Find the last two digit of `(545454)^(380)`

Options

A. a) `01`
B. b) 67
C. c) 76
D. d) 34

Answer

To find the last two digits of \( (545454)^{380} \), we need to calculate \( (545454)^{380} \mod 100 \).

### Step 1: Simplify the Base
First, we simplify \( 545454 \mod 100 \):
\[ 545454 \mod 100 = 54 \]
So, we need to calculate \( 54^{380} \mod 100 \).

### Step 2: Use Euler's Theorem
To apply Euler's theorem, we first need to find \( \phi(100) \):
\[ 100 = 2^2 \times 5^2 \]
Using the formula for Euler's totient function:
\[ \phi(100) = 100 \left(1 - \frac{1}{2}\right)\left(1 - \frac{1}{5}\right) = 100 \times \frac{1}{2} \times \frac{4}{5} = 40 \]
Since \( \gcd(54, 100) = 2 \neq 1 \), we cannot directly use Euler's theorem. Instead, we can use the Chinese Remainder Theorem.

### Step 3: Calculate \( 54^{380} \mod 4 \) and \( 54^{380} \mod 25 \)

#### Calculate \( 54^{380} \mod 4 \):
\[ 54 \mod 4 = 2 \]
\[ 2^{380} \mod 4 = 0 \quad (\text{since } 2^2 = 0 \mod 4) \]

#### Calculate \( 54^{380} \mod 25 \):
Next, we calculate \( 54 \mod 25 \):
\[ 54 \mod 25 = 4 \]
Now we need to find \( 4^{380} \mod 25 \). We can use Euler's theorem here since \( \gcd(4, 25) = 1 \):
\[ \phi(25) = 25 \left(1 - \frac{1}{5}\right) = 20 \]
Now we reduce the exponent \( 380 \mod 20 \):
\[ 380 \mod 20 = 0 \]
Thus, \( 4^{380} \mod 25 = 1 \) (since \( 4^{20} \equiv 1 \mod 25 \)).

### Step 4: Solve the System of Congruences
Now we have the two congruences:
1. \( x \equiv 0 \mod 4 \)
2. \( x \equiv 1 \mod 25 \)

We can solve this system using the method of substitution. Let \( x = 25k + 1 \) for some integer \( k \):
\[ 25k + 1 \equiv 0 \mod 4 \]
Calculating \( 25 \mod 4 \):
\[ 25 \equiv 1 \mod 4 \]
So, we have:
\[ k + 1 \equiv 0 \mod 4 \implies k \equiv -1 \equiv 3 \mod 4 \]
Let \( k = 4m + 3 \) for some integer \( m \):
\[ x = 25(4m + 3) + 1 = 100m + 76 \]
Thus, \( x \equiv 76 \mod 100 \).

### Conclusion
The last two digits of \( (545454)^{380} \) are \( \boxed{76} \).