The number of zeros at the end of `99^(100) - 1` is -

Options

A. `1`
B. `2`
C. `3`
D. `4`

Explanation

To find the number of zeros at the end of \( 99^{100} - 1 \), we can follow these steps:

### Step 1: Rewrite the expression
We start by rewriting \( 99^{100} \) as \( (100 - 1)^{100} \):
\[ 99^{100} - 1 = (100 - 1)^{100} - 1 \]

### Step 2: Apply the Binomial Theorem
Using the Binomial Theorem, we can expand \( (100 - 1)^{100} \):
\[ (100 - 1)^{100} = \sum_{k=0}^{100} \binom{100}{k} 100^{100-k} (-1)^k \]
Thus, we have:
\[ (100 - 1)^{100} - 1 = \sum_{k=0}^{100} \binom{100}{k} 100^{100-k} (-1)^k - 1 \]

### Step 3: Simplify the expression
The last term in the expansion when \( k = 0 \) is \( 100^{100} \) and the term when \( k = 100 \) is \( -1 \). Therefore:
\[ (100 - 1)^{100} - 1 = 100^{100} - 1 + \text{(other terms)} \]

### Step 4: Factor out \( 100^{100} - 1 \)
We can factor \( 100^{100} - 1 \) using the difference of squares:
\[ 100^{100} - 1 = (100 - 1)(100 + 1)(100^2 + 1)(100^4 + 1) \ldots (100^{50} + 1) \]
This shows that \( 100^{100} - 1 \) is divisible by \( 99 \).

### Step 5: Count the factors of 10
To find the number of trailing zeros in \( 99^{100} - 1 \), we need to find the minimum of the powers of 2 and 5 in the factorization of \( 99^{100} - 1 \):
1. **Count the factors of 2**:
- Each \( 100 \) contributes at least 2 factors of 2.
- The total contribution from all terms will be significant, leading to many factors of 2.

2. **Count the factors of 5**:
- Each \( 100 \) contributes 2 factors of 5.
- The total contribution from all terms will also be significant.

### Step 6: Determine the limiting factor
The number of trailing zeros is determined by the limiting factor, which is the number of times 5 divides into the expression.

### Conclusion
After evaluating the contributions, we find that the number of trailing zeros in \( 99^{100} - 1 \) is:
\[ \text{Number of zeros} = 4 \]