The last digit of the LCM of `(3^(2003)-1)and(3^(2003)+1)` is

Options

A. 8
B. 2
C. 4
D. 6

Answer

To find the last digit of the LCM of \(3^{2003} - 1\) and \(3^{2003} + 1\), we can follow these steps:

### Step 1: Identify the two numbers
We have two numbers:
- \(a = 3^{2003} - 1\)
- \(b = 3^{2003} + 1\)

### Step 2: Calculate the difference
The difference between these two numbers is:
\[ b - a = (3^{2003} + 1) - (3^{2003} - 1) = 2 \]
Since the difference is 2, this means that \(a\) and \(b\) are consecutive even and odd numbers.

### Step 3: Determine if the numbers are even or odd
Since \(3^{2003}\) is odd (as any power of an odd number is odd), we can conclude:
- \(a = 3^{2003} - 1\) is even (odd - 1 = even)
- \(b = 3^{2003} + 1\) is odd (odd + 1 = even)

Thus, both \(a\) and \(b\) are even.

### Step 4: Find the LCM
The LCM of two numbers can be calculated using the formula:
\[ \text{LCM}(a, b) = \frac{a \cdot b}{\text{GCD}(a, b)} \]
Since the difference is 2, the GCD of \(a\) and \(b\) is 2. Therefore:
\[ \text{LCM}(a, b) = \frac{(3^{2003} - 1)(3^{2003} + 1)}{2} \]

### Step 5: Simplify the expression
Using the difference of squares:
\[ (3^{2003} - 1)(3^{2003} + 1) = (3^{2003})^2 - 1^2 = 3^{4006} - 1 \]
Thus, we have:
\[ \text{LCM}(a, b) = \frac{3^{4006} - 1}{2} \]

### Step 6: Find the last digit
Now we need to find the last digit of \(\frac{3^{4006} - 1}{2}\).

1. **Find the last digit of \(3^{4006}\)**:
The last digits of powers of 3 cycle every 4:
- \(3^1 \equiv 3\)
- \(3^2 \equiv 9\)
- \(3^3 \equiv 7\)
- \(3^4 \equiv 1\)
- This cycle repeats.

To find \(3^{4006}\), we calculate \(4006 \mod 4\):
\[ 4006 \mod 4 = 2 \]
Thus, the last digit of \(3^{4006}\) is the same as that of \(3^2\), which is 9.

2. **Calculate \(3^{4006} - 1\)**:
The last digit of \(3^{4006} - 1\) is:
\[ 9 - 1 = 8 \]

3. **Divide by 2**:
Now, we find the last digit of \(\frac{8}{2}\):
\[ \frac{8}{2} = 4 \]

### Conclusion
The last digit of the LCM of \(3^{2003} - 1\) and \(3^{2003} + 1\) is **4**.