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 \( (81)^{100} \times (121)^{100} - 1 \), we can follow these steps:

### Step 1: Rewrite the expression
We start by rewriting the bases in a more manageable form:
\[ (81)^{100} = (9^2)^{100} = 9^{200} \]
\[ (121)^{100} = (11^2)^{100} = 11^{200} \]
Thus, we can rewrite the original expression as:
\[ 9^{200} \times 11^{200} - 1 \]

### Step 2: Combine the bases
Using the property of exponents, we can combine the bases:
\[ 9^{200} \times 11^{200} = (9 \times 11)^{200} = 99^{200} \]
So now we have:
\[ 99^{200} - 1 \]

### Step 3: Apply the Binomial Theorem
Next, we can express \( 99^{200} \) using the binomial expansion:
\[ 99^{200} = (100 - 1)^{200} \]
Using the Binomial Theorem, we expand this:
\[ (100 - 1)^{200} = \sum_{k=0}^{200} \binom{200}{k} 100^{200-k} (-1)^k \]
This gives us:
\[ = \binom{200}{0} 100^{200} (-1)^0 + \binom{200}{1} 100^{199} (-1)^1 + \binom{200}{2} 100^{198} (-1)^2 + \ldots + \binom{200}{200} 100^{0} (-1)^{200} \]

### Step 4: Identify relevant terms
We are interested in \( 99^{200} - 1 \):
\[ 99^{200} - 1 = \left( \sum_{k=0}^{200} \binom{200}{k} 100^{200-k} (-1)^k \right) - 1 \]
The constant term (when \( k=0 \)) is \( 100^{200} \) and the term for \( k=1 \) is \( -200 \times 100^{199} \).

### Step 5: Simplify the expression
The first few terms of the expansion are:
\[ 100^{200} - 200 \times 100^{199} + \binom{200}{2} 100^{198} - \ldots \]
Subtracting 1 gives:
\[ 100^{200} - 200 \times 100^{199} + \binom{200}{2} 100^{198} - 1 \]

### Step 6: Calculate the tens digit
To find the tens digit, we need to focus on the last two digits of the expression. The first term \( 100^{200} \) contributes 00, the second term \( -200 \times 100^{199} \) contributes 00, and the third term \( \binom{200}{2} \times 100^{198} \) also contributes 00.

The significant term that affects the last two digits will be the \( -1 \):
\[ 00 - 1 = -1 \]
This means the last two digits are 99.

Thus, the tens digit is:
\[ \text{Tens digit} = 9 \]

### Final Answer
The tens digit of \( (81)^{100} \times (121)^{100} - 1 \) is **9**.

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