The tens digit of `(81)^(100)(121)^(100)-1` is

Options

A. 1
B. 0
C. 9
D. 8

Answer

To find the tens digit of the expression \( (81)^{100} \cdot (121)^{100} - 1 \), we can simplify the problem using the binomial theorem and properties of powers.

### Step-by-Step Solution:

1. **Rewrite the Expression:**
\[ (81)^{100} \cdot (121)^{100} - 1 = ((81) \cdot (121))^{100} - 1 \]
Here, we can combine the bases since they are both raised to the same power.

2. **Calculate the Product:**
\[ 81 \cdot 121 = 9801 \]
Therefore, we rewrite the expression as:
\[ (9801)^{100} - 1 \]

3. **Use the Binomial Theorem:**
We can expand \( (9801)^{100} \) using the binomial theorem:
\[ (9800 + 1)^{100} = \sum_{k=0}^{100} \binom{100}{k} (9800)^k (1)^{100-k} \]
This simplifies to:
\[ \binom{100}{0} (9800)^{100} + \binom{100}{1} (9800)^{99} + \binom{100}{2} (9800)^{98} + \ldots + \binom{100}{100} \]

4. **Identify Significant Terms:**
The first term \( (9800)^{100} \) will dominate, and we need to find the last two digits of \( (9800)^{100} \) to determine the tens digit of \( (9801)^{100} - 1 \).

5. **Calculate Last Two Digits of \( (9800)^{100} \):**
Since \( 9800 = 98 \cdot 100 \), the last two digits of \( (9800)^{100} \) are determined by \( (00)^{100} \), which is \( 00 \).

6. **Subtract 1:**
Now, we calculate:
\[ (9801)^{100} - 1 = 00 - 1 = 99 \]

7. **Identify the Tens Digit:**
The tens digit of \( 99 \) is \( 9 \).

### Final Answer:
Thus, the tens digit of \( (81)^{100} \cdot (121)^{100} - 1 \) is \( 9 \).