The last two digit in the expansion of `(1989)^(91)` are :

Options

A. 9,1
B. 8,1
C. 6,9
D. 8,9

Answer

To find the last two digits of \( (1989)^{91} \), we can simplify the problem by focusing on the last two digits of the base number, which is \( 1989 \). The last two digits of \( 1989 \) are \( 89 \). Therefore, we need to find the last two digits of \( (89)^{91} \).

### Step-by-Step Solution:

1. **Identify the Base**:
We take the last two digits of \( 1989 \), which gives us:
\[ 1989 \equiv 89 \mod 100 \]
So, we need to compute \( (89)^{91} \mod 100 \).

**Hint**: Focus on the last two digits of the base number.

2. **Use Euler's Theorem**:
To simplify \( (89)^{91} \mod 100 \), we can use Euler's theorem. First, we need to calculate \( \phi(100) \):
\[ 100 = 2^2 \times 5^2 \implies \phi(100) = 100 \left(1 - \frac{1}{2}\right)\left(1 - \frac{1}{5}\right) = 100 \times \frac{1}{2} \times \frac{4}{5} = 40 \]
Since \( \gcd(89, 100) = 1 \), we can apply Euler's theorem:
\[ 89^{40} \equiv 1 \mod 100 \]

**Hint**: Calculate \( \phi(n) \) to use Euler's theorem effectively.

3. **Reduce the Exponent**:
Now, we reduce \( 91 \) modulo \( 40 \):
\[ 91 \mod 40 = 11 \]
Thus, we need to compute \( (89)^{11} \mod 100 \).

**Hint**: Always reduce the exponent using \( \phi(n) \) when using Euler's theorem.

4. **Calculate \( (89)^{11} \mod 100 \)**:
We can compute \( (89)^{11} \) using successive squaring:
- First, calculate \( (89)^2 \):
\[ 89^2 = 7921 \implies 7921 \mod 100 = 21 \]
- Next, calculate \( (89^4) \):
\[ (89^2)^2 = 21^2 = 441 \implies 441 \mod 100 = 41 \]
- Then, calculate \( (89^8) \):
\[ (89^4)^2 = 41^2 = 1681 \implies 1681 \mod 100 = 81 \]
- Now combine to find \( (89^{11}) \):
\[ 89^{11} = 89^8 \cdot 89^2 \cdot 89 = 81 \cdot 21 \cdot 89 \]
- First, calculate \( 81 \cdot 21 \):
\[ 81 \cdot 21 = 1701 \implies 1701 \mod 100 = 1 \]
- Finally, multiply by \( 89 \):
\[ 1 \cdot 89 = 89 \]

**Hint**: Use successive squaring to simplify calculations of large powers.

5. **Conclusion**:
The last two digits of \( (1989)^{91} \) are:
\[ \boxed{89} \]