Find the product of $218_9 \cdot 5_9$.  Express your answer in base 9.
We begin by multiplying the units digit: $8_9 \times 5_9 = 40_{10} = 44_9$. So, we write down a $4$ and carry-over another $4$. Moving on to the next digit, we need to evaluate $1_9 \times 5_9 + 4_9 = 9_{10} = 10_{9}$. Thus, the next digit is a $0$ and a $1$ is carried over. Finally, the leftmost digits are given by the operation $2_9 \times 5_9 + 1_9 = 11_{10} = 12_9$. Writing this out, we have  $$\begin{array}{@{}c@{\;}c@{}c@{}c@{}c@{}c@{}c}
& & & & \stackrel{1}{2} & \stackrel{4}{1} & \stackrel{}{8}_9 \\
& & & \times & & & 5_9 \\
\cline{4-7} & & & 1 & 2 & 0 & 4_9 \\
\end{array}$$So our final answer is $\boxed{1204_9}$.
1204_9
Level 4
Number Theory
Problem: Find the product of $218_9 \cdot 5_9$.  Express your answer in base 9.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the product of $218_9$ and $5_9$, first convert both numbers from base 9 to base 10. 

For $218_9$:
- The digit 2 is in the 81's place (9^2), contributing $2 \cdot 81 = 162$.
- The digit 1 is in the 9's place (9^1), contributing $1 \cdot 9 = 9$.
- The digit 8 is in the 1's place (9^0), contributing $8 \cdot 1 = 8$.
Adding these contributions gives $162 + 9 + 8 = 179_{10}$.

For $5_9$:
- The digit 5 is in the 1's place (9^0), contributing $5 \cdot 1 = 5_{10}$.

Now, multiply the two base 10 numbers:
$179_{10} \cdot 5_{10} = 895_{10}$.

Next, convert $895_{10}$ back to base 9. To do this, divide 895 by 9 and keep track of the remainders:
1. $895 \div 9 = 99$ with a remainder of 4.
2. $99 \div 9 = 11$ with a remainder of 0.
3. $11 \div 9 = 1$ with a remainder of 2.
4. $1 \div 9 = 0$ with a remainder of 1.

Now, read the remainders from bottom to top to get the base 9 representation: $1204_9$.

Thus, the final answer is $1204_9$.<|END_OF_LATENT|> The final answer is $1204_9$.