What is the remainder when `7^(81)` is divided by 5

Answer

To find the remainder when \( 7^{81} \) is divided by 5, we can use modular arithmetic. Here’s a step-by-step solution:

### Step 1: Simplify \( 7 \) modulo \( 5 \)
First, we find \( 7 \mod 5 \):
\[ 7 \div 5 = 1 \quad \text{(remainder 2)} \]
Thus, \( 7 \equiv 2 \mod 5 \).

### Step 2: Rewrite the expression
We can now rewrite \( 7^{81} \) in terms of \( 2 \):
\[ 7^{81} \equiv 2^{81} \mod 5 \]

### Step 3: Use Fermat's Little Theorem
Fermat's Little Theorem states that if \( p \) is a prime and \( a \) is an integer not divisible by \( p \), then:
\[ a^{p-1} \equiv 1 \mod p \]
In our case, \( p = 5 \) and \( a = 2 \). Thus:
\[ 2^{4} \equiv 1 \mod 5 \]

### Step 4: Reduce the exponent modulo \( 4 \)
Now we need to reduce \( 81 \) modulo \( 4 \):
\[ 81 \div 4 = 20 \quad \text{(remainder 1)} \]
So, \( 81 \equiv 1 \mod 4 \).

### Step 5: Substitute back into the expression
Using the result from Fermat's theorem:
\[ 2^{81} \equiv 2^{1} \mod 5 \]

### Step 6: Calculate the final result
Now we compute \( 2^{1} \):
\[ 2^{1} \equiv 2 \mod 5 \]

### Conclusion
Thus, the remainder when \( 7^{81} \) is divided by \( 5 \) is:
\[ \boxed{2} \]