A box of chocolates in the shape of a cuboid was full of chocolates arranged in rows and columns. Misha ate some of them and rearranged the remaining ones so that they filled (without gaps) up to one place three whole rows. The remaining chocolates from the next incomplete row were eaten. Then the remaining chocolates were rearranged and filled five columns without one place. The chocolates in the incomplete column were eaten again. Thus, only a third of the original number of chocolates remained in the box. Find:

a) how many chocolates the entire box had,

b) how many chocolates Misha ate before he did the first rearrangement.

(Bednářrová)
Given that in the end only four columns were filled, which was only a third of the original number of candies, the box must contain 12 columns. (1 point)

Now we know that each row contains 12 candies. It is therefore clear that before the second rearrangement, there were 24 candies in the box. (1 point)

Since one was missing to complete five full columns, it is clear that in five columns there are $24+1=$ $=25$ candies, so one column has 5 candies. A full box therefore contains $12 \cdot 5=60$ candies. (2 points)

Since before the first rearrangement, one candy was missing to complete three full rows, at that time there were 35 candies in the box, or Misha ate $60-35=25$.

The entire box contains 60 candies, and before the first rearrangement, Misha ate 25.

(Complete solution 6 points)