6.1. On September 2nd, Robin Bobin ate 12 chickens, and starting from September 3rd, he ate every day as many as he had already eaten on average in September. On September 15th, Robin Bobin ate 32 chickens. How many chickens did he eat on September 1st?
Answer: 52

Solution. Let at the beginning of the $k$-th day he on average ate $x$ chickens. This means that by this point, he had eaten $(k-1) x$ chickens. Then on the $k$-th day, he also ate $x$ chickens, which means that over $k$ days, he ate $k x$ chickens. This means that after the $k$-th day, the average number of chickens eaten did not change, and thus, on the $k+1$-th day, the same number will be eaten. This means that this average number is 32 chickens. Therefore, in the first two days, Robin Bobin ate 64 chickens, i.e., on the first day -52 chickens.