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

Answer

To find the last two digits of the number \( 19^{9^4} \), we can use the concept of modular arithmetic, specifically calculating \( 19^{9^4} \mod 100 \).

### Step-by-step solution:

1. **Rewrite the expression**:
We want to find \( 19^{9^4} \mod 100 \).

2. **Use the Binomial Theorem**:
We can express \( 19 \) as \( 20 - 1 \). Therefore, we have:
\[ 19^{9^4} = (20 - 1)^{9^4} \]

3. **Apply the Binomial Expansion**:
Using the binomial theorem, we expand \( (20 - 1)^{9^4} \):
\[ (20 - 1)^{9^4} = \sum_{k=0}^{9^4} \binom{9^4}{k} 20^k (-1)^{9^4-k} \]

4. **Focus on the last two digits**:
We are interested in the last two digits, which means we need to consider the terms of the expansion that contribute to \( \mod 100 \).

5. **Identify significant terms**:
The terms where \( k = 0 \) and \( k = 1 \) will be significant:
- For \( k = 0 \):
\[ \binom{9^4}{0} 20^0 (-1)^{9^4} = 1 \cdot 1 \cdot (-1) = -1 \]
- For \( k = 1 \):
\[ \binom{9^4}{1} 20^1 (-1)^{9^4-1} = 9^4 \cdot 20 \cdot 1 = 9^4 \cdot 20 \]

6. **Calculate \( 9^4 \mod 5 \) and \( 9^4 \mod 4 \)**:
- \( 9 \equiv 4 \mod 5 \) so \( 9^4 \equiv 4^4 \equiv 1 \mod 5 \).
- \( 9 \equiv 1 \mod 4 \) so \( 9^4 \equiv 1 \mod 4 \).

7. **Use the Chinese Remainder Theorem**:
Since \( 9^4 \equiv 1 \mod 5 \) and \( 9^4 \equiv 1 \mod 4 \), we have:
\[ 9^4 \equiv 1 \mod 20 \]

8. **Substitute back**:
Now substituting back into our earlier expression:
\[ 19^{9^4} \equiv -1 + 20 \cdot 1 \mod 100 \]
\[ \equiv -1 + 20 \equiv 19 \mod 100 \]

9. **Final result**:
Thus, the last two digits of \( 19^{9^4} \) are \( \boxed{19} \).