# Problem 1. 

Consider the sequence defined as $a_{1}=3$, and $a_{n+1}=a_{n}+a_{n}^{2}$.

Determine the last two digits of $\mathrm{a}_{2} 000$.
Solution:

We have $a_{1}=3$ and $a_{n+1}=a_{n}+a_{n}^{2}=a_{n}(1+a_{n})$.

We write the first terms of the sequence:

$$
3,12,156,156.157=24492,24492.24493=\ldots . .56, \ldots \ldots
$$

Suppose that $a_{n}$ ends in 56. Then, $a_{n}=100a+56$, and we have

$$
a_{n+1}=(100a+56)(100a+57)=100b+56.57=100b+100c+92=100d+92
$$

that is, the last digits of $a_{n+1}$ are 92.

Similarly, if $a_{n}$ ends in 92, it is proven that $a_{n+1}$ ends in 56.

Since 2000 is even, then $a_{2000}$ ends in 92.