1. An army of mice attacked a grain warehouse. To fight the mice, one cat was released into the warehouse on the first day, a second cat on the next day, and each subsequent day, one more cat was released. A mouse eats 100 grams of grain per day, and a cat eats 10 mice per day. After several days, there were no mice left. After calculating the losses, it was found that the mice had eaten 55 kg of grain over these few days. How many mice were there? (A mouse manages to eat 100 grams of grain even on the day it is caught by a cat.)
Solution. Let the entire process last $n$ days, then by its end, there were $n$ cats on the warehouse, with the first cat eating $10 n$ mice, the second cat eating 10( $n-1)$ mice, $\ldots, n$-th cat - 10 mice, meaning there were a total of $10(1+2+\ldots+n)$ mice. Let's calculate how much grain the mice ate.

Ten mice caught by the cat on the first day ate $10 \cdot 0.1=1$ kg of grain. Twenty mice caught on the second day (by two cats) ate $20 \cdot 2 \cdot 0.1=4$ kg of grain over 2 days.

Thirty mice caught on the third day ate $30 \cdot 3 \cdot 0.1=9$ kg of grain over 3 days.

Forty mice caught on the fourth day ate $40 \cdot 4 \cdot 0.1=16$ kg of grain over 4 days.

Fifty mice caught on the fifth day ate $50 \cdot 5 \cdot 0.1=25$ kg of grain over 5 days.

In total, over five days, $1+4+9+16+25=55$ kg of grain was eaten.

Therefore, the process lasted 5 days, and there were 150 mice.

Evaluation criteria. A positive score is allowed if the solution path is correct but not completed to the answer due to arithmetic errors, but no more than 3 points. In particular, if it is correctly established that the process lasted 5 days but the number of mice is calculated incorrectly, -3 points.