Statement-1 `4^(101)` when divided by 101 leaves the remainder4.
Statement-2 ` (n^(p) -n)` when divided by 'p' leaves
remainder zero when ` n ge 2, n in N ` is a prime number .

Answer

To solve the problem, we need to analyze both statements provided:

### Statement 1:
**4^(101) when divided by 101 leaves the remainder 4.**

To evaluate this statement, we can use Fermat's Little Theorem, which states that if \( p \) is a prime number and \( a \) is an integer not divisible by \( p \), then:

\[ a^{p-1} \equiv 1 \ (\text{mod} \ p) \]

In our case:
- \( a = 4 \)
- \( p = 101 \)

Since 101 is a prime number and 4 is not divisible by 101, we can apply Fermat's theorem:

\[ 4^{100} \equiv 1 \ (\text{mod} \ 101) \]

Now, we need to find \( 4^{101} \mod 101 \):

\[ 4^{101} = 4^{100} \cdot 4 \equiv 1 \cdot 4 \equiv 4 \ (\text{mod} \ 101) \]

Thus, when \( 4^{101} \) is divided by 101, the remainder is indeed 4.

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

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### Statement 2:
**(n^p - n) when divided by 'p' leaves remainder zero when n ≥ 2, n ∈ N is a prime number.**

Again, we can apply Fermat's Little Theorem here. According to the theorem:

\[ n^{p} \equiv n \ (\text{mod} \ p) \]

This implies:

\[ n^{p} - n \equiv 0 \ (\text{mod} \ p) \]

Thus, \( n^{p} - n \) is divisible by \( p \), which means that when \( n^{p} - n \) is divided by \( p \), it leaves a remainder of 0.

### Conclusion for Statement 2:
The statement is **true**.

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