\begin{align*} 3 & | \underline{2025} \\ 3 & | \underline{675} \\ 3 & | \underline{225} \\ 3 & | \underline{75} \\ 5 & | \underline{25} \\ 5 & | \underline{5} \\ & | \underline{1} \\ \\ 2025 & = 3 \times 3 \times 3 \times 3 \times 5 \times 5 \\ & = 3^4 \times 5^2 \end{align*}

\begin{align*} 1170 & = 2 \times 3^2 \times 5 \times 13 \\ 1485 & = 3^3 \times 5 \times 11 \\ \\ \text{HCF} & = 3^2 \times 5 \phantom{000000} [\text{Common factors + Smallest power}] \\ & = 45 \end{align*}

\begin{align*} 36 & = 2^2 \times 3^2 \\ 45 & = 3^2 \times 5 \\ 112 & = 2^4 \times 7 \\ \\ \text{LCM} & = 2^4 \times 3^2 \times 5 \times 7 \\ & = 5040 \end{align*}

\begin{align*} 1012 & = 2^2 \times 11 \times 23 \phantom{000000} [\text{Factor } 2 \text{ is ok, need one more } 11 \text{ and } 23] \\ \\ 1012p & = 2^2 \times 11 \times 23 \times (11 \times 23) \\ & = 2^2 \times 11^2 \times 23^2 \phantom{0000.} [\text{Perfect square ✓}] \\ \\ \therefore p & = 11 \times 23 = 253 \end{align*}

\begin{align*} 3240 & = 2^3 \times 3^4 \times 5 \phantom{00000} [\text{Factor } 2 \text{ is ok, need to remove one } 3 \text{ and } 5] \\ \\ {3240 \over q} & = {2^3 \times 3^4 \times 5 \over 3 \times 5} \\ & = 2^3 \times 3^3 \phantom{00000000.} [\text{Perfect cube ✓}] \\ \\ \therefore q & = 3 \times 5 = 15 \end{align*}

\begin{align*} \text{Surface area of cuboid} & = 2lb + 2lh + 2bh \\ \\ \text{Volume of cuboid} & = l \times b \times h \\ \\ \text{Volume} = 72 & = 2^3 \times 3^2 \phantom{00000} [l \times b \times h \text{ must be a combination of these factors}] \\ \\ \text{Case 1: Square base of } & 2 \times 2, \text{ height} = 2 \times 3^2 = 18 \\ \\ \text{Surface area} & = 2(2 \times 2) + 2(2 \times 18) + 2(2 \times 18) \\ & = 152 \text{ cm}^2 \text{ (Reject)} \\ \\ \text{Case 2: Square base of } & 3 \times 3, \text{ height} = 2^3 = 8 \\ \\ \text{Surface area} & = 2(3 \times 3) + 2(3 \times 8) + 2(3 \times 8) \\ & = 114 \text{ cm}^2 \text{ (Reject)} \\ \\ \text{Case 3: Square base of } & 6 \times 6, \text{ height} = 2 \\ \\ \text{Surface area} & = 2 (6 \times 6) + 2(6 \times 2) + 2(6 \times 2) \\ & = 120 \text{ cm}^2 \phantom{0} \checkmark \\ \\ \therefore \text{Height} & = 2 \text{ cm} \end{align*}

\begin{align*} 70 & = 2 \times 5 \times 7 \\ \\ 85 & = 5 \times 17 \\ \\ \text{Greatest divisor} & = \text{Highest common factor} \\ & = 5 \end{align*}

\begin{align*} 65 & = 5 \times 13 \\ \\ 100 & = 2^2 \times 5^2 \\ \\ 190 & = 2 \times 5 \times 19 \\ \\ \text{LCM} & = 2^2 \times 5^2 \times 13 \times 19 \phantom{000000} [\text{This number can be divided by 65, 100 or 190}] \\ & = 24 \phantom{.} 700 \end{align*}

\begin{align*} 120 & = 2^3 \times 3 \times 5 \\ \\ 150 & = 2 \times 3 \times 5^2 \\ \\ \text{HCF} & = 2 \times 3 \times 5 \\ & = 30 \\ \\ \text{Side length} & = 30 \text{ cm} \end{align*}

