A pigeon owner, when asked how many pigeons he had, replied: "The number of my pigeons is more than 300 but less than 900. If I release them in pairs, one remains; if in threes, two remain; if in fours, three remain; if in fives, four remain; if in sixes, five remain, but if I release them in sevens, none remain." How many pigeons did he have?
I. Solution: The number of pigeons is divisible by 7, so it is of the form $7 x$, and it is 1 less than a common multiple of $2,3,4,5,6$. The least common multiple of 2, 3, 4, 5, 6 is 60, so the sought number is also of the form $60 y-1$.

Thus,

$$
\begin{aligned}
& \quad 7 x=60 y-1 \\
& x=\frac{60 y-1}{7}=8 y+\frac{4 y-1}{7}=8 y+t \\
& y=\frac{7 t+1}{4}=\frac{8 t-(t-1)}{4}=2 t-\frac{t-1}{4}=2 t-u \\
& t=4 u+1
\end{aligned}
$$

Substituting back

$$
\begin{aligned}
y & =8 u+2-u=7 u+2 \\
x & =56 u+16+4 u+1=60 u+17 \\
7 x & =420 u+119
\end{aligned}
$$

According to the problem,

$$
300900$, so 539 is the only solution.

Sándor Réti (Esztergom, Rákóczi Military Secondary School II. grade)