B3. When the third-year student Benjamin calculated the sum $1+2+3+\ldots+2012$, he omitted some terms and got an incorrect sum that was divisible by 2011. Anika, when calculating the sum $A=1+2+3+\ldots+2013$, omitted exactly the same terms as Benjamin and got an incorrect sum $N$, which was divisible by 2014. What is the ratio of the sums $\frac{N}{A}$?

## Problems for 2nd Year

Time for solving: 120 minutes. In section A, a correct answer will be awarded two points, while a wrong answer will result in the deduction of one point. Enter your answers for section A in the left table.
![](https://cdn.mathpix.com/cropped/2024_06_07_6340f9febbabfa1859cdg-06.jpg?height=164&width=802&top_left_y=547&top_left_x=640)
B3. Let the sum of the terms that Benjamin omitted be denoted by $x$. Since $1+2+3+\ldots+2012 = \frac{2012 \cdot 2013}{2} = 1006 \cdot 2013$, Benjamin's result is $1006 \cdot 2013 - x$. Therefore, there exists a non-negative integer $m$ such that $1006 \cdot 2013 - x = 2011 m$. Since $A = 1 + 2 + 3 + \ldots + 2013 = \frac{2013 \cdot 2014}{2} = 2013 \cdot 1007$, Anika's result is $N = 2013 \cdot 1007 - x$. Therefore, there exists a non-negative integer $n$ such that $2013 \cdot 1007 - x = 2014 n$. If we express $x$ from both equations and equate the results, we get $1006 \cdot 2013 - 2011 m = 2013 \cdot 1007 - 2014 n$ or $2014 n - 2011 m - 2013 = 0$. We can rearrange the last equation to $2011(n - m) = 2013 - 3n$. Since $2014 n = 2013 \cdot 1007 - x \leq 2013 \cdot 1007$, we have $n \leq \frac{2013 \cdot 1007}{2014} < 1007$. Therefore, $-1008 < 2013 - 3n \leq 2013$. At the same time, $2013 - 3n$ is divisible by 3, and from the equation, it is also divisible by 2011. The only possibility is $2013 - 3n = 0$ or $n = 671$. From this, we calculate $\frac{N}{A} = \frac{2014 n}{2013 \cdot 1007} = \frac{2014 \cdot 671}{2013 \cdot 1007} = \frac{2}{3}$.

Writing Benjamin's result as a multiple of 2011 2 points

Writing Anika's result as a multiple of 2014 1 point

Writing the equation $2014 n - 2011 m - 2013 = 0$ 1 point

Determining that $n = 671$ 1 point

![](https://cdn.mathpix.com/cropped/2024_06_07_6340f9febbabfa1859cdg-19.jpg?height=59&width=1576&top_left_y=1047&top_left_x=277)
(If the contestant only writes the ratio, 2 points are awarded.)

## Solutions for 2nd Year

Within section A, a correct answer will be awarded 2 points, while an incorrect answer will result in the deduction of 1 point. To avoid a potential negative final score, each contestant is given 3 initial points.

| 1 | 2 | 3 |
| :--- | :--- | :--- |
| C | B | E |

Justifications: