Find the remainder when $3^{1999}$ is divided by $13$.
Since $3^3=27=2\cdot13+1$ we find that  \[3^3\equiv1\pmod{13}.\] Therefore \[3^{1999}\equiv3^{3\cdot666+1}\equiv1^{666}\cdot3\equiv3\pmod{13}.\] The remainder when $3^{1999}$ is divided by 13 is $\boxed{3}$.
3
Level 4
Number Theory
Problem: Find the remainder when $3^{1999}$ is divided by $13$.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the remainder of $3^{1999}$ when divided by $13$, we can use Fermat's Little Theorem. This theorem states that if $p$ is a prime number and $a$ is an integer not divisible by $p$, then $a^{p-1} \equiv 1 \mod p$. Here, $p = 13$ and $a = 3$, which is not divisible by $13$. 

According to Fermat's Little Theorem, we have:

$3^{12} \equiv 1 \mod 13$.

Next, we need to express $1999$ in terms of $12$ to utilize this theorem. We calculate the division of $1999$ by $12$:

$1999 \div 12 = 166$ remainder $7$.

This means:

$1999 = 12 \times 166 + 7$.

Using this, we can rewrite $3^{1999}$ as:

$3^{1999} = 3^{12 \times 166 + 7} = (3^{12})^{166} \cdot 3^7$.

Applying Fermat's theorem, we know that $(3^{12})^{166} \equiv 1^{166} \equiv 1 \mod 13$. Therefore, we simplify:

$3^{1999} \equiv 1 \cdot 3^7 \mod 13$.

Now, we need to compute $3^7 \mod 13$. We can calculate this step-by-step:

$3^1 \equiv 3 \mod 13$,

$3^2 \equiv 9 \mod 13$,

$3^3 \equiv 27 \equiv 1 \mod 13$ (since $27 - 2 \times 13 = 1$),

$3^4 \equiv 3 \cdot 3^3 \equiv 3 \cdot 1 \equiv 3 \mod 13$,

$3^5 \equiv 3 \cdot 3^4 \equiv 3 \cdot 3 \equiv 9 \mod 13$,

$3^6 \equiv 3 \cdot 3^5 \equiv 3 \cdot 9 \equiv 27 \equiv 1 \mod 13$,

$3^7 \equiv 3 \cdot 3^6 \equiv 3 \cdot 1 \equiv 3 \mod 13$.

Thus, we find that:

$3^{1999} \equiv 3^7 \equiv 3 \mod 13$.

The final answer is $3$.<|END_OF_LATENT|> The final answer is $3$.