Problem 3.7. Katya passes the time while her parents are at work. On a piece of paper, she absentmindedly drew Cheburashkas in two rows (at least one Cheburashka was drawn in each row).

Then, after some thought, between every two adjacent Cheburashkas in a row, she drew a Gena the Crocodile. And then to the left of each Cheburashka, she drew a Old Lady Shapoklyak. And finally, between every two characters in a row, she drew a Krakozyabra.

Looking carefully at the drawing, she realized that only the Krakozyabras looked good, and she angrily erased all the others. In the end, her parents saw two rows of Krakozyabras: a total of 29. How many Cheburashkas were erased?
Answer: 11.

Solution. The Krokodilofes are drawn exactly in the gaps between the other characters. In each of the two rows of gaps, there is 1 less gap than there are characters, so

There are 2 fewer Krokodilofes than all the other characters. Therefore, Cheburashkas, Crocodile Gens, and Old Ladies Shapoklyak total $29+2=31$. We will now consider only them, forgetting about the Krokodilofes.

Each Cheburashka has one neighboring Old Lady Shapoklyak to the left, so there are an equal number of them. Crocodile Gens are drawn exactly in the gaps between Cheburashkas, so by the previously described principle, there are 2 fewer of them than Cheburashkas.

Let's mentally add 2 Crocodile Gens. Then the total number of characters will be $31+2=33$. In this case, all three characters are now equal, so there were 11 Cheburashkas.