 504 and 810} & = 2^3 \times 3^4 \times 5 \times 7 \\ & = 22 \, 680 \\ \\ 504n & = 22 \, 680 \\ n & = \frac{22 \, 680}{504} \\ n & = 45 \end{align*}

(ii)

\begin{align*} \frac{504}{m} & \text{ must be a factor of both 504 and 810.} \\ \text{To minimise } m, & \text{ maximise } \frac{504}{m} \text{ by using their HCF.} \\ \\ \text{HCF of 504 and 810} & = 2 \times 3^2 \\ & = 18 \\ \\ \frac{504}{m} & = 18 \\ \\ \implies 18 \times m & = 504 \\ m & = 504 \div 18 \\ m & = 28 \end{align*}

(i)

\begin{align*} 15 & = 3 \times 5 \\ \\ & \begin{array}{r|r} 3 & \underline{375} \\ 5 & \underline{125} \\ 5 & \underline{25} \\ 5 & \underline{5} \\ & 1 \end{array} \\ \\ 375 & = 3 \times 5^3 \\ \\ 15 \times 375 & = 3^2 \times 5^4 \\ \\ \text{Since the powers of factors } 3 \text{ and } 5 & \text{ are multiples of 2, } 15 \times 375 \text{ is a perfect square} \end{align*}

(ii)

\begin{align*} 15 \times 375 & = 3^2 \times 5^4 \\ \\ 15 \times 375 \times k & = 3^3 \times 5^6 \qquad [\text{Perfect cube} ] \\ \\ k & = 3 \times 5^2 \\ k & = 75 \end{align*}

(iii)

\begin{align*} 15 \times 375 & = 3^2 \times 5^4 \\ \\ 15 \times 375 \times \frac{p}{q} & = 3^3 \times 5^3 \qquad [\text{Perfect cube} ] \\ \\ p & = 3, q = 5 \qquad [\checkmark \text{Both are prime numbers}] \end{align*}