3. We will build a sequence of numbers in the following way. We will place the number 7 in the first position, and after each number, we will place the sum of the digits of its square, increased by one. For example, the second position will hold the number 14, since $7^{2}=49$, and $4+9+1=14$. The third position will hold the number 17, since $14^{2}=196$, and $1+9+6+1=17$, and so on. What number stands in the 2016th position?
Solution Let's continue finding the first few terms of the sequence: $7 ; 14 ; 17 ; 20 ; 5 ; 8 ; 11 ; 5 ; \ldots$ - the number 5 is repeated. This means the sequence has a period of length 3: the numbers 5; 8; 11 will repeat subsequently. The number 8 is in the sixth position, so for any $\mathrm{k}>0$, the number 8 will also be in the $3 \mathrm{k}$-th position. Since $2016=3 \times 672$, the number 8 is in the 2016-th position.

Answer: 8.