Task B-3.5. A group of children were competing who would eat more strawberries. The winner ate $n$ strawberries, and each subsequent one ate two strawberries less, all the way to the last child in the $k$-th place, who ate $n+2-2k$ strawberries. The total number of strawberries eaten in the competition is 2014. How many strawberries did the winner eat at least?
## Solution.

The winner ate $n$ strawberries, the second place ate $n-2$ strawberries, the third place ate $n-4$ strawberries, ..., the $k$-th place ate $n+2-2k$ strawberries. The total sum is

$$
n+(n-2)+(n-4)+\ldots+(n-2(k-1))=2014
$$

It follows that

$$
\begin{aligned}
& n \cdot k-2(1+2+3+\ldots+k-1)=2014 \\
& n \cdot k-2 \cdot \frac{k(k-1)}{2}=2014 \\
& k(n-k+1)=2 \cdot 19 \cdot 53
\end{aligned}
$$

From this, we get $(k, n-k+1) \in\{(1,2014),(2,1007),(19,106),(38,53),(53,38),(106,19),(1007,2),(2014,1)\}$. It follows that

$$
\begin{array}{lll}
k=1, n-k+1=2014 & \Rightarrow & n=2014 \\
k=2, n-k+1=1007 & \Rightarrow & n=1008 \\
k=19, n-k+1=106 & \Rightarrow & n=124 \\
k=38, n-k+1=53 & \Rightarrow & n=90 \\
k=53, n-k+1=38 & \Rightarrow & n=90 \\
k=106, n-k+1=19 & \Rightarrow & n=124 \\
k=1007, n-k+1=2 & \Rightarrow & n=1008 \\
k=2014, n-k+1=1 & \Rightarrow & n=2014
\end{array}
$$

We conclude that the smallest value of the number $n$ is 90.

## NATIONAL MATHEMATICS COMPETITION

## 4th grade - high school - B variant

Šibenik, April 3, 2014.

IF A STUDENT HAS A DIFFERENT APPROACH TO SOLVING THE PROBLEM, THE COMMITTEE IS OBLIGED TO EVALUATE THAT APPROACH AND GRADE IT ACCORDINGLY.