It is evident that if a number appears for the second time in the sequence, the segment bounded by these two numbers will repeat indefinitely. Therefore, we will list the numbers in the sequence until they start repeating.

- 1st number: 52,
- 2nd number: 14,
- 3rd number: 18,
- 4th number: $8^{2}+2 \cdot 1=66$,
- 5th number: $6^{2}+2 \cdot 6=48$,

- 6th number: $8^{2}+2 \cdot 4=72$,

- 7th number: $2^{2}+2 \cdot 7=18$,
- 8th number: $8^{2}+2 \cdot 1=66$,
- etc.

From the third number, the sequence regularly repeats the numbers 18, 66, 48, and 72. The number 18 is therefore on the 3rd, 7th, 11th, 15th, 19th, 23rd positions, etc. This number will also appear on every position that differs from any of the already mentioned positions by a multiple of 4. Since $2011=11+500 \cdot 4$, the number 18 will also be on the 2011th position.

Evaluation. 3 points for finding the numbers that repeat; 1 point for determining that the 2011th number is 18; 2 points for explaining why this number is on the 2011th position.
Martina came up with a method for generating a numerical sequence. She started with the number 52. From it, she derived the next term in the sequence as follows: $2^{2}+2 \cdot 5=4+10=14$. Then she continued in the same way and from the number 14, she got $4^{2}+2 \cdot 1=16+2=18$. Each time, she would take a number, detach the digit in the units place, square this detached digit, and add twice the number that remains from the original number after detaching the last digit. What is the 2011th number in this sequence?

(M. Dillingerová)