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 the number \( 19^{9^4} \), we can use modular arithmetic, specifically modulo 100, since we are interested in the last two digits.

### Step-by-step Solution:

1. **Express the problem in terms of modulo 100**:
We need to calculate \( 19^{9^4} \mod 100 \).

2. **Calculate \( 9^4 \)**:
First, we compute \( 9^4 \):
\[ 9^2 = 81 \]
\[ 9^4 = 81^2 = 6561 \]

3. **Now we need to find \( 19^{6561} \mod 100 \)**:
To simplify this calculation, we can use Euler's theorem, which states that if \( a \) and \( n \) are coprime, then:
\[ a^{\phi(n)} \equiv 1 \mod n \]
where \( \phi(n) \) is Euler's totient function.

4. **Calculate \( \phi(100) \)**:
The prime factorization of 100 is \( 2^2 \times 5^2 \). Thus:
\[ \phi(100) = 100 \left(1 - \frac{1}{2}\right)\left(1 - \frac{1}{5}\right) = 100 \times \frac{1}{2} \times \frac{4}{5} = 40 \]

5. **Reduce the exponent \( 6561 \mod 40 \)**:
Now we need to reduce \( 6561 \) modulo \( 40 \):
\[ 6561 \div 40 = 164.025 \quad \Rightarrow \quad 164 \text{ full cycles of } 40 \]
\[ 6561 - 164 \times 40 = 6561 - 6560 = 1 \]
Thus, \( 6561 \mod 40 = 1 \).

6. **Calculate \( 19^{6561} \mod 100 \)**:
Since \( 6561 \equiv 1 \mod 40 \), we have:
\[ 19^{6561} \equiv 19^1 \mod 100 \]
Therefore:
\[ 19^{6561} \mod 100 = 19 \]

7. **Conclusion**:
The last two digits of \( 19^{9^4} \) are \( 19 \).

### Final Answer:
The last two digits of \( 19^{9^4} \) are **19**.