Find the last two digits of `54^(380)`.

Answer

To find the last two digits of \(54^{380}\), we can use modular arithmetic, specifically modulo \(100\). Here’s a step-by-step solution:

### Step 1: Factor the base
We start by factoring \(54\):
\[ 54 = 2 \times 3^3 \]
So we can rewrite \(54^{380}\) as:
\[ 54^{380} = (2 \times 3^3)^{380} = 2^{380} \times 3^{1140} \]

### Step 2: Calculate \(2^{380} \mod 100\)
To find \(2^{380} \mod 100\), we can use the Chinese Remainder Theorem (CRT). We will calculate \(2^{380} \mod 4\) and \(2^{380} \mod 25\).

- **Calculate \(2^{380} \mod 4\)**:
\[ 2^{380} \equiv 0 \mod 4 \]

- **Calculate \(2^{380} \mod 25\)**:
Using Euler's theorem, since \(\phi(25) = 20\):
\[ 2^{20} \equiv 1 \mod 25 \]
Now, we find \(380 \mod 20\):
\[ 380 \div 20 = 19 \quad \text{(remainder 0)} \]
Thus,
\[ 2^{380} \equiv (2^{20})^{19} \equiv 1^{19} \equiv 1 \mod 25 \]

### Step 3: Combine results using CRT
We have:
\[ 2^{380} \equiv 0 \mod 4 \]
\[ 2^{380} \equiv 1 \mod 25 \]
Let \(x\) be the solution. We can set up the system of congruences:
1. \(x \equiv 0 \mod 4\)
2. \(x \equiv 1 \mod 25\)

To solve this, we can express \(x\) in terms of \(25\):
\[ x = 25k + 1 \]
Substituting into the first congruence:
\[ 25k + 1 \equiv 0 \mod 4 \]
Calculating \(25 \mod 4\):
\[ 25 \equiv 1 \mod 4 \]
So we have:
\[ k + 1 \equiv 0 \mod 4 \implies k \equiv 3 \mod 4 \]
Let \(k = 4m + 3\). Substituting back:
\[ x = 25(4m + 3) + 1 = 100m + 76 \]
Thus,
\[ x \equiv 76 \mod 100 \]

### Step 4: Calculate \(3^{1140} \mod 100\)
Next, we calculate \(3^{1140} \mod 100\) using CRT again.

- **Calculate \(3^{1140} \mod 4\)**:
\[ 3 \equiv -1 \mod 4 \implies 3^{1140} \equiv (-1)^{1140} \equiv 1 \mod 4 \]

- **Calculate \(3^{1140} \mod 25\)**:
Using Euler's theorem, since \(\phi(25) = 20\):
\[ 3^{20} \equiv 1 \mod 25 \]
Now, find \(1140 \mod 20\):
\[ 1140 \div 20 = 57 \quad \text{(remainder 0)} \]
Thus,
\[ 3^{1140} \equiv (3^{20})^{57} \equiv 1^{57} \equiv 1 \mod 25 \]

### Step 5: Combine results using CRT for \(3^{1140}\)
We have:
\[ 3^{1140} \equiv 1 \mod 4 \]
\[ 3^{1140} \equiv 1 \mod 25 \]
Thus,
\[ 3^{1140} \equiv 1 \mod 100 \]

### Step 6: Combine \(2^{380}\) and \(3^{1140}\)
Now, we combine the results:
\[ 54^{380} = 2^{380} \times 3^{1140} \equiv 76 \times 1 \equiv 76 \mod 100 \]

### Final Answer
The last two digits of \(54^{380}\) are:
\[ \boxed{76} \]