In `n in N`, then `121^(n)-25^(n)+ 1900^(n)- (-4)^(n)` is divisible by

Options

A. 1904
B. 2000
C. 2002
D. 2006

Answer

To solve the question, we need to analyze the expression \(121^n - 25^n + 1900^n - (-4)^n\) and determine what it is divisible by for \(n \in \mathbb{N}\).

### Step 1: Break down the expression
We start with the expression:
\[ 121^n - 25^n + 1900^n - (-4)^n \]

### Step 2: Evaluate for \(n = 1\)
Let's first evaluate the expression for \(n = 1\):
\[ 121^1 - 25^1 + 1900^1 - (-4)^1 \]
Calculating each term:
\[ 121 - 25 + 1900 + 4 \]
Now, simplify:
\[ 121 - 25 = 96 \]
\[ 96 + 1900 = 1996 \]
\[ 1996 + 4 = 2000 \]

### Step 3: Check divisibility of 2000
Now we need to find the factors of 2000 to determine what it is divisible by.

First, we can factor 2000:
\[ 2000 = 2^4 \times 5^3 \]
The divisors of 2000 can be calculated from its prime factorization.

### Step 4: Generalize for \(n\)
Next, we can analyze the expression for general \(n\). Notice that:
- \(121 = 11^2\)
- \(25 = 5^2\)
- \(1900 = 19 \times 100 = 19 \times 10^2\)
- \(-(-4)^n = 4^n\)

Thus, the expression can be rewritten as:
\[ (11^2)^n - (5^2)^n + (19 \times 10^2)^n - (2^2)^n \]

### Step 5: Check for divisibility by 4
For \(n \geq 2\), we can check the expression modulo 4:
- \(121^n \equiv 1^n \equiv 1 \mod 4\)
- \(25^n \equiv 1^n \equiv 1 \mod 4\)
- \(1900^n \equiv 0 \mod 4\) (since \(1900\) is divisible by \(4\))
- \((-4)^n \equiv 0 \mod 4\)

Thus:
\[ 1 - 1 + 0 - 0 \equiv 0 \mod 4 \]
This shows that the expression is divisible by 4 for \(n \geq 2\).

### Step 6: Conclusion
From our calculations, we can conclude that:
\[ 121^n - 25^n + 1900^n - (-4)^n \text{ is divisible by } 2000 \text{ for } n = 1 \text{ and by } 4 \text{ for } n \geq 2. \]