Problem 4. Consider the set $A=\left\{x \in \mathbb{N} \mid x=n^{2}+5, n \in \mathbb{N}, n \leq 2014\right\}$.

a) Find the four smallest elements of the set $A$.

b) Show that $261 \in A$ and that $2014 \notin A$.

c) Let the set $B=\{y \in \mathbb{N} \mid y=n(n+1)+5, n \in \mathbb{N}, n \leq 2014\}$.

Calculate the difference $S(B)-S(A)$, where $S(A)$ is the sum of the elements of set $A$, and $S(B)$ is the sum of the elements of set $B$.

Florin Năsui, Slatina

NOTE. All subjects are mandatory. Time for work: 2 hours and 30 minutes.

## MATHEMATICS OLYMPIAD

## LOCAL STAGE - FEBRUARY 16, 2014  5th Grade

Solutions and grading criteria
Problem 4. The smallest four elements are $5,6,9,14$ (2p)
a) $n^{2}+5=261 \Rightarrow n^{2}=256 \Rightarrow n=16$, so $261 \in A$ (1p)

$n^{2}+5=2014 \Rightarrow n^{2}=2009$

Since $2009=7^{2} \cdot 41 \Rightarrow 2009$ is not a perfect square, so 2014 is not a term of the sequence $(2 p)$

c)S(B)-S(A)=0+1+2+3+...+2014=2029105 (2p)