Find the last two digits of $$ 53^{51^{49^{\cdots 1}}} $$

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To find the last two digits of $$ 53^{51^{49^{\cdots 1}}} $$, we need to compute $$ 53^{51^{49^{\cdots 1}}} \mod 100 $$. Since $$ 53 \equiv 53 \mod 100 $$, we can simplify our problem to finding $$ 53^n \mod 100 $$ where $$ n = 51^{49^{\cdots 1}} $$.

To apply the Chinese Remainder Theorem, we will compute $$ 53^n \mod 4 $$ and $$ 53^n \mod 25 $$ separately.

### Step 1: Compute $$ 53^n \mod 4 $$

Since $$ 53 \equiv 1 \mod 4 $$, we have:
$$
53^n \equiv 1^n \equiv 1 \mod 4.
$$

### Step 2: Compute $$ 53^n \mod 25 $$

Next, we compute $$ 53 \mod 25 $$:
$$
53 \equiv 3 \mod 25.
$$
Thus, we need to find $$ 3^n \mod 25 $$.

To do this, we first need to determine $$ n \mod \phi(25) $$, where $$ \phi(25) = 25(1 - \frac{1}{5}) = 20 $$. Therefore, we need to compute $$ n \mod 20 $$.

### Step 3: Compute $$ n = 51^{49^{\cdots 1}} \mod 20 $$

Next, we compute $$ 51 \mod 20 $$:
$$
51 \equiv 11 \mod 20.
$$
Thus, we need to find $$ 11^{49^{\cdots 1}} \mod 20 $$.

Now, we compute $$ 49 \mod \phi(20) $$. We have $$ \phi(20) = 20(1 - \frac{1}{4})(1 - \frac{1}{5}) = 8 $$. Therefore, we need to compute $$ 49 \mod 8 $$:
$$
49 \equiv 1 \mod 8.
$$
This means $$ 49^{\cdots 1} \equiv 1 \mod 8 $$.

### Step 4: Compute $$ 11^1 \mod 20 $$

Now we compute:
$$
n \equiv 11^1 \equiv 11 \mod 20.
$$

### Step 5: Compute $$ 3^n \mod 25 $$

Now we need to compute $$ 3^{11} \mod 25 $$. We can calculate this using successive squaring:
$$
3^1 \equiv 3,
$$
$$
3^2 \equiv 9,
$$
$$
3^4 \equiv 9^2 = 81 \equiv 6 \mod 25,
$$
$$
3^8 \equiv 6^2 = 36 \equiv 11 \mod 25.
$$
Now we combine these:
$$
3^{11} = 3^8 \cdot 3^2 \cdot 3^1 \equiv 11 \cdot 9 \cdot 3 \mod 25.
$$
Calculating $$ 11 \cdot 9 = 99 \equiv 24 \mod 25 $$, and then:
$$
24 \cdot 3 = 72 \equiv 22 \mod 25.
$$
Thus, we have:
$$
3^{11} \equiv 22 \mod 25.
$$

### Step 6: Combine results using the Chinese Remainder Theorem

We have:
$$
53^n \equiv 1 \mod 4,
$$
$$
53^n \equiv 22 \mod 25.
$$

Let $$ x $$ be the solution to the system:
$$
x \equiv 1 \mod 4,
$$
$$
x \equiv 22 \mod 25.
$$

We can express $$ x $$ in terms of the second congruence:
$$
x = 25k + 22.
$$
Substituting into the first congruence:
$$
25k + 22 \equiv 1 \mod 4.
$$
Calculating $$ 25 \mod 4 $$:
$$
25 \equiv 1 \mod 4,
$$
thus:
$$
k + 22 \equiv 1 \mod 4 \implies k \equiv 1 - 22 \equiv 1 - 2 \equiv -1 \equiv 3 \mod 4.
$$
So we can write $$ k = 4m + 3 $$ for some integer $$ m $$. Substituting back:
$$
x = 25(4m + 3) + 22 = 100m + 75 + 22 = 100m + 97.
$$
Thus:
$$
x \equiv 97 \mod 100.
$$

### Final Result

The last two digits of $$ 53^{51^{49^{\cdots 1}}} $$ are:
$$
\boxed{97}.
$$
The last two digits of $$ 53^{51^{49^{\cdots 1}}} $$ are 97.

Find the last two digits of \( 53^{51^{49^{\cdots 1}}} \)

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