B1. The product of two consecutive natural numbers after the number $n$ is 600 greater than the product of two consecutive natural numbers before the number $n$. Determine the number $n$. Write down the solution.
B1. Let the natural number $n$ be sought. Then $(n-2) \cdot(n-1)+600=(n+1) \cdot(n+2)$, from which we get $n^{2}-3 n+2+600=n^{2}+3 n+2$ and express $n=100$.

The written sequence of natural numbers: $n-2, n-1, n, n+1, n+2 \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots$

![](https://cdn.mathpix.com/cropped/2024_06_07_ce887fced8b2a2bf0894g-10.jpg?height=52&width=1641&top_left_y=1913&top_left_x=276)

Removing parentheses.........................................................................................................................................

![](https://cdn.mathpix.com/cropped/2024_06_07_ce887fced8b2a2bf0894g-10.jpg?height=52&width=1639&top_left_y=2007&top_left_x=274)

Answer . .........................................................................................................