What is the $2012th$ number in this pattern?

Question

This is question 30 from Australian Maths 2012

$(0,1,2,1,2,3,2,3,4,1,2,3,2,3,4,3,4,5,2,3,4,...)$

What is the $2012th $ number in this list?

What I did:

I broke up the first few numbers into smaller parts to identify the pattern

$(0,1,2,1,2,3,2,3,4)$

$(1,2,3,2,3,4,3,4,5)$

From this,I can see that every 9 numbers,the pattern restarts.

So, the $2012th$ number should be in the $\frac {2012}{9} th $ group which is the $223 \frac {5}{9} th$ group.

So,the first number in the $223 th $ group is $(223-1)$.

And to further break it up to make it easier to find the $2012th$ number.

From,$(0,1,2,1,2,3,2,3,4)$

$(0,1,2)$-$x$

$(1,2,3)$-$y$

$(2,3,4)$-$z$

The last number number in $x$ is the $2nd$ number in $y$,which is also the first number in $z$.

So,$222+2 =224$ is my answer.

However,checking the answer key I got,the answer is $9$.

Where did I go wrong?

Answer 1

First you write the nonnegative integers in base $3$
$0,1,2,10,11,12,20,21,22,100,101,102,110,111,112,120,\dots$
Then you add up the (base 3) digits in each number (presumably in base $10$) $0,1,2,1,2,3,2,3,4,1,2,3,2,3,4,3,\dots$
Since we start from zero, we express $2011$ in base $3$, getting $2202111_3$, then add the digits to get $9$.

Answer 2

If you graph this sequence of numbers, you will find an interesting curve:

(0,1,2), (1,2,3), (2,3,4), -> first cycle

(1,2,3), (2,3,4), (3,4,5), -> second cycle

(2,3,4), ...

Use Excel or squared paper.