## Problem T-1.

Three strictly increasing sequences

$$
a_{1}, a_{2}, a_{3}, \ldots, \quad b_{1}, b_{2}, b_{3}, \ldots, \quad c_{1}, c_{2}, c_{3}, \ldots
$$

of positive integers are given. Every positive integer belongs to exactly one of the three sequences. For every positive integer $n$, the following conditions hold:

(i) $c_{a_{n}}=b_{n}+1$;

(ii) $a_{n+1}>b_{n}$;

(iii) the number $c_{n+1} c_{n}-(n+1) c_{n+1}-n c_{n}$ is even.

Find $a_{2010}, b_{2010}$, and $c_{2010}$.


Solution. Since $\left\{c_{n}\right\}$ is a strictly increasing sequence of positive integers, it is clear that $c_{n} \geq n, n \in \mathbb{N}$. Hence, $c_{a_{n}} \geq a_{n}, n \in \mathbb{N}$. However, the given sequences do not contain equal terms, so $c_{a_{n}}>a_{n}$ and $b_{n}=c_{a_{n}}-1>a_{n}, n \in \mathbb{N}$. Similarly, from (ii) and (iii), $a_{n+1}>b_{n}+1=c_{a_{n}}, n \in \mathbb{N}$. It is also easy to see that $b_{n}1$ then $c_{1}=1$ or $b_{1}=1$. The latter case is impossible because $b_{1}>a_{1}$. Then we must have $c_{1}=1$, and either $c_{2}=2$, or $a_{1}=2$ and $c_{a_{1}}=c_{2}=a_{1}+2=4$. In both cases we obtain a contradiction by setting $n=1$ in (iv). This proves that $a_{1}=1$, and, together with (2), defines a unique sequence $\left\{a_{n}\right\}$ :

$a_{n}=a_{n-1}+(2 n-1)=a_{n-2}+(2 n-3)+(2 n-1)=\cdots=a_{1}+3+5+\ldots+(2 n-1)=n^{2}, \quad n \in \mathbb{N}$.

Hence,

$$
\begin{aligned}
a_{2010} & =2010^{2} \\
b_{2010} & =c_{a_{2010}}-1=a_{2010}+2 \cdot 2010-1=2011^{2}-2 \\
c_{1936} & =c_{44^{2}}=c_{a_{44}}=a_{44}+2 \cdot 44=44^{2}+88=2024 \\
a_{45} & =45^{2}=2025
\end{aligned}
$$

and all the integers between $a_{45}$ and $b_{45}=c_{a_{45}}-1=a_{45}+2 \cdot 45-1=a_{45}+89$ belong to the sequence $\left\{c_{n}\right\}$. Hence, these integers have the form

$$
c_{1936+k}=a_{45}+k, \quad k=1,2, \ldots, 88
$$

and $c_{2010}=c_{1936+74}=a_{45}+74=2099$.

Answer. $a_{2010}=2010^{2}, b_{2010}=2011^{2}-2, c_{2010}=2099$.
