How many steps are there in a staircase where if you step on it in twos, one step remains; if in threes, two; if in fours, three; if in fives, four; if in sixes, five; and finally, if in sevens, none remain.
Let the number of steps be $x$. This can be expressed in the following forms:

$$
x=2 x_{1}+1=3 x_{2}+2=4 x_{4}+3=5 x_{4}+4=6 x_{5}+5=7 x_{6}
$$

We will solve the second of the six indeterminate equations:

$$
\begin{gathered}
2 x_{1}+1=3 x_{2}+2 \\
2 x_{1}=3 x_{2}+1 \\
x_{1}=x_{2}+\frac{x_{2}+1}{2}, \frac{x_{2}+1}{2}=y, x_{2}=2 y-1 \\
x_{1}=2 y-1+y=3 y-1 \\
x=6 y-2+1=6 y-1
\end{gathered}
$$

Next, we need to solve the following indeterminate equation:

$$
\begin{gathered}
6 y-1=4 x_{3}+3 \\
4 x_{3}=6 y-4 \\
x_{3}=y-1+\frac{y}{2}, \frac{y}{2}=y^{\prime}, y=2 y^{\prime} \\
x_{3}=2 y^{\prime}-1+y^{\prime}=3 y^{\prime}-1 \\
x=12 y^{\prime}-4+3=12 y^{\prime}-1
\end{gathered}
$$

The next indeterminate equation to solve is:

$$
\begin{gathered}
12 y^{\prime}-1+3=5 x_{4}+4 \\
5 x_{4}=12 y^{\prime}-5 \\
x_{4}=2 y^{\prime}-1+\frac{2 y^{\prime}}{5}, \frac{2 y^{\prime}}{5}=2 y^{\prime \prime}, y^{\prime}=5 y^{\prime \prime} \\
x_{4}=10 y^{\prime \prime}-1+2 y^{\prime \prime}=12 y^{\prime \prime}-1 \\
x=60 y^{\prime \prime}-5+4=60 y^{\prime \prime}-1
\end{gathered}
$$

The fourth indeterminate equation is:

$$
\begin{gathered}
60 y^{\prime \prime}-1=6 x_{5}+5 \\
6 x_{5}=60 y^{\prime \prime}-6
\end{gathered}
$$

Since this equation is divisible by 6 without a remainder, it does not introduce a new unknown for determining $x$. Therefore, we can write the fifth equation:

$$
\begin{gathered}
7 x_{6}=60 y^{\prime \prime}-1 \\
x_{6}=8 y^{\prime \prime}+\frac{4 y^{\prime \prime}-1}{7}, \frac{4 y^{\prime \prime}-1}{7}=y^{\prime \prime \prime} \\
4 y^{\prime \prime}=7 y^{\prime \prime \prime}+1, y^{\prime \prime}=y^{\prime \prime \prime}+\frac{3 y^{\prime \prime \prime}+1}{4} \\
\frac{3 y^{\prime \prime \prime}+1}{4}=y^{I V}, \quad y^{\prime \prime \prime}=\frac{4 y^{I V}-1}{3}=y^{I V}+\frac{y^{I V}-1}{3} \\
\frac{y^{I V}-1}{3}=y^{V} \cdot \quad y^{I V}=3 y^{V}+1 \\
y^{\prime \prime \prime}=3 y^{V}+1+y^{V}=4 y^{V}+1 \\
y^{\prime \prime}=4 y^{V}+1+3 y^{V}+1=7 y^{V}+2 \\
x_{6}=56 y^{V}+16+4 y^{V}+1=60 y^{V}+17
\end{gathered}
$$

And finally,

$$
x=420 y^{V}+119
$$

The number of steps can therefore be: $119, 539, 959$, and so on.

(István Sztrapkovits, 8th grade student, S.-A.-Ujhely.)

The problem was also solved by: Mihály Seidner, Losoncz; K. Koncz, Kisujszállás; Miklós Malesevits, high school teacher, Zombor.