QUESTION 3 3.1 Write down the LCM of the following numbers, give your answer in exponential form. \[ \begin{array}{l} A=2^{9} \times 3^{6} \times 5^{4} \times 11^{3} \\ B=2^{2} \times 3^{7} \times 7^{2} \\ C=2^{7} \times 3^{3} \times 13^{3} \end{array} \] \( \qquad \) 3.2 Use prime factors to determine the HCF of 28; 196 and 280 \( \qquad \) \( \qquad \) \( \qquad \) \( \qquad \) 3.3 Two buses, bus A and bus B, both use the same bus stop. - Bus A runs every 10 minutes. - Bus B runs every 14 minutes. - Both buses are at the bus stop at 11 am. What time will both buses next be at the bus stop? \( \qquad \) \( \qquad \) \( \qquad \) \( \qquad \)

**3.1 LCM of A, B, and C** Write each number in its prime factor form: \[ A = 2^{9} \times 3^{6} \times 5^{4} \times 11^{3} \] \[ B = 2^{2} \times 3^{7} \times 7^{2} \] \[ C = 2^{7} \times 3^{3} \times 13^{3} \] For the LCM, take each prime factor raised to the highest power present: - For \(2\): highest exponent is \(9\) (from \(A\)). - For \(3\): highest exponent is \(7\) (from \(B\)). - For \(5\): appears as \(5^4\) in \(A\). - For \(7\): appears as \(7^2\) in \(B\). - For \(11\): appears as \(11^3\) in \(A\). - For \(13\): appears as \(13^3\) in \(C\). Thus, the LCM is: \[ \text{LCM} = 2^{9} \times 3^{7} \times 5^{4} \times 7^{2} \times 11^{3} \times 13^{3} \] --- **3.2 HCF of 28, 196, and 280** First, express each number in its prime factorisation: \[ 28 = 2^2 \times 7 \] \[ 196 = 2^2 \times 7^2 \quad \text{(since }196 = 14^2 \text{ and }14 = 2 \times 7\text{)} \] \[ 280 = 2^3 \times 5 \times 7 \quad \text{(since }280 = 8 \times 35 \text{ and }35 = 5 \times 7\text{)} \] For the HCF, take each common prime factor raised to the lowest power: - For \(2\): lowest power is \(2^2\) (from 28 and 196). - For \(7\): lowest power is \(7^1\) (present as exponent 1 in 28 and 280). Thus, the HCF is: \[ \text{HCF} = 2^2 \times 7 = 4 \times 7 = 28 \] --- **3.3 Next time both buses will be at the bus stop** Bus A runs every 10 minutes and Bus B runs every 14 minutes. We need to find the LCM of 10 and 14. Express each in prime factorisation: \[ 10 = 2 \times 5 \] \[ 14 = 2 \times 7 \] Take each prime raised to the highest exponent: - For \(2\): \(2^1\) - For \(5\): \(5^1\) - For \(7\): \(7^1\) Thus, the LCM is: \[ \text{LCM} = 2 \times 5 \times 7 = 70 \text{ minutes} \] Since both buses are at the bus stop at 11:00 am, after 70 minutes they will next both be at the bus stop. Calculate the time: \[ 11:00 \text{ am} + 70 \text{ minutes} = 12:10 \text{ pm} \] --- Final Answers: 3.1: \(2^{9} \times 3^{7} \times 5^{4} \times 7^{2} \times 11^{3} \times 13^{3}\) 3.2: \(28\) 3.3: \(12:10 \text{ pm}\)