Find the remainder when `(3)^(162)` is divided by 162.

Options

A. 1
B. 81
C. 150
D. 100

Answer

To find the remainder when \(3^{162}\) is divided by 162, we can use the Chinese Remainder Theorem (CRT) since 162 can be factored into its prime components.

1. **Factor 162**:
\[ 162 = 2 \times 81 = 2 \times 3^4 \]

2. **Find \(3^{162} \mod 2\)**:
- Since \(3\) is odd, \(3^{162} \mod 2 = 1\).

3. **Find \(3^{162} \mod 81\)**:
- We can use Euler's theorem here. First, we need to find \(\phi(81)\):
\[ \phi(81) = 81 \left(1 - \frac{1}{3}\right) = 81 \times \frac{2}{3} = 54 \]
- According to Euler's theorem, since \(3\) and \(81\) are coprime:
\[ 3^{54} \equiv 1 \mod 81 \]
- Now, we need to reduce \(162\) modulo \(54\):
\[ 162 \mod 54 = 0 \]
- Therefore:
\[ 3^{162} \equiv (3^{54})^3 \equiv 1^3 \equiv 1 \mod 81 \]

4. **Now we have the two congruences**:
\[ 3^{162} \equiv 1 \mod 2 \]
\[ 3^{162} \equiv 1 \mod 81 \]

5. **Using the Chinese Remainder Theorem**:
- We need to solve the system:
\[ x \equiv 1 \mod 2 \]
\[ x \equiv 1 \mod 81 \]
- The solution to this system is:
\[ x \equiv 1 \mod 162 \]

6. **Final Remainder**:
- Since both congruences give us the same result, the remainder when \(3^{162}\) is divided by \(162\) is:
\[ \text{Remainder} = 1 \]

Thus, the final answer is that the remainder when \(3^{162}\) is divided by \(162\) is **1**.