Find the product of $218_9 \cdot 5_9$.  Express your answer in base 9.
To multiply two numbers in base 9, I can use the standard algorithm of multiplying each digit by each digit and adding the results, but I have to be careful with the arithmetic and the carry-overs, since they are different in base 9. First, I'll multiply 218 by 5, starting from the rightmost digit. 8 times 5 is 40 in base 10, but in base 9, 40 is equal to 44, since 9 times 4 is 36 and 4 more is 40. So I write 4 in the ones place and carry over 4 to the next column. Next, I multiply 1 by 5, which is 5, and add the carry-over of 4, which gives 9.
[*,] But in base 9, 9 is equal to 10, since 9 times 1 is 9 and 1 more is 10. So I write 0 in the nines place and carry over 1 to the next column. Finally, I multiply 2 by 5, which is 10, and add the carry-over of 1, which gives 11. But in base 9, 11 is equal to 12, since 9 times 1 is 9 and 2 more is 11. So far, I have 1204 as the partial product of 218 and 5 in base 9. The final product is 1204 in base 9. 