## Task 1 - 310831

A box $B$ is filled with blue balls, and another box $R$ with red balls. The number of red balls is $\frac{15}{17}$ of the number of blue balls.

From box $B$, $\frac{2}{5}$ of its contents can be removed; afterwards, it still contains more than 1000 balls. From box $R$, $\frac{3}{7}$ of its contents can be removed; afterwards, it contains fewer than 1000 balls.

Determine whether these conditions uniquely determine the original number of balls in the boxes! If so, state these two numbers!
For the quantities $b$ and $r$ of blue and red balls, respectively, it holds that $r=\frac{15 b}{17}$, so $15 b$ is divisible by 17. Since 15 is coprime to 17, it follows that $b$ is also divisible by 17; i.e., $b=17 n$ with a natural number $n$. This further implies

$$
r=\frac{15 \cdot 17 n}{17}=15 n
$$

Since $\frac{2 b}{5}$ is assumed to be a (removable from $B$) quantity, $2 b$ is divisible by 5; given $2 b=2 \cdot 17 n$ and the coprimality of $2 \cdot 17$ to 5, it follows that $n$ is divisible by 5; i.e., with a natural number $m$ we have $n=5 m$, thus

$$
b=17 \cdot 5 m \quad ; \quad r=15 \cdot 5 m
$$

Similarly, from the assumption that $\frac{3 r}{7}$ is a quantity and from $3 r=3 \cdot 15 \cdot 5 m$, it follows that $m$ is divisible by 7. With a natural number $k$ we thus have $m=7 k$,

$$
b=17 \cdot 5 \cdot 7 k \quad ; \quad r=15 \cdot 5 \cdot 7 k
$$

After removing $\frac{2}{5} b$ from $B$ and $\frac{3}{7} r$ from $R$, there remain $\frac{3}{5} b=3 \cdot 17 \cdot 7 k=357 k$ balls in $B$ and $\frac{4}{7} r=4 \cdot 15 \cdot 5 k=300 k$ balls in $R$. According to the assumption, for these quantities it holds that

$$
\begin{array}{lll}
357 k>1000 & , & 300 k<1000 \\
k>\frac{1000}{357}>2 & , & k<\frac{1000}{300}<4
\end{array}
$$

Thus, it is uniquely determined: $k=3, b=17 \cdot 5 \cdot 7 \cdot 3=1785, r=15 \cdot 5 \cdot 7 \cdot 3=1575$.