Grade 6 HCF-LCM

Step 1: Find the prime factors of 24: Divide by smallest prime numbers repeatedly.

\(24 \div 2 = 12\), \(12 \div 2 = 6\), \(6 \div 2 = 3\), \(3 \div 3 = 1\).

So, \(24 = 2^3 \times 3^1\).

Step 2: Find the prime factors of 36:

\(36 \div 2 = 18\), \(18 \div 2 = 9\), \(9 \div 3 = 3\), \(3 \div 3 = 1\).

So, \(36 = 2^2 \times 3^2\).

Step 3: Identify common prime factors and take the lowest power:

Common primes: 2 and 3.

For 2: Lowest power is \(2^2\). For 3: Lowest power is \(3^1\).

Step 4: Multiply the common factors: \(HCF = 2^2 \times 3^1 = 4 \times 3 = 12\).

Step 5: Therefore, the HCF of 24 and 36 is 12.

Step 1: Find the prime factors of 15:

\(15 \div 3 = 5\), \(5 \div 5 = 1\).

So, \(15 = 3^1 \times 5^1\).

Step 2: Find the prime factors of 25:

\(25 \div 5 = 5\), \(5 \div 5 = 1\).

So, \(25 = 5^2\).

Step 3: Take all prime factors with the highest power:

Primes: 3 (from 15), 5 (from both).

For 3: Highest power is \(3^1\). For 5: Highest power is \(5^2\).

Step 4: Multiply these factors: \(LCM = 3^1 \times 5^2 = 3 \times 25 = 75\).

Step 5: Therefore, the LCM of 15 and 25 is 75.

Step 1: To find when the bells ring together, calculate the LCM of their intervals: 12 and 18.

Step 2: Find the prime factors of 12:

\(12 \div 2 = 6\), \(6 \div 2 = 3\), \(3 \div 3 = 1\).

So, \(12 = 2^2 \times 3^1\).

Step 3: Find the prime factors of 18:

\(18 \div 2 = 9\), \(9 \div 3 = 3\), \(3 \div 3 = 1\).

So, \(18 = 2^1 \times 3^2\).

Step 4: Take all prime factors with the highest power:

For 2: Highest power is \(2^2\). For 3: Highest power is \(3^2\).

Step 5: Multiply: \(LCM = 2^2 \times 3^2 = 4 \times 9 = 36\).

Step 6: Therefore, the bells will ring together again after 36 seconds.