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**Simplified Answers:**
1. **Q5:** Cannot determine the number without clarification on "Dotn."
2. **Q6:** Arran owns **156** action figures.
3. **Q7:** Jess and Seamus will swim together again in **105 days**.
4. **Q8a:**
- \( 30,140 = 2^8 \times 5 \times 13 \times 233 \)
- \( 210 = 2 \times 3 \times 5 \times 7 \)
5. **Q8b:** The HCF of 30,140 and 210 is **10**.
6. **Q9a:**
- \( 121,280 = 2^8 \times 5 \times 19 \)
- \( 550 = 2 \times 5^2 \times 11 \)
7. **Q9b:** The LCM of 121,280 and 550 is \( 2^8 \times 5^2 \times 11 \times 19 \).
8. **Q10a:**
- HCF: **1**
- LCM: \( 3 \times 5 \times 7 \times 11 \times 13 \)
9. **Q10b:**
- HCF: **35**
- LCM: **1225**
10. **Q10c:**
- HCF: **26**
- LCM: \( 2^3 \times 7 \times 13^2 \)
11. **Q11a:**
- One possible pair is **12 and 120**.
12. **Q11b:**
- One possible pair is **20 and 300**.
**Summary:**
- **Q5:** Cannot determine without clarification.
- **Q6:** 156 action figures.
- **Q7:** 105 days.
- **Q8:** 30,140 = \( 2^8 \times 5 \times 13 \times 233 \), 210 = \( 2 \times 3 \times 5 \times 7 \), HCF = 10.
- **Q9:** 121,280 = \( 2^8 \times 5 \times 19 \), 550 = \( 2 \times 5^2 \times 11 \), LCM = \( 2^8 \times 5^2 \times 11 \times 19 \).
- **Q10:**
- a) HCF = 1, LCM = \( 3 \times 5 \times 7 \times 11 \times 13 \)
- b) HCF = 35, LCM = 1225
- c) HCF = 26, LCM = \( 2^3 \times 7 \times 13^2 \)
- **Q11:**
- a) Pair with HCF=12 and LCM=120: 12 and 120
- b) Pair with HCF=20 and LCM=300: 20 and 300
Solución
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 \( LCM \):
\[
HCF(a, b) \times LCM(a, b) = a \times b
\]
\[
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