Belarus — 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:
$$2029 \dots d\ c\ b\ a \dots d_1\ c_1\ b_1\ a_1 \dots$$
(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 $d\ c\ b\ a$ and $d_1\ c_1\ b_1\ a_1$, respectively, then $e = e_1$:
$$2029 \dots e\ d\ c\ b\ a \dots e_1\ d_1\ c_1\ b_1\ a_1 \dots$$
Hence the quadruple $2029$ should occur in the given sequence somewhere between $d\ c\ b\ a$ and $d_1\ c_1\ b_1\ a_1$:
$$2029 \dots d\ c\ b\ a \dots 2029 \dots d_1\ c_1\ b_1\ a_1 \dots$$
Now proceeding to the left we can easily see which digits precede the second quadruple $2029$:
$$2029 \dots d\ c\ b\ a \dots 2015\ 84\ 85\ 52\ 029\ d_1\ c_1\ b_1\ a_1 \dots$$
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 $2024$ ($\equiv 2029 \pmod{5}$) 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.