2. What is the minimum number of factors that need to be crossed out from the number 99! (99! is the product of all numbers from 1 to 99) so that the product of the remaining factors ends in 2?
Answer: 20.

Solution. From the number 99! all factors that are multiples of 5 must be removed, otherwise the product will end in 0. There are 19 such factors in total.

The product of the remaining factors ends in 6. Indeed, the product $1 \cdot 2 \cdot 3 \cdot 4 \cdot 6 \cdot 7 \cdot 8 \cdot 9$ ends in 6, and similar