## Task B-1.5.

Determine the digits $x, y, z$ and $t$ such that $\overline{x y z t}+\overline{y z t}+\overline{z t}+\bar{t}=2018$.
## Solution.

From the conditions of the problem, we have

$$
\begin{gathered}
1000 x + 100 y + 10 z + t + 100 y + 10 z + t + 10 z + t + t = 2018 \\
1000 x + 200 y + 30 z + 4 t = 2018 \\
500 x + 100 y + 15 z + 2 t = 1009
\end{gathered}
$$

Obviously, \( x \leqslant 2 \).

1 point

For \( x = 1 \), we have \( 100 y + 15 z + 2 t = 509 \).

Since \( z \) and \( t \) are at most 9 (in which case \( 15 z + 2 t = 153 \)), \( y \) must be at least 4 and at most 5.

1 point

For \( y = 4 \), we have \( 15 z + 2 t = 109 \), so \( z \) must be at least 6 and at most 7.

1 point

For \( z = 6 \), we get \( 2 t = 19 \), which is not possible.

For \( z = 7 \), we get \( 2 t = 4 \), so \( t = 2 \). 1 point

For \( y = 5 \), we get \( 15 z + 2 t = 9 \), which is not possible for any choice of digits \( z \) and \( t \). 1 point

If \( x = 2 \), we have \( 100 y + 5 z + 2 t = 9 \), which is not possible for any choice of digits.

1 point

The only solution is \( x = 1 \), \( y = 4 \), \( z = 7 \), and \( t = 2 \).

1 point