The remainder 'R' when `3^(37)+ 4^(37)` is divided by 7 is :

Options

A. 0
B. 1
C. 3
D. none of these

Answer

To find the remainder \( R \) when \( 3^{37} + 4^{37} \) is divided by 7, we can use Fermat's Little Theorem, which states that if \( p \) is a prime and \( a \) is an integer not divisible by \( p \), then \( a^{p-1} \equiv 1 \mod p \).

### Step 1: Identify the prime and the integers
Here, \( p = 7 \) (a prime number) and our integers are \( 3 \) and \( 4 \).

### Step 2: Apply Fermat's Little Theorem
According to Fermat's Little Theorem:
- For \( a = 3 \):
\[ 3^{6} \equiv 1 \mod 7 \]
- For \( a = 4 \):
\[ 4^{6} \equiv 1 \mod 7 \]

### Step 3: Reduce the exponent modulo \( p-1 \)
Since \( 6 = 7 - 1 \), we can reduce \( 37 \) modulo \( 6 \):
\[ 37 \mod 6 = 1 \]
Thus,
\[ 3^{37} \equiv 3^{1} \mod 7 \]
\[ 4^{37} \equiv 4^{1} \mod 7 \]

### Step 4: Calculate \( 3^{1} \) and \( 4^{1} \)
Now we can compute:
\[ 3^{1} \equiv 3 \mod 7 \]
\[ 4^{1} \equiv 4 \mod 7 \]

### Step 5: Add the results
Now, we add the two results:
\[ 3 + 4 = 7 \]

### Step 6: Find the remainder when divided by 7
Finally, we find the remainder when \( 7 \) is divided by \( 7 \):
\[ 7 \mod 7 = 0 \]

### Conclusion
Thus, the remainder \( R \) when \( 3^{37} + 4^{37} \) is divided by \( 7 \) is:
\[ \boxed{0} \]