What is the remainder obtained on dividing `34^(43)+43^(34)` by 7 ?

Options

A. 4
B. 3
C. 1
D. 0

Answer

To find the remainder of \( 34^{43} + 43^{34} \) when divided by 7, we can follow these steps:

### Step 1: Reduce the bases modulo 7
First, we reduce \( 34 \) and \( 43 \) modulo \( 7 \):
- \( 34 \div 7 = 4 \) remainder \( 6 \) (since \( 34 = 7 \times 4 + 6 \))
- \( 43 \div 7 = 6 \) remainder \( 1 \) (since \( 43 = 7 \times 6 + 1 \))

Thus, we have:
\[ 34 \equiv 6 \mod 7 \]
\[ 43 \equiv 1 \mod 7 \]

### Step 2: Substitute the reduced values into the expression
Now we can substitute these values into our expression:
\[ 34^{43} + 43^{34} \equiv 6^{43} + 1^{34} \mod 7 \]

### Step 3: Simplify \( 1^{34} \)
Since \( 1^{34} = 1 \):
\[ 1^{34} \equiv 1 \mod 7 \]

### Step 4: Calculate \( 6^{43} \mod 7 \)
Next, we need to calculate \( 6^{43} \mod 7 \). 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 \( a = 6 \), so:
\[ 6^{6} \equiv 1 \mod 7 \]

### Step 5: Reduce the exponent modulo 6
Now, we need to reduce \( 43 \) modulo \( 6 \):
\[ 43 \div 6 = 7 \text{ remainder } 1 \]
Thus,
\[ 43 \equiv 1 \mod 6 \]

### Step 6: Calculate \( 6^{43} \mod 7 \)
Now we can simplify \( 6^{43} \):
\[ 6^{43} \equiv 6^{1} \equiv 6 \mod 7 \]

### Step 7: Combine the results
Now we can combine our results:
\[ 34^{43} + 43^{34} \equiv 6 + 1 \mod 7 \]
\[ \equiv 7 \mod 7 \]

### Step 8: Find the remainder
Since \( 7 \) is divisible by \( 7 \):
\[ \text{Remainder} = 0 \]

### Final Answer
The remainder obtained on dividing \( 34^{43} + 43^{34} \) by \( 7 \) is \( \boxed{0} \).