Statement-1: `11^25 + 12^25` when divided by 23 leaves the remainder zero.Statement-2: `(a + b)^n` is divisible by `(a + b)` for all values of `n epsilon`N

Answer

To solve the problem, we need to analyze both statements separately and determine their validity.

### Step 1: Analyze Statement 1
We need to check if \( 11^{25} + 12^{25} \) is divisible by 23.

**Using Fermat's Little Theorem:**
Fermat's Little Theorem states that if \( p \) is a prime number and \( a \) is an integer not divisible by \( p \), then:
\[ a^{p-1} \equiv 1 \mod p \]
Here, \( p = 23 \).

1. **Calculate \( 11^{25} \mod 23 \):**
- Since \( 11 \) is not divisible by \( 23 \), we can apply Fermat's theorem:
\[ 11^{22} \equiv 1 \mod 23 \]
- Therefore, \( 11^{25} = 11^{22} \cdot 11^3 \equiv 1 \cdot 11^3 \mod 23 \).
- Now calculate \( 11^3 \):
\[ 11^2 = 121 \equiv 6 \mod 23 \quad (\text{since } 121 - 5 \cdot 23 = 6) \]
\[ 11^3 = 11 \cdot 11^2 = 11 \cdot 6 = 66 \equiv 20 \mod 23 \quad (\text{since } 66 - 2 \cdot 23 = 20) \]

2. **Calculate \( 12^{25} \mod 23 \):**
- Again, since \( 12 \) is not divisible by \( 23 \):
\[ 12^{22} \equiv 1 \mod 23 \]
- Therefore, \( 12^{25} = 12^{22} \cdot 12^3 \equiv 1 \cdot 12^3 \mod 23 \).
- Now calculate \( 12^3 \):
\[ 12^2 = 144 \equiv 6 \mod 23 \quad (\text{since } 144 - 6 \cdot 23 = 6) \]
\[ 12^3 = 12 \cdot 12^2 = 12 \cdot 6 = 72 \equiv 3 \mod 23 \quad (\text{since } 72 - 3 \cdot 23 = 3) \]

3. **Combine the results:**
\[ 11^{25} + 12^{25} \equiv 20 + 3 \equiv 23 \equiv 0 \mod 23 \]
Thus, \( 11^{25} + 12^{25} \) is divisible by 23.

### Conclusion for Statement 1:
**Statement 1 is true.**

### Step 2: Analyze Statement 2
We need to check if \( (a + b)^n \) is divisible by \( (a + b) \) for all \( n \in \mathbb{N} \).

1. **Consider the Binomial Expansion:**
\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \]
- The first term when \( k=0 \) is \( a^n \) and the last term when \( k=n \) is \( b^n \).
- The remaining terms involve products of \( a \) and \( b \).

2. **Divisibility Check:**
- For \( n=1 \):
\[ (a + b)^1 = a + b \]
- This is trivially divisible by \( a + b \).
- For \( n=2 \):
\[ (a + b)^2 = a^2 + 2ab + b^2 \]
- This is also divisible by \( a + b \).
- However, for odd \( n \), the term \( a^n + b^n \) does not necessarily yield a remainder of zero when divided by \( a + b \).

3. **Counterexample:**
- For \( n=3 \):
\[ (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 \]
- Here, \( a^3 + b^3 \) is not divisible by \( a + b \) unless \( a = -b \).

### Conclusion for Statement 2:
**Statement 2 is false.**

### Final Conclusion:
- Statement 1 is true.
- Statement 2 is false.