Write last two digits of the number `3^(400)dot`

Explanation

To find the last two digits of \(3^{400}\), we can use the concept of modular arithmetic, specifically calculating \(3^{400} \mod 100\).

### Step-by-Step Solution:

1. **Use the Chinese Remainder Theorem**:
We can break down the problem by finding \(3^{400} \mod 4\) and \(3^{400} \mod 25\), and then combine the results using the Chinese Remainder Theorem.

2. **Calculate \(3^{400} \mod 4\)**:
\[ 3 \equiv -1 \mod 4 \]
Therefore,
\[ 3^{400} \equiv (-1)^{400} \equiv 1 \mod 4 \]

3. **Calculate \(3^{400} \mod 25\)**:
We can use Euler's theorem here. First, we find \(\phi(25)\):
\[ \phi(25) = 25 \left(1 - \frac{1}{5}\right) = 25 \cdot \frac{4}{5} = 20 \]
Since \(3\) and \(25\) are coprime, we can apply Euler's theorem:
\[ 3^{20} \equiv 1 \mod 25 \]
Now, we need to find \(400 \mod 20\):
\[ 400 \div 20 = 20 \quad \text{(exactly, so remainder is 0)} \]
Thus,
\[ 3^{400} \equiv (3^{20})^{20} \equiv 1^{20} \equiv 1 \mod 25 \]

4. **Combine the results using the Chinese Remainder Theorem**:
We have:
\[ 3^{400} \equiv 1 \mod 4 \]
\[ 3^{400} \equiv 1 \mod 25 \]
Since both congruences yield the same result, we can conclude:
\[ 3^{400} \equiv 1 \mod 100 \]

5. **Final Result**:
The last two digits of \(3^{400}\) are \(01\).