Task 1 - 151231
Someone solved a division problem \(A\); in this, a natural number, which (in decimal notation) is written with five identical digits, was to be divided by a four-digit natural number, which (in decimal notation) is written with four identical digits.

In this division, the number 16 and a certain remainder resulted.

Subsequently, someone formed a new division problem \(A^{\prime}\) from this problem \(A\) by omitting one digit in both the dividend and the divisor.

When dividing the resulting numbers, the number 16 and a remainder 2000 smaller than in problem \(A\) resulted.

State (by specifying the dividend and divisor) all division problems \(A\) that have this property.
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A five-digit natural number \(Z\) with identical digits can be represented as follows

\[
Z=x+10 x+100 x+1000 x+10000 x=11111 x
\]

where \(x\) is a single-digit natural number \((0<x<10)\). Since analogous relationships also apply to three- and four-digit natural numbers with identical digits, the division problem \(A\) can be written as follows

\[
11111 x=16 \cdot 1111 y+R
\]

where \(0<x<10,0<y<10,0<R<1111 y \quad\) (3) holds and both \(x\) and \(y\) are single-digit natural numbers. The remainder \(R\) is a natural number. An equivalent relationship to (2) can also be established for the division problem \(A^{\prime}\)

\[
1111 x=16 \cdot 111 y+(R-2000)
\]

where, in addition to (3), \(0<(R-2000)<111 y\) (5) must also be satisfied. If we now solve both (2) and (4) for \(R\) and set the resulting relationships equal, we obtain

\[
(11111-1111) x=16 \cdot(1111-111) y+2000
\]

This equation can be solved for \(x\)

\[
x=\frac{8}{5} y+\frac{1}{5}
\]

If we now substitute the 9 different values for \(y\) [compare (3)], we find that only for \(y=3\) and for \(y=8\) a natural number for \(x\) results. Since the result for \(x\) when \(y=8\) is not a single digit, there is only one possible combination: \(x=5, y=3\).

Substituting into (2) immediately yields

\[
55555=16 \cdot 3333+R \Rightarrow R=2227
\]

which shows that all requirements from (3) and (5) are met.

Thus, it is shown that there is only one division problem \(A\); the dividend is 55555 and the divisor is 3333.