What is the number of integral solutions of the equations HCF (a, b) = 5 and a + b = 65?

Options

A. None
B. Infinitely many
C. Less than 65
D. Exactly one

Answer

To find the number of integral solutions for the equations HCF(a, b) = 5 and a + b = 65, we can follow these steps:

### Step 1: Express a and b in terms of their HCF
Since HCF(a, b) = 5, we can express a and b as:
- \( a = 5m \)
- \( b = 5n \)
where m and n are integers such that HCF(m, n) = 1 (because 5 is the common factor).

### Step 2: Substitute into the second equation
Now, substitute \( a \) and \( b \) into the equation \( a + b = 65 \):
\[ 5m + 5n = 65 \]
Dividing the entire equation by 5 gives:
\[ m + n = 13 \]

### Step 3: Find the integral solutions for m and n
Now we need to find the pairs of integers (m, n) such that:
- \( m + n = 13 \)
- HCF(m, n) = 1

### Step 4: List possible pairs (m, n)
The pairs (m, n) that satisfy \( m + n = 13 \) are:
- (1, 12)
- (2, 11)
- (3, 10)
- (4, 9)
- (5, 8)
- (6, 7)
- (7, 6)
- (8, 5)
- (9, 4)
- (10, 3)
- (11, 2)
- (12, 1)

### Step 5: Check HCF condition
Now we check which of these pairs have HCF(m, n) = 1:
- (1, 12) → HCF = 1
- (2, 11) → HCF = 1
- (3, 10) → HCF = 1
- (4, 9) → HCF = 1
- (5, 8) → HCF = 1
- (6, 7) → HCF = 1
- (7, 6) → HCF = 1
- (8, 5) → HCF = 1
- (9, 4) → HCF = 1
- (10, 3) → HCF = 1
- (11, 2) → HCF = 1
- (12, 1) → HCF = 1

All pairs have HCF(m, n) = 1.

### Step 6: Count the valid pairs
Since all pairs satisfy the condition, we count the pairs:
There are 13 pairs in total.

### Final Answer
Thus, the number of integral solutions for the equations HCF(a, b) = 5 and a + b = 65 is **13**.

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