1. In an urn, there are only white and black balls, the number of which, rounded to the nearest hundred, is 1000. The probability of drawing two black balls is $\frac{17}{43}$ greater than the probability of drawing two white balls. How many white and how many black balls are in the urn? (The probability of drawing any ball is the same.)
SOLUTION. Let there be $n$ balls in the urn, of which $b$ are white (and $n-b$ are black). Then the probability of drawing two white balls is equal to the ratio

$$
\frac{\left(\begin{array}{l}
b \\
2
\end{array}\right)}{\left(\begin{array}{l}
n \\
2
\end{array}\right)}=\frac{b(b-1)}{n(n-1)}
$$

while the probability of drawing two black balls is the ratio

$$
\frac{\left(\begin{array}{c}
n-b \\
2
\end{array}\right)}{\left(\begin{array}{l}
n \\
2
\end{array}\right)}=\frac{(n-b)(n-b-1)}{n(n-1)} .
$$

According to the problem, the equation

$$
\frac{(n-b)(n-b-1)}{n(n-1)}=\frac{b(b-1)}{n(n-1)}+\frac{17}{43}
$$

holds, which can be simplified algebraically to the form $43 b=13 n$ (for $n \notin\{0,1\}$ these are equivalent equations). From this, given the coprimality of the numbers 13 and 43, it follows that the natural numbers $n$ and $b$ are of the form $n=43 k$ and $b=13 k$, where $k$ is a suitable natural number. According to the problem, the number $n$ satisfies the estimates $950 \leqq n<1050$, from which we find that $k \in\{23,24\}$. For $k=23$, we get $n=989$ and $b=299$ (then $n-b=690$), while the value $k=24$ corresponds to $n=1032$ and $b=312$ (then $n-b=720$).

Answer: The problem has two solutions - there are either 299 white and 690 black, or 312 white and 720 black balls in the urn.