If ` 6^(83) + 8^(83)` is divided by 49 , the raminder is

Options

A. 0
B. 14
C. 35
D. 42

Explanation

To find the remainder when \( 6^{83} + 8^{83} \) is divided by 49, we can use the Binomial Theorem and properties of modular arithmetic. Here’s a step-by-step solution:

### Step 1: Rewrite the terms
We can express \( 6 \) and \( 8 \) in terms of \( 7 \):
\[ 6 = 7 - 1 \quad \text{and} \quad 8 = 7 + 1 \]
Thus, we can rewrite the expression:
\[ 6^{83} + 8^{83} = (7 - 1)^{83} + (7 + 1)^{83} \]

### Step 2: Apply the Binomial Theorem
Using the Binomial Theorem, we expand both terms:
\[ (7 - 1)^{83} = \sum_{k=0}^{83} \binom{83}{k} 7^{83-k} (-1)^k \]
\[ (7 + 1)^{83} = \sum_{k=0}^{83} \binom{83}{k} 7^{83-k} (1)^k \]

### Step 3: Combine the expansions
Adding the two expansions together:
\[ (7 - 1)^{83} + (7 + 1)^{83} = \sum_{k=0}^{83} \binom{83}{k} 7^{83-k} \left((-1)^k + 1\right) \]
Notice that \( (-1)^k + 1 \) is zero for odd \( k \) and two for even \( k \). Therefore, we only consider the even \( k \):
\[ = 2 \sum_{j=0}^{41} \binom{83}{2j} 7^{83-2j} \]

### Step 4: Analyze the terms modulo 49
Now, we need to find the value of this sum modulo 49. The terms \( 7^{83-2j} \) for \( j \geq 1 \) will be multiples of \( 49 \) (since \( 7^2 = 49 \)), and thus contribute 0 to the remainder when divided by 49.

The only term we need to consider is when \( j = 0 \):
\[ 2 \cdot \binom{83}{0} \cdot 7^{83} = 2 \cdot 1 \cdot 7^{83} \]

### Step 5: Calculate \( 7^{83} \mod 49 \)
Since \( 7^2 = 49 \), we have:
\[ 7^{83} \equiv 0 \mod 49 \]
Thus, the contribution from \( j = 0 \) is also 0.

### Step 6: Combine results
Since all terms contribute 0 modulo 49, we conclude:
\[ 6^{83} + 8^{83} \equiv 0 \mod 49 \]

### Final Answer
The remainder when \( 6^{83} + 8^{83} \) is divided by 49 is:
\[ \boxed{35} \]