SELECTION and TRAINING SESSION

In the given sequence consider all possible quadruple of successive digits. There are finite number of possible quadruples (no more than $10^4$). So if we proceed the sequence sufficiently long, some of the quadruple will occur more than once: $$2029\dots dcba\dots d_1{c}_1{b}_1{a}_1\dots$$ (here $d=d_1$, $c=c_1$, $b=b_1$, $a=a_1$). Note that any quadruple uniquely defines the preceding digits. That is, if $e$ and $e_1$ precede to $dcba$ and $d_1{c}_1{b}_1{a}_1$, respectively, then $e=e_1$: $$2029\dots edcba\dots e_1{d}_1{c}_1{b}_1{a}_1\dots$$ Hence the quadruple $2029$ should occur in the given sequence somewhere between $dcba$ and $d_1{c}_1{b}_1{a}_1$: $$2029\dots dcba\dots 2029\dots d_1{c}_1{b}_1{a}_1\dots$$ Now proceeding to the left we can easily see which digits precede the second quadruple $2029$: $$2029\dots dcba\dots 2015848552029d_1{c}_1{b}_1{a}_1\dots$$ We see that the quadruple $2015$ occurs in the sequence.

Remark. Considering the given sequence modulo $2$ we see that it is periodic with $5$ as a period. Further, considering the sequence modulo $5$, starting from $2024$ ($\equiv 2029(\mathrm{mod}5)$) after $78$ moves (it takes about ten minutes) we can get $1012$ which are $3\cdot 2,3\cdot 0,3\cdot 2,3\cdot 4$ modulo $5$. So after $4\cdot 78$ moves we get $3^4\cdot 2,3^4\cdot 0,3^2,3^4\cdot 4$ which is the starting quadruple $2024$. Thus we conclude that the sequence modulo $5$ has period $4\cdot 78=312$. So the initial sequence is periodic with period $5\cdot 312=1560$. So to get the quadruple $2015$ in the sequence we need to perform $1560-9=1551$ moves.