2. Find the LCM for these sets of numbers. \( \quad \) b) 18 and 48 \( \begin{array}{ll}\text { a) } 24 \text { and } 90 & \text { c) } 90,105 \text { and } 150 \\ \text { Mr Abrahams is able to divide the learners in his Mathematics class into equal } \\ \text { groups of } 2,5 \text { or } 6 \text { without leaving any learners out. What is the smallest } \\ \text { possible number of learners in the class? } \\ \text { In a village, the bell rings every } 30 \text { minutes at the Junior School and every } 40 \\ \text { minutes at the High School. Both schools start lessons at 8:00 a.m. } \\ \text { a) At what times between 8:00 a.m. and 2:30 p.m. do the bells both ring at } \\ \text { the same time? }\end{array} \) If the High School bell rings every 45 minutes, how many times during school day would this bell ring at the same time as the Junior School

To find the **Least Common Multiple (LCM)** of **18** and **48**, we'll use the **prime factorization method**. This method involves breaking down each number into its prime factors and then selecting the highest power of each prime number present in the factorizations. ### Step 1: Prime Factorization 1. **Factorizing 18:** - 18 ÷ 2 = 9 - 9 ÷ 3 = 3 - 3 ÷ 3 = 1 - **Prime Factors of 18:** \(2 \times 3 \times 3 = 2 \times 3^2\) 2. **Factorizing 48:** - 48 ÷ 2 = 24 - 24 ÷ 2 = 12 - 12 ÷ 2 = 6 - 6 ÷ 2 = 3 - 3 ÷ 3 = 1 - **Prime Factors of 48:** \(2 \times 2 \times 2 \times 2 \times 3 = 2^4 \times 3\) ### Step 2: Identify the Highest Powers of All Prime Factors - **For 2:** The highest power between \(2^1\) (from 18) and \(2^4\) (from 48) is \(2^4\). - **For 3:** The highest power between \(3^2\) (from 18) and \(3^1\) (from 48) is \(3^2\). ### Step 3: Multiply These Highest Powers Together to Find the LCM \[ \text{LCM} = 2^4 \times 3^2 = 16 \times 9 = 144 \] ### **Conclusion** The **Least Common Multiple (LCM)** of **18** and **48** is **144**. --- If you have other parts of the problem you'd like assistance with, feel free to ask!