Problem 3. A sequence of numbers is formed in the following way: the first is the number 7, then each subsequent term is obtained from the sum of the digits of its square increased by 1. Thus, the second term is the number 14, because

$$
7^{2}=49,4+9+1=14
$$

the third term is the number 17, because

$$
14^{2}=196,1+9+6+1=17 \text { and so on. }
$$

Which number is at the 2002nd position?
Solution. The first few members of the sequence are: $7,14,17,20,5,8,11,5,8$, $11, \ldots$. From the first four members, the triplet 5, 8, 11 repeats periodically, so we have: $2002-4=1998=3 \cdot 666$, which means that the number at the 2002nd position is 11.