Sure, let's break down each problem step by step:
12. If the product of two numbers is 4032 and their HCF is 12, find their LCM.
Step-by-Step Solution:
1. Given:
- The product of the two numbers is 4032.
- The HCF (Highest Common Factor) of the two numbers is 12.
2. We know the relationship between HCF, LCM (Least Common Multiple), and the product of the two numbers:
[tex]\[
\text{HCF} \times \text{LCM} = \text{Product of the two numbers}
\][/tex]
3. Substitute the given values into the equation:
[tex]\[
12 \times \text{LCM} = 4032
\][/tex]
4. To find the LCM, divide the product by the HCF:
[tex]\[
\text{LCM} = \frac{4032}{12} = 336
\][/tex]
So, the LCM of the two numbers is 336.
---
13. The HCF and LCM of two numbers are 9 and 270 respectively. If one of the numbers is 45, find the other number.
Step-by-Step Solution:
1. Given:
- The HCF is 9.
- The LCM is 270.
- One of the numbers is 45.
2. Let the other number be \( x \).
3. We know the relationship between the numbers, their HCF, and LCM:
[tex]\[
\text{HCF} \times \text{LCM} = \text{Product of the two numbers}
\][/tex]
4. Substitute the given values into the equation:
[tex]\[
9 \times 270 = 45 \times x
\][/tex]
5. Solve for \( x \):
[tex]\[
2430 = 45 \times x
\][/tex]
[tex]\[
x = \frac{2430}{45} = 54
\][/tex]
So, the other number is 54.
---
14. Find the HCF of 180 and 336.
Step-by-Step Solution:
1. To find the HCF (Highest Common Factor) of 180 and 336, we can use the Euclidean algorithm:
2. Apply the Euclidean algorithm:
- Divide 336 by 180, and find the remainder:
[tex]\[
336 \div 180 = 1 \, \text{remainder} \, 156
\][/tex]
- Now, divide 180 by the remainder 156:
[tex]\[
180 \div 156 = 1 \, \text{remainder} \, 24
\][/tex]
- Finally, divide 156 by 24:
[tex]\[
156 \div 24 = 6 \, \text{remainder} \, 0
\][/tex]
3. Since the remainder is now 0, the last non-zero remainder is the HCF.
So, the HCF of 180 and 336 is 24.
