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

Options

A. `10^(2)`
B. `10^(3)`
C. `10^(4)`
D. `10^(5)`

Answer

To determine the divisibility of the expression \( 101^{100} - 1 \), we can use the Binomial Theorem to expand \( (100 + 1)^{100} \).

### Step-by-Step Solution:

1. **Rewrite the Expression**:
\[ 101^{100} - 1 = (100 + 1)^{100} - 1 \]

2. **Apply the Binomial Theorem**:
The Binomial Theorem states that:
\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \]
In our case, \( a = 100 \), \( b = 1 \), and \( n = 100 \):
\[ (100 + 1)^{100} = \sum_{k=0}^{100} \binom{100}{k} 100^{100-k} \cdot 1^k \]

3. **Expand the Expression**:
Expanding this gives:
\[ (100 + 1)^{100} = \binom{100}{0} 100^{100} + \binom{100}{1} 100^{99} + \binom{100}{2} 100^{98} + \ldots + \binom{100}{100} 1 \]
Thus, we have:
\[ (100 + 1)^{100} - 1 = \binom{100}{0} 100^{100} + \binom{100}{1} 100^{99} + \binom{100}{2} 100^{98} + \ldots + \binom{100}{99} 100 + \binom{100}{100} - 1 \]

4. **Simplify the Expression**:
The last term \( -1 \) cancels with the \( \binom{100}{100} \) term:
\[ = \binom{100}{0} 100^{100} + \binom{100}{1} 100^{99} + \binom{100}{2} 100^{98} + \ldots + \binom{100}{99} 100 \]

5. **Factor Out 100**:
Notice that every term in the expansion (except for the last one which was canceled) contains a factor of \( 100 \):
\[ = 100 \left( \binom{100}{0} 100^{99} + \binom{100}{1} 100^{98} + \ldots + \binom{100}{99} \right) \]
This shows that \( 101^{100} - 1 \) is divisible by \( 100 \).

### Conclusion:
Thus, we conclude that \( 101^{100} - 1 \) is divisible by \( 100 \).