The number `101^(100)` -1 is divisible by

Options

A. 100
B. 1000
C. 10000
D. all the above

Explanation

To determine the divisibility of \( 101^{100} - 1 \), we can utilize the Binomial Theorem. Here's a step-by-step solution:

### Step 1: Rewrite \( 101^{100} \)
We can express \( 101 \) as \( 1 + 100 \). Therefore, we can rewrite \( 101^{100} \) as:
\[ 101^{100} = (1 + 100)^{100} \]

### Step 2: Apply the Binomial Theorem
According to the Binomial Theorem, we have:
\[ (1 + x)^n = \sum_{k=0}^{n} \binom{n}{k} x^k \]
Applying this to our expression:
\[ (1 + 100)^{100} = \sum_{k=0}^{100} \binom{100}{k} 100^k \]

### Step 3: Expand the expression
The expansion gives us:
\[ (1 + 100)^{100} = \binom{100}{0} 100^0 + \binom{100}{1} 100^1 + \binom{100}{2} 100^2 + \ldots + \binom{100}{100} 100^{100} \]
This simplifies to:
\[ 1 + 100 \cdot 100 + \frac{100 \cdot 99}{2} \cdot 100^2 + \ldots + 100^{100} \]

### Step 4: Consider \( 101^{100} - 1 \)
Now, we need to find \( 101^{100} - 1 \):
\[ 101^{100} - 1 = \left( \sum_{k=0}^{100} \binom{100}{k} 100^k \right) - 1 \]
This simplifies to:
\[ 100 \cdot 100 + \frac{100 \cdot 99}{2} \cdot 100^2 + \ldots + 100^{100} \]

### Step 5: Factor out \( 100 \)
Notice that every term in this expansion, except for the first term, has \( 100 \) as a factor. Therefore, we can factor out \( 100 \):
\[ = 100 \left( 100 + \frac{99}{2} \cdot 100^2 + \ldots + 100^{99} \right) \]

### Step 6: Check for higher divisibility
Since \( 100 = 10^2 \), we can see that \( 101^{100} - 1 \) is divisible by \( 100 \). Furthermore, since \( 100 = 10^2 \), it is also divisible by \( 1000 \) (which is \( 10^3 \)) and \( 10000 \) (which is \( 10^4 \)).

### Conclusion
Thus, we conclude that:
\[ 101^{100} - 1 \text{ is divisible by } 10000 \]