B4. The product of two consecutive integers is 128 more than twice their sum. What are these numbers? Write down all possible options.

10th Mathematics Knowledge Competition for Students of Secondary Technical and Vocational Schools, Regional Competition, March 31, 2010

# PROBLEMS FOR THE SECOND YEAR

Time for solving: 90 minutes. In section A, a correct answer will be worth two points, while a wrong answer will result in a deduction of half a point. Write your answers for section A in the left table.
![](https://cdn.mathpix.com/cropped/2024_06_07_ec35f4a613e63419eaedg-04.jpg?height=162&width=1192&top_left_y=632&top_left_x=448)
B4. Let $n$ and $n+1$ be consecutive integers. Considering the problem statement, we write the equation $n(n+1)=2(n+n+1)+128$. Removing the parentheses and simplifying the equation, we get $n^{2}-3 n-130=0$. We solve the equation using Vieta's rule $(n-13)(n+10)$, and read the solutions $n_{1}=-10$ and $n_{2}=13$. The sought numbers are 13 and 14, and -10 and -9.

![](https://cdn.mathpix.com/cropped/2024_06_07_ec35f4a613e63419eaedg-11.jpg?height=57&width=1687&top_left_y=2644&top_left_x=230)

![](https://cdn.mathpix.com/cropped/2024_06_07_ec35f4a613e63419eaedg-11.jpg?height=51&width=1688&top_left_y=2693&top_left_x=228)
Factoring the left side of the equation $(n-13)(n+10)$ ..... 1 point
Solutions to the equation $n_{1}=-10$ and $n_{2}=13$ ..... 1 point
Writing the consecutive numbers $n_{1}+1=-9$ and $n_{2}+1=14$ ..... 1 point
Writing the answer ..... 1 point
NOTE: Consecutive numbers can also be written as $n-1$ and $n$.
In the case of writing only one pair of solutions, the contestant loses 1 point.

## Second Year

## Set A

| $\mathrm{A} 1$ | $\mathrm{~A} 2$ | $\mathrm{~A} 3$ | $\mathrm{~A} 4$ | $\mathrm{~A}$ | $\mathrm{~A} 6$ |
| :---: | :---: | :---: | :---: | :---: | :---: |
| $\mathrm{C}$ | $\mathrm{C}$ | $\mathrm{D}$ | $\mathrm{C}$ | $\mathrm{A}$ | $\mathrm{D}$ |