HCF of two numbers is 15, while their product is 2025. Numbers are __

Options

A. 15175
B. 45,75
C. 75,75
D. 45,45

Answer

To find the two numbers given that their HCF is 15 and their product is 2025, we can follow these steps:

### Step 1: Define the two numbers
Let the two numbers be \( a \) and \( b \). Given that their HCF is 15, we can express these numbers in terms of their HCF:
\[ a = 15x \quad \text{and} \quad b = 15y \]
where \( x \) and \( y \) are co-prime numbers (i.e., their HCF is 1).

### Step 2: Set up the equation for the product
According to the problem, the product of the two numbers is given as:
\[ a \times b = 2025 \]
Substituting the expressions for \( a \) and \( b \):
\[ (15x) \times (15y) = 2025 \]
This simplifies to:
\[ 225xy = 2025 \]

### Step 3: Solve for \( xy \)
To isolate \( xy \), divide both sides of the equation by 225:
\[ xy = \frac{2025}{225} \]
Calculating the right side:
\[ xy = 9 \]

### Step 4: Find co-prime pairs of \( x \) and \( y \)
Now, we need to find pairs of co-prime numbers \( x \) and \( y \) such that their product is 9. The pairs of factors of 9 that are co-prime are:
1. \( (1, 9) \)
2. \( (3, 3) \)

### Step 5: Calculate the numbers
Using the pairs found:
1. For \( (1, 9) \):
- \( x = 1 \), \( y = 9 \):
\[ a = 15 \times 1 = 15, \quad b = 15 \times 9 = 135 \]
So one pair of numbers is \( 15 \) and \( 135 \).

2. For \( (3, 3) \):
- \( x = 3 \), \( y = 3 \):
\[ a = 15 \times 3 = 45, \quad b = 15 \times 3 = 45 \]
So another pair of numbers is \( 45 \) and \( 45 \).

### Conclusion
The two numbers can be either \( 15 \) and \( 135 \) or \( 45 \) and \( 45 \).

Thus, the answer is:
- The numbers are \( 15 \) and \( 135 \) or \( 45 \) and \( 45 \).