Statement-1: The remainder when `(128)^((128)^128` is divided by 7 is 3. because Statement-2: `(128)^128` when divided by 3 leaves the remainder 1.

Options

A. Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.
B. Statement-1 is true, statement-2 is true and statement-2 is not the correct explanation for statement-1.
C. Statement-1 is true, statement-2 is false.
D. Statement-1 is false, statement-2 is true.

Answer

To solve the problem, we need to analyze both statements step by step.

### Step-by-Step Solution:

**Step 1: Evaluate Statement 1**

We need to find the remainder when \( 128^{(128^{128})} \) is divided by 7.

1. First, we calculate \( 128 \mod 7 \):
\[ 128 \div 7 = 18 \quad \text{(integer part)} \]
\[ 128 - (18 \times 7) = 128 - 126 = 2 \]
So, \( 128 \equiv 2 \mod 7 \).

2. Therefore, we can rewrite \( 128^{(128^{128})} \) as:
\[ 128^{(128^{128})} \equiv 2^{(128^{128})} \mod 7 \]

3. Now, we need to find \( 128^{128} \mod 6 \) (since \( \phi(7) = 6 \)):
\[ 128 \div 6 = 21 \quad \text{(integer part)} \]
\[ 128 - (21 \times 6) = 128 - 126 = 2 \]
So, \( 128 \equiv 2 \mod 6 \).

4. Now we need to calculate \( 2^{128} \mod 6 \):
\[ 2^1 \equiv 2 \mod 6 \]
\[ 2^2 \equiv 4 \mod 6 \]
\[ 2^3 \equiv 2 \mod 6 \]
The pattern repeats every 2 terms: \( 2, 4, 2, 4, \ldots \)

5. Since \( 128 \) is even, we have:
\[ 2^{128} \equiv 4 \mod 6 \]

6. Therefore, we need to evaluate \( 2^4 \mod 7 \):
\[ 2^4 = 16 \]
\[ 16 \div 7 = 2 \quad \text{(integer part)} \]
\[ 16 - (2 \times 7) = 16 - 14 = 2 \]
So, \( 2^{(128^{128})} \equiv 2 \mod 7 \).

**Conclusion for Statement 1**: The remainder is 2, not 3. Thus, Statement 1 is **false**.

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**Step 2: Evaluate Statement 2**

We need to find the remainder when \( 128^{128} \) is divided by 3.

1. First, we calculate \( 128 \mod 3 \):
\[ 128 \div 3 = 42 \quad \text{(integer part)} \]
\[ 128 - (42 \times 3) = 128 - 126 = 2 \]
So, \( 128 \equiv 2 \mod 3 \).

2. Therefore, we can rewrite \( 128^{128} \) as:
\[ 128^{128} \equiv 2^{128} \mod 3 \]

3. Now we need to calculate \( 2^{128} \mod 3 \):
\[ 2^1 \equiv 2 \mod 3 \]
\[ 2^2 \equiv 1 \mod 3 \]
The pattern repeats every 2 terms: \( 2, 1, 2, 1, \ldots \)

4. Since \( 128 \) is even, we have:
\[ 2^{128} \equiv 1 \mod 3 \]

**Conclusion for Statement 2**: The remainder is indeed 1. Thus, Statement 2 is **true**.

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