Find the last two digits of LCM `7^(24)-1,7^(24)+1` ?

Options

A. `00`
B. 11
C. 55
D. 88

Answer

To find the last two digits of the LCM of \(7^{24} - 1\) and \(7^{24} + 1\), we can follow these steps:

### Step 1: Identify the numbers
Let \(x = 7^{24} - 1\) and \(y = 7^{24} + 1\).

### Step 2: Use the LCM formula
The LCM of two numbers can be calculated using the formula:
\[ \text{LCM}(x, y) = \frac{x \cdot y}{\text{HCF}(x, y)} \]

### Step 3: Find the HCF of \(x\) and \(y\)
Since \(x\) and \(y\) are consecutive integers (one is odd and the other is even), their HCF is:
\[ \text{HCF}(x, y) = 1 \quad \text{(since consecutive integers are coprime)} \]

### Step 4: Calculate the LCM
Using the HCF we found:
\[ \text{LCM}(x, y) = x \cdot y = (7^{24} - 1)(7^{24} + 1) \]
This can be simplified using the difference of squares:
\[ \text{LCM}(x, y) = 7^{48} - 1 \]

### Step 5: Find the last two digits of \(7^{48} - 1\)
To find the last two digits, we need to calculate \(7^{48} \mod 100\).

### Step 6: Use Euler's theorem
First, we calculate \(\phi(100)\):
\[ \phi(100) = 100 \left(1 - \frac{1}{2}\right)\left(1 - \frac{1}{5}\right) = 100 \cdot \frac{1}{2} \cdot \frac{4}{5} = 40 \]
According to Euler's theorem, since \(7\) and \(100\) are coprime:
\[ 7^{40} \equiv 1 \mod 100 \]
Thus, we can reduce \(48\) modulo \(40\):
\[ 48 \mod 40 = 8 \]
So, we need to calculate \(7^8 \mod 100\).

### Step 7: Calculate \(7^8\)
We can calculate \(7^8\) step by step:
\[ 7^2 = 49 \]
\[ 7^4 = 49^2 = 2401 \quad \Rightarrow \quad 2401 \mod 100 = 1 \]
\[ 7^8 = (7^4)^2 \equiv 1^2 \equiv 1 \mod 100 \]

### Step 8: Find \(7^{48} - 1 \mod 100\)
Now we can find:
\[ 7^{48} \equiv 1 \mod 100 \]
Thus,
\[ 7^{48} - 1 \equiv 1 - 1 \equiv 0 \mod 100 \]

### Conclusion
The last two digits of the LCM \(7^{24} - 1\) and \(7^{24} + 1\) are:
\[ \boxed{00} \]

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