When `(3)^1989` is divided by 7, then the remainder will be?

Options

A. 4
B. 6
C. 5
D. 3

Answer

To find the remainder when \(3^{1989}\) is divided by 7, we can use Fermat's Little Theorem, which states that if \(p\) is a prime number and \(a\) is an integer not divisible by \(p\), then:

\[ a^{p-1} \equiv 1 \ (\text{mod} \ p) \]

In this case, \(a = 3\) and \(p = 7\). Since 3 is not divisible by 7, we can apply the theorem.

### Step 1: Calculate \(p - 1\)
First, we calculate \(p - 1\):
\[ p - 1 = 7 - 1 = 6 \]

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

### Step 3: Reduce the exponent modulo \(p - 1\)
Now we need to reduce the exponent 1989 modulo 6:
\[ 1989 \mod 6 \]

To find \(1989 \mod 6\), we can divide 1989 by 6:
\[ 1989 \div 6 = 331.5 \quad \text{(take the integer part, which is 331)} \]
\[ 331 \times 6 = 1986 \]
\[ 1989 - 1986 = 3 \]
Thus,
\[ 1989 \mod 6 = 3 \]

### Step 4: Substitute back into the expression
Now we substitute back into our expression using the reduced exponent:
\[ 3^{1989} \equiv 3^3 \ (\text{mod} \ 7) \]

### Step 5: Calculate \(3^3\)
Now we calculate \(3^3\):
\[ 3^3 = 27 \]

### Step 6: Find the remainder when \(27\) is divided by \(7\)
Now we find the remainder of \(27\) when divided by \(7\):
\[ 27 \div 7 = 3 \quad \text{(remainder is 6)} \]
So,
\[ 27 \mod 7 = 6 \]

### Final Answer
Thus, the remainder when \(3^{1989}\) is divided by \(7\) is:
\[ \boxed{6} \]