Find the greatest integer which divides the number `101^100 - 1 `

Options

A. `100^2`
B. `100^3`
C. `100^4`
D. `100^5`

Answer

To find the greatest integer that divides \( 101^{100} - 1 \), we can use the properties of binomial expansion and factorization. Here’s a step-by-step solution:

### Step 1: Recognize the Expression
We need to evaluate \( 101^{100} - 1 \). This expression can be factored using the difference of squares and the properties of exponents.

### Step 2: Factor the Expression
The expression \( a^n - b^n \) can be factored as:
\[ a^n - b^n = (a - b)(a^{n-1} + a^{n-2}b + a^{n-3}b^2 + \ldots + b^{n-1}) \]
For our case, let \( a = 101 \) and \( b = 1 \):
\[ 101^{100} - 1^{100} = (101 - 1)(101^{99} + 101^{98} \cdot 1 + 101^{97} \cdot 1^2 + \ldots + 1^{99}) \]
This simplifies to:
\[ 101^{100} - 1 = 100(101^{99} + 101^{98} + \ldots + 1) \]

### Step 3: Evaluate the Sum
The sum \( 101^{99} + 101^{98} + \ldots + 1 \) is a geometric series with \( n = 100 \) terms, first term \( a = 1 \), and common ratio \( r = 101 \). The sum of a geometric series can be calculated using the formula:
\[ S_n = \frac{a(r^n - 1)}{r - 1} \]
Substituting the values:
\[ S_{100} = \frac{1(101^{100} - 1)}{101 - 1} = \frac{101^{100} - 1}{100} \]

### Step 4: Combine the Results
Thus, we can express \( 101^{100} - 1 \) as:
\[ 101^{100} - 1 = 100 \cdot S_{100} = 100 \cdot \frac{101^{100} - 1}{100} = 100 \cdot (101^{99} + 101^{98} + \ldots + 1) \]

### Step 5: Identify the Greatest Integer Divisor
From the factorization, we see that \( 101^{100} - 1 \) is divisible by \( 100 \). Now, we need to find out if there are any larger factors.

### Step 6: Check for Higher Powers of 10
Since \( 100 = 10^2 \), we check if \( 10^3 = 1000 \) divides \( 101^{100} - 1 \). We can use the Lifting The Exponent Lemma (LTE) which states that for odd primes \( p \) and integers \( a \) and \( b \):
\[ v_p(a^n - b^n) = v_p(a - b) + v_p(a + b) + v_p(n) - 1 \]
In our case, \( p = 2 \) and \( p = 5 \) both yield \( v_2(101^{100} - 1) = 2 + 1 + 2 - 1 = 4 \) and \( v_5(101^{100} - 1) = 1 + 1 + 2 - 1 = 3 \).

### Conclusion
Thus, the highest power of \( 10 \) that divides \( 101^{100} - 1 \) is \( 10^3 = 1000 \).

### Final Answer
The greatest integer which divides \( 101^{100} - 1 \) is:
\[ \boxed{10000} \]