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 to expand both terms.

### Step-by-step Solution:

1. **Rewrite the Terms**:
\[ 6^{83} = (7 - 1)^{83} \quad \text{and} \quad 8^{83} = (7 + 1)^{83} \]

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

3. **Combine the Expansions**:
Now, we add the two expansions:
\[ 6^{83} + 8^{83} = (7 - 1)^{83} + (7 + 1)^{83} \]
This results in:
\[ = \sum_{k=0}^{83} \binom{83}{k} 7^{83-k} (-1)^k + \sum_{k=0}^{83} \binom{83}{k} 7^{83-k} (1)^k \]

4. **Simplify the Combined Series**:
Notice that when \( k \) is even, the terms will add up, and when \( k \) is odd, they will cancel out. Therefore, we only need to consider the even terms:
\[ = 2 \sum_{j=0}^{41} \binom{83}{2j} 7^{83-2j} \]

5. **Focus on the Remainder**:
We can factor out \( 49 \) from the terms, as \( 7^2 = 49 \). Thus, we can express the sum as:
\[ = 2 \cdot \text{(some integer)} + \text{(remaining terms)} \]
The remaining terms will be those where \( k = 1 \):
\[ 2 \cdot \binom{83}{1} \cdot 7^{82} = 2 \cdot 83 \cdot 7^{82} \]

6. **Calculate the Remainder**:
Now we focus on the last term:
\[ 2 \cdot 83 \cdot 7 = 1162 \]
When we divide \( 1162 \) by \( 49 \):
\[ 1162 \div 49 = 23 \quad \text{(integer part)} \]
\[ 1162 - 49 \cdot 23 = 1162 - 1127 = 35 \]

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