Statement-1: `2^(33)-1` is divisible by 7.
Statement-2: `x^(n)-a^(n)` is divisibel by x-a, for all `n in N` and `x ne a`.

Answer

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

### Step 1: Analyze Statement 2
Statement 2 states that \( x^n - a^n \) is divisible by \( x - a \) for all \( n \in \mathbb{N} \) and \( x \neq a \).

**Proof:**
1. **Base Case (n = 1):**
\[ x^1 - a^1 = x - a \]
Clearly, \( x - a \) is divisible by \( x - a \).

2. **Inductive Step:**
Assume that the statement is true for \( n = k \), i.e., \( x^k - a^k \) is divisible by \( x - a \). We need to show it is true for \( n = k + 1 \):
\[ x^{k+1} - a^{k+1} = x \cdot x^k - a \cdot a^k = x^k \cdot x - a^k \cdot a \]
We can rewrite this as:
\[ = x^k \cdot x - a^k \cdot a = (x^k - a^k) \cdot x + a^k \cdot (x - a) \]
By the inductive hypothesis, \( x^k - a^k \) is divisible by \( x - a \), and \( a^k \cdot (x - a) \) is also divisible by \( x - a \). Thus, \( x^{k+1} - a^{k+1} \) is divisible by \( x - a \).

By the principle of mathematical induction, Statement 2 is true for all \( n \in \mathbb{N} \).

### Step 2: Analyze Statement 1
Statement 1 states that \( 2^{33} - 1 \) is divisible by 7.

**Using Statement 2:**
We can express \( 2^{33} - 1 \) in the form of \( x^n - a^n \):
Let \( x = 2 \), \( a = 1 \), and \( n = 33 \). According to Statement 2, since \( x - a = 2 - 1 = 1 \), we can conclude:
\[ 2^{33} - 1^{33} \text{ is divisible by } 2 - 1 \]
However, we need to check if \( 2^{33} - 1 \) is divisible by 7.

### Step 3: Check Divisibility by 7
To check if \( 2^{33} - 1 \) is divisible by 7, we can use Fermat's Little Theorem, which states that if \( p \) is a prime and \( a \) is not divisible by \( p \), then:
\[ a^{p-1} \equiv 1 \mod p \]
Here, \( p = 7 \) and \( a = 2 \):
\[ 2^{6} \equiv 1 \mod 7 \]
Now, we need to find \( 2^{33} \mod 7 \):
\[ 33 \mod 6 = 3 \quad (\text{since } 33 = 6 \cdot 5 + 3) \]
Thus,
\[ 2^{33} \equiv 2^3 \mod 7 \]
Calculating \( 2^3 \):
\[ 2^3 = 8 \]
Now, find \( 8 \mod 7 \):
\[ 8 \equiv 1 \mod 7 \]
Therefore,
\[ 2^{33} - 1 \equiv 1 - 1 \equiv 0 \mod 7 \]
This shows that \( 2^{33} - 1 \) is indeed divisible by 7.

### Conclusion
Both statements are true, and Statement 2 provides a correct explanation for Statement 1.

### Final Answer
Both Statement 1 and Statement 2 are true, and Statement 2 is the correct explanation for Statement 1.

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