\begin{align*} \text{No. of tiles on the shorter side} & = {120 \over 30} = 4 \\ \\ \text{No. of tiles on the longer side} & = {150 \over 30} = 5 \\ \\ \text{No. of tiles required} & = 4 \times 5 \\ & = 20 \end{align*}

\begin{align*} 15 & = 3 \times 5 \\ \\ 20 & = 2^2 \times 5 \\ \\ 45 & = 3^2 \times 5 \\ \\ \text{LCM} & = 2^2 \times 3^2 \times 5 \\ & = 180 \\ \\ \therefore \text{They will depart } & \text{at the same time after 180 mins = 3 hours} \\ \\ \text{Time: } & \text{11.00 am} \end{align*}

\begin{align*} 28 & = 2^2 \times 7 \\ \\ \text{HCF, } 14 & = 2 \times 7 \\ \\ \\ \text{Case 1: } x & = 2 \times 7 = 14 \phantom{0} \text{ (Reject, } x > 14) \\ \\ \text{Case 2: } x & = 2 \times 3 \times 7 = 42 \phantom{0} (\checkmark, \text{ since HCF} = 14) \\ \\ \text{Case 3: } x & = 2 \times 5 \times 7 = 70 \phantom{0} (\checkmark, \text{ since HCF} = 14) \\ \\ \therefore x & = 42 \text{ or } 70 \end{align*}

\begin{align*} 12 & = 2^2 \times 3 \\ \\ \text{HCF, } 6 & = 2 \times 3 \\ \\ \text{LCM, } 36 & = 2^2 \times 3^2 \\ \\ x & = 2 \times 3^2 \phantom{0} (\checkmark \text{HCF} = 2 \times 3, \text{ LCM} = 2^2 \times 3^2) \\ x & = 18 \end{align*}

\begin{align*} \text{HCF, } 12 & = 2^2 \times 3 \\ \\ \text{LCM, } 72 & = 2^3 \times 3^2 \\ \\ \text{Case 1: } x & = 2^2 \times 3 = 12, y = 2^3 \times 3^2 = 72 \phantom{0} \text{ (Reject)} \\ \\ \text{Case 2: } x & = 2^3 \times 3 = 24, y = 2^2 \times 3^2 = 36 \phantom{0} \checkmark \\ \\ \text{Case 3: } x & = 2^2 \times 3^2 = 36, y = 2^3 \times 3 = 24 \phantom{0} (\text{Same as case 2}) \\ \\ \therefore x & = 24, y = 36 \end{align*}

\begin{align*} & 5184 = 2^6 \times 3^4 \\ \\ & \text{The powers of its prime factors are divisible by 2.} \\ & \text{Since 6 and 4 are both even, 5184 is a perfect square.} \\ \\ & \text{The power of factor 3, 4, is not divisible by 3.} \\ & \text{Thus, 5184 is not a perfect cube.} \end{align*}

\begin{align*} 5184 & = 2^6 \times 3^4 \\ \\ 5184k & = 2^6 \times 3^4 \times 3^2 \\ & = 2^6 \times 3^6 \phantom{000000} [6 \text{ is divisible by 2 and by 3}] \\ \\ \therefore k & = 3^2 = 9 \end{align*}

\begin{align*} 5184 & = 2^6 \times 3^4 \\ \\ {5184 \over p} & = {2^6 \times 3^4 \over 3^4} \\ & = 2^6 \\ \\ \therefore p & = 3^4 = 81 \end{align*}

\begin{align*} 2200 & = 2^3 \times 5^2 \times 11 \\ \\ {2200a \over b} & = {(2^3 \times 5^2 \times 11) \times 11 \over 2} \phantom{00000} [a > b, \text{ factor 5 is ok}] \\ & = 2^2 \times 5^2 \times 11^2 \phantom{00000000000} [\text{Perfect square}] \\ \\ \therefore a & = 11, b = 2 \end{align*}