Find the HCF of `3^(2002)+1,3^(2002)-1` ?

Options

A. A)2
B. B)3
C. C)5
D. D)7

Answer

To find the HCF (Highest Common Factor) of the numbers \(3^{2002} + 1\) and \(3^{2002} - 1\), we can follow these steps:

### Step 1: Identify the Numbers
We have two numbers:
- \(a = 3^{2002} + 1\)
- \(b = 3^{2002} - 1\)

### Step 2: Calculate the Difference
Calculate the difference between the two numbers:
\[ a - b = (3^{2002} + 1) - (3^{2002} - 1) = 2 \]

### Step 3: Use the Property of HCF
According to the property of HCF, the HCF of two numbers cannot be greater than their difference. Therefore, we have:
\[ \text{HCF}(a, b) \leq 2 \]

### Step 4: Determine the Parity of the Numbers
Next, we need to determine whether both numbers are even or odd:
- \(3^{2002}\) is an odd number (since any power of an odd number is odd).
- Thus, \(3^{2002} + 1\) is even (odd + even = even).
- Similarly, \(3^{2002} - 1\) is even (odd - odd = even).

### Step 5: Identify the HCF of Two Even Numbers
Since both \(a\) and \(b\) are even, we can conclude that their HCF is at least 2.

### Step 6: Check for Consecutive Even Numbers
The two even numbers \(3^{2002} + 1\) and \(3^{2002} - 1\) are consecutive even numbers. The HCF of any two consecutive even numbers is always 2.

### Conclusion
Thus, the HCF of \(3^{2002} + 1\) and \(3^{2002} - 1\) is:
\[ \text{HCF}(3^{2002} + 1, 3^{2002} - 1) = 2 \]