What is the remainder when $9^{1995}$ is divided by 7?
Note that $9^{1995} \equiv 2^{1995} \pmod{7}$.  Also, note that $2^3 = 8 \equiv 1 \pmod{7}$.  Therefore, \[2^{1995} = 2^{3 \cdot 665} = (2^3)^{665} \equiv \boxed{1} \pmod{7}.\]
1
Level 5
Number Theory
Problem: What is the remainder when $9^{1995}$ is divided by 7?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the remainder of $9^{1995}$ when divided by 7, we can simplify the base first. Since $9 \equiv 2 \mod 7$, we can replace $9$ with $2$. Thus, we need to calculate $2^{1995} \mod 7$. 

Next, 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 $a^{p-1} \equiv 1 \mod p$. Here, $p = 7$ and $a = 2$. According to the theorem, $2^{6} \equiv 1 \mod 7$.

Now, we need to reduce the exponent $1995$ modulo $6$ (since $6$ is $7-1$). We calculate $1995 \mod 6$. Dividing $1995$ by $6$ gives a quotient of $332$ and a remainder of $3$. Therefore, $1995 \equiv 3 \mod 6$.

This means we can simplify our calculation to $2^{1995} \equiv 2^{3} \mod 7$. Now we compute $2^{3}$, which equals $8$. Finally, we find the remainder of $8$ when divided by $7$, which is $1$. 

Thus, the final answer is $1$.<|END_OF_LATENT|> The final answer is $1$.