Like an Accordion

Since $3^{2}=9$ , we can do a 'shortcut':

$\begin{aligned} 121212121212_{3} &= 1 \times 3^{11} + 2 \times 3^{10} + 1 \times 3^{9} + 2 \times 3^{8} + 1 \times 3^{7} + 2 \times 3^{6} + 1 \times 3^{5} + 2 \times 3^{4} + 1 \times 3^{3} + 2 \times 3^{2} + 1 \times 3^{1} + 2 \times 3^{0} \\ &=(1 \times 3 + 2) \times 9^{5} + (1 \times 3 + 2) \times 9^{4} + (1 \times 3 + 2) \times 9^{3} + (1 \times 3 + 2) \times 9^{2} + (1 \times 3 + 2) \times 9^{1} + (1 \times 3 + 2) \times 9^{0} \\ &=5 \times 9^{5} + 5 \times 9^{4} + 5 \times 9^{3} + 5 \times 9^{2} + 5 \times 9^{1} + 5 \times 9^{0} \end{aligned}$

Now, we can easily see that $121212121212_{3}=555555_{9}$ .

Essentially, you're just pairing the digits in the number in base 3 from right to left, converting the pairs to base 10 (yes, base 10), and that's the number in base 9.

There is a little shortcut: $\begin{aligned}121212121212_{3} &= 10101010101_{3} \times 12_{3} = 111111_{9} \times 5_{9} = 555555_{9}\end{aligned}$

The second equality is true since in decimal representation of $\begin{aligned}10101010101_{3}\end{aligned}$ , we have only even powers of 3 that in turn are equal to correspondent consecutive powers of 9.

Since $3^2 = 9$ , we can take the number $121212121212_3$ and split it into groups of 2-digits each, and convert each 2 digits in base 3 into 1 digit in base 9:

In our example we got 6 groups of $12_3$ so we convert each group seperately from base 3 to base 9:

$12_3 = 5_9$

so:

$121212121212_3 = 555555_9$

Let's call the first digit (singles digit) of the outcome X. We can simply find X by adding the first digits of the given number until they exceed 8.

$X = 2\times3^0 + 1 \times 3^1 = 5$

If we add $2 \times 3^2$ we will get $5+18 = 23$ which is more than one digit in base 9, therefore the first digit has to be 5.