The sum of two numbers is 36 and their HCF is 4. How many pairs of such number are possible?

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the problem step by step, we need to find pairs of numbers whose sum is 36 and whose highest common factor (HCF) is 4.

### Step-by-Step Solution:

1. **Define the Numbers**:
Let the two numbers be \( n_1 \) and \( n_2 \). According to the problem, we know:
\[ n_1 + n_2 = 36 \]
and their HCF is given as 4.

2. **Express the Numbers in Terms of HCF**:
Since the HCF of \( n_1 \) and \( n_2 \) is 4, we can express \( n_1 \) and \( n_2 \) as:
\[ n_1 = 4a \quad \text{and} \quad n_2 = 4b \]
where \( a \) and \( b \) are co-prime integers (i.e., their HCF is 1).

3. **Substitute into the Sum Equation**:
Substituting \( n_1 \) and \( n_2 \) into the sum equation gives:
\[ 4a + 4b = 36 \]
Dividing the entire equation by 4, we get:
\[ a + b = 9 \]

4. **Find Pairs of Co-prime Integers**:
Now we need to find pairs of integers \( (a, b) \) such that:
\[ a + b = 9 \]
and \( \text{HCF}(a, b) = 1 \).

The possible pairs of \( (a, b) \) that satisfy \( a + b = 9 \) are:
- (1, 8)
- (2, 7)
- (3, 6)
- (4, 5)
- (5, 4)
- (6, 3)
- (7, 2)
- (8, 1)

5. **Check for Co-primality**:
Now we check the HCF of each pair:
- For (1, 8): HCF = 1 (valid)
- For (2, 7): HCF = 1 (valid)
- For (3, 6): HCF = 3 (not valid)
- For (4, 5): HCF = 1 (valid)

The pairs that are valid (co-prime) are:
- (1, 8)
- (2, 7)
- (4, 5)

6. **Count the Valid Pairs**:
The valid pairs of \( (a, b) \) that satisfy both conditions are:
- (1, 8)
- (2, 7)
- (4, 5)

Therefore, the total number of pairs \( (n_1, n_2) \) is 3.

### Final Answer:
The number of pairs of such numbers is **3**.

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