The last two digits of the number ` 19^(9^(4))`is

Options

A. 19
B. 29
C. 39
D. 81

Explanation

To find the last two digits of \( 19^{9^4} \), we can use the Binomial Theorem and properties of modular arithmetic. Here’s a step-by-step solution:

### Step 1: Rewrite the base
We can express \( 19 \) in a form that is easier to work with for finding the last two digits:
\[ 19 = 20 - 1 \]
Thus, we can rewrite the expression as:
\[ 19^{9^4} = (20 - 1)^{9^4} \]

### Step 2: Apply the Binomial Theorem
Using the Binomial Theorem, we expand \( (20 - 1)^{9^4} \):
\[ (20 - 1)^{9^4} = \sum_{k=0}^{9^4} \binom{9^4}{k} 20^k (-1)^{9^4 - k} \]
We are interested in the last two digits, which means we need to find this expression modulo \( 100 \).

### Step 3: Identify significant terms
In the expansion, the terms with \( k \geq 2 \) will contain \( 20^2 \) or higher powers of \( 20 \), which are multiples of \( 100 \) and will not affect the last two digits. Therefore, we only need to consider the first two terms of the expansion:
- For \( k = 0 \):
\[ \binom{9^4}{0} 20^0 (-1)^{9^4} = 1 \cdot 1 \cdot (-1)^{9^4} = -1 \]
- For \( k = 1 \):
\[ \binom{9^4}{1} 20^1 (-1)^{9^4 - 1} = 9^4 \cdot 20 \cdot (-1) = -9^4 \cdot 20 \]

### Step 4: Calculate \( 9^4 \)
Now we need to calculate \( 9^4 \):
\[ 9^4 = (9^2)^2 = 81^2 = 6561 \]

### Step 5: Substitute back into the expression
Now we substitute \( 9^4 \) back into our expression:
\[ (20 - 1)^{9^4} \equiv -1 - 6561 \cdot 20 \mod 100 \]

### Step 6: Simplify the expression modulo \( 100 \)
Calculating \( 6561 \cdot 20 \mod 100 \):
\[ 6561 \mod 100 = 61 \quad \text{(since } 6561 = 65 \cdot 100 + 61\text{)} \]
Thus,
\[ 6561 \cdot 20 \mod 100 = 61 \cdot 20 = 1220 \mod 100 = 20 \]
Now substituting back:
\[ -1 - 20 \equiv -21 \mod 100 \equiv 79 \mod 100 \]

### Step 7: Final result
Thus, the last two digits of \( 19^{9^4} \) are:
\[ \boxed{79} \]