The remainder obtained when $17^{200}$ is divided by 18, is:

Q: The remainder obtained when $17^{200}$ is divided by 18, is:

To determine the remainder when $17^{200}$ is divided by 18, we can use Fermat's Little Theorem and properties of modular arithmetic. ### Explanation Supporting Option 1: 1. **Fermat's Little Theorem**: This theorem states that if $p$ is a prime number and $a$ is an integer not divisible by $p$, then $a^{p-1} \equiv 1 \pmod{p}$. Here, 18 is not a prime number, but we can still use modular arithmetic to simplify the problem. 2. **Simplifying $17^{200} \mod 18$**: - First, observe that $17 \equiv -1 \pmod{18}$. This is because $17 + 1 = 18$, which is divisible by 18. - Therefore, $17^{200} \equiv (-1)^{200} \pmod{18}$. 3. **Calculating $(-1)^{200}$**: - Since 200 is an even number, $(-1)^{200} = 1$. - Thus, $17^{200} \equiv 1 \pmod{18}$. 4. **Conclusion**: - The remainder when $17^{200}$ is divided by 18 is 1. ### Explanation for Why Other Options are Incorrect: - **Option 2 (2)**: If $17^{200} \equiv 2 \pmod{18}$, it would imply that $17^{200} - 2$ is divisible by 18. However, as shown, $17^{200} \equiv 1 \pmod{18}$, not 2. - **Option 3 (16)**: Similarly, if $17^{200} \equiv 16 \pmod{18}$, it would mean $17^{200} - 16$ is divisible by 18. This contradicts our calculation where $17^{200} \equiv 1 \pmod{18}$. - **Option 4 (17)**: If $17^{200} \equiv 17 \pmod{18}$, it would imply that $17^{200} - 17$ is divisible by 18. Again, this is incorrect as $17^{200} \equiv 1 \pmod{18}$. In conclusion, the remainder when $17^{200}$ is divided by 18 is indeed 1, making Option 1 the correct answer.

Question

Accepted Answer

To determine the remainder when $17^{200}$ is divided by 18, we can use Fermat's Little Theorem and properties of modular arithmetic.

### Explanation Supporting Option 1:

1. **Fermat's Little Theorem**: This theorem states that if $p$ is a prime number and $a$ is an integer not divisible by $p$, then $a^{p-1} \equiv 1 \pmod{p}$. Here, 18 is not a prime number, but we can still use modular arithmetic to simplify the problem.

2. **Simplifying $17^{200} \mod 18$**:
   - First, observe that $17 \equiv -1 \pmod{18}$. This is because $17 + 1 = 18$, which is divisible by 18.
   - Therefore, $17^{200} \equiv (-1)^{200} \pmod{18}$.

3. **Calculating $(-1)^{200}$**:
   - Since 200 is an even number, $(-1)^{200} = 1$.
   - Thus, $17^{200} \equiv 1 \pmod{18}$.

4. **Conclusion**:
   - The remainder when $17^{200}$ is divided by 18 is 1.

### Explanation for Why Other Options are Incorrect:

- **Option 2 (2)**: If $17^{200} \equiv 2 \pmod{18}$, it would imply that $17^{200} - 2$ is divisible by 18. However, as shown, $17^{200} \equiv 1 \pmod{18}$, not 2.

- **Option 3 (16)**: Similarly, if $17^{200} \equiv 16 \pmod{18}$, it would mean $17^{200} - 16$ is divisible by 18. This contradicts our calculation where $17^{200} \equiv 1 \pmod{18}$.

- **Option 4 (17)**: If $17^{200} \equiv 17 \pmod{18}$, it would imply that $17^{200} - 17$ is divisible by 18. Again, this is incorrect as $17^{200} \equiv 1 \pmod{18}$.

In conclusion, the remainder when $17^{200}$ is divided by 18 is indeed 1, making Option 1 the correct answer.