What are the last two digits of `7^(7^(7^(7)))` ?

Answer

To find the last two digits of \( 7^{7^{7^{7}}} \), we need to compute \( 7^{7^{7^{7}}} \mod 100 \).

### Step 1: Use the Chinese Remainder Theorem
We will find \( 7^{7^{7^{7}}} \mod 4 \) and \( 7^{7^{7^{7}}} \mod 25 \), and then combine the results using the Chinese Remainder Theorem.

### Step 2: Calculate \( 7^{7^{7^{7}}} \mod 4 \)
Since \( 7 \equiv -1 \mod 4 \):
\[ 7^{7^{7^{7}}} \equiv (-1)^{7^{7^{7}}} \mod 4 \]
Now, \( 7^{7^{7}} \) is odd, therefore:
\[ (-1)^{\text{odd}} \equiv 3 \mod 4 \]
Thus,
\[ 7^{7^{7^{7}}} \mod 4 \equiv 3 \]

### Step 3: Calculate \( 7^{7^{7^{7}}} \mod 25 \)
Next, we need to find \( 7^{7^{7^{7}}} \mod 25 \). We will use Euler's theorem here. First, we compute \( \phi(25) \):
\[ \phi(25) = 25 \left(1 - \frac{1}{5}\right) = 20 \]
So, we need \( 7^{7^{7}} \mod 20 \).

### Step 4: Calculate \( 7^{7^{7}} \mod 20 \)
Now, we find \( 7^{7} \mod 20 \):
\[ \phi(20) = 20 \left(1 - \frac{1}{4}\right)\left(1 - \frac{1}{5}\right) = 8 \]
We need \( 7^{7} \mod 8 \):
\[ 7 \equiv -1 \mod 8 \implies 7^{7} \equiv (-1)^{7} \equiv 7 \mod 8 \]
Now we find \( 7^{7} \mod 20 \):
\[ 7^{7} \mod 20 \]
Calculating \( 7^1, 7^2, 7^3, \ldots \):
- \( 7^1 \equiv 7 \)
- \( 7^2 = 49 \equiv 9 \)
- \( 7^3 = 7 \times 9 = 63 \equiv 3 \)
- \( 7^4 = 7 \times 3 = 21 \equiv 1 \)

Thus, \( 7^{7} \mod 20 \) can be computed as:
\[ 7^{7} \equiv 7^{(7 \mod 4)} \equiv 7^3 \equiv 3 \mod 20 \]

### Step 5: Calculate \( 7^{3} \mod 25 \)
Now we compute \( 7^{3} \mod 25 \):
\[ 7^1 \equiv 7 \]
\[ 7^2 = 49 \equiv 24 \mod 25 \]
\[ 7^3 = 7 \times 24 = 168 \equiv 18 \mod 25 \]

### Step 6: Combine results using CRT
We have:
\[ 7^{7^{7^{7}}} \equiv 3 \mod 4 \]
\[ 7^{7^{7^{7}}} \equiv 18 \mod 25 \]

Let \( x \) be the solution. We can write:
\[ x \equiv 18 \mod 25 \]
\[ x \equiv 3 \mod 4 \]

### Step 7: Solve the system of congruences
We can express \( x \) in terms of \( k \):
\[ x = 25k + 18 \]
Now substitute into the second congruence:
\[ 25k + 18 \equiv 3 \mod 4 \]
Since \( 25 \equiv 1 \mod 4 \):
\[ k + 2 \equiv 3 \mod 4 \implies k \equiv 1 \mod 4 \]
Thus, \( k = 4m + 1 \) for some integer \( m \):
\[ x = 25(4m + 1) + 18 = 100m + 43 \]
Therefore,
\[ x \equiv 43 \mod 100 \]

### Final Answer
The last two digits of \( 7^{7^{7^{7}}} \) are \( \boxed{43} \).