Q5 Use prime factors to find the lowest number that divides exactiy by Dotn Q6 One day, Arran divides his action figures into equal groups of 26 . The next day, he divides them up into equal groups of 12 . Use prime factors to find the lowest possible number af action figures he owns. Q7 Jess goes swimming every 21 days. Seamus goes swimming every 35 days. They both went swimming today. Use prime factors to find the number of days it will be until they both go swimming on the same day again. Q8 a) Write 30,140 and 210 as products of their prime factors. b) Hence find the HCF of 30,140 and 210. Q9 a) Write 121,280 and 550 as products of their prime factors. b) Hence find the LCM of 121,280 and 550 . Q10 Use prime factors to find (i) the HCF and (ii) the LCM of each of the following sets of numbers. a) 65,143 and 231 b) 175,245 and 1225 c) 104,338 and 1078 Q11 a) Use prime factors to find a pair of numbers that have \( H C F=12 \) and \( L C M=120 \). b) Use prime factors to find a pair of numbers that have \( H C F=20 \) and \( L C M=300 \). Section 1 Arithmetic, Multiples and Factors

Let's solve each question step by step, focusing on prime factorization and the concepts of Highest Common Factor (HCF) and Lowest Common Multiple (LCM). ### Q5 **Find the lowest number that divides exactly by Dotn.** It seems like there might be a typo in "Dotn." If you meant a specific number, please clarify. Otherwise, I cannot proceed with this question. ### Q6 **Arran divides his action figures into equal groups of 26 and 12. Find the lowest possible number of action figures he owns.** 1. **Prime Factorization:** - \( 26 = 2 \times 13 \) - \( 12 = 2^2 \times 3 \) 2. **Finding the LCM:** - The LCM is found by taking the highest power of each prime factor: - For \( 2 \): \( 2^2 \) - For \( 3 \): \( 3^1 \) - For \( 13 \): \( 13^1 \) Therefore, the LCM is: \[ LCM = 2^2 \times 3^1 \times 13^1 = 4 \times 3 \times 13 \] Let's calculate this: \[ 4 \times 3 = 12 \] \[ 12 \times 13 = 156 \] So, the lowest possible number of action figures Arran owns is **156**. ### Q7 **Jess goes swimming every 21 days, and Seamus every 35 days. Find the number of days until they both go swimming on the same day again.** 1. **Prime Factorization:** - \( 21 = 3 \times 7 \) - \( 35 = 5 \times 7 \) 2. **Finding the LCM:** - For \( 3 \): \( 3^1 \) - For \( 5 \): \( 5^1 \) - For \( 7 \): \( 7^1 \) Therefore, the LCM is: \[ LCM = 3^1 \times 5^1 \times 7^1 = 3 \times 5 \times 7 \] Let's calculate this: \[ 3 \times 5 = 15 \] \[ 15 \times 7 = 105 \] So, they will both go swimming on the same day again in **105 days**. ### Q8 **a) Write 30,140 and 210 as products of their prime factors.** 1. **Prime Factorization:** - \( 30,140 = 2 \times 5 \times 13 \times 233 \) - \( 210 = 2 \times 3 \times 5 \times 7 \) **b) Find the HCF of 30,140 and 210.** - Common prime factors: \( 2 \) and \( 5 \) - HCF = \( 2^1 \times 5^1 = 10 \) ### Q9 **a) Write 121,280 and 550 as products of their prime factors.** 1. **Prime Factorization:** - \( 121,280 = 2^8 \times 5 \times 19 \) - \( 550 = 2 \times 5^2 \times 11 \) **b) Find the LCM of 121,280 and 550.** - Highest powers of each prime factor: - For \( 2 \): \( 2^8 \) - For \( 5 \): \( 5^2 \) - For \( 11 \): \( 11^1 \) - For \( 19 \): \( 19^1 \) Therefore, the LCM is: \[ LCM = 2^8 \times 5^2 \times 11^1 \times 19^1 \] ### Q10 **Use prime factors to find (i) the HCF and (ii) the LCM of the following sets of numbers.** **a) 65, 143, and 231** 1. **Prime Factorization:** - \( 65 = 5 \times 13 \) - \( 143 = 11 \times 13 \) - \( 231 = 3 \times 7 \times 11 \) **HCF:** No common factors, so HCF = 1. **LCM:** - Highest powers: \( 3^1, 5^1, 7^1, 11^1, 13^1 \) - LCM = \( 3 \times 5 \times 7 \times 11 \times 13 \) **b) 175, 245, and 1225** 1. **Prime Factorization:** - \( 175 = 5^2 \times 7 \) - \( 245 = 5 \times 7^2 \) - \( 1225 = 5^2 \times 7^2 \) **HCF:** \( 5^1 \times 7^1 = 35 \) **LCM:** - Highest powers: \( 5^2, 7^2 \) - LCM = \( 5^2 \times 7^2 = 1225 \) **c) 104, 338, and 1078** 1. **Prime Factorization:** - \( 104 = 2^3 \times 13 \) - \( 338 = 2 \times 13 \times 13 \) - \( 1078 = 2 \times 7 \times 77 \) **HCF:** \( 2^1 \times 13^1 = 26 \) **LCM:** - Highest powers: \( 2^3, 7^1, 13^2 \) - LCM = \( 2^3 \times 7^1 \times 13^2 \) ### Q11 **a) Use prime factors to find a pair of numbers that have \( HCF=12 \) and \( LCM=120 \).** 1. **HCF = 12 = \( 2^2 \times 3^1 \)** 2. **LCM = 120 = \( 2^3 \times 3^1 \times 5^1 \)** Let \( a = 12 \times m \) and \( b = 12 \times n \) where \( m \) and \( n \) are coprime. From \( HCF \) and \( LCM \): \[ HCF(a, b) \times LCM(a, b) = a \times b \] \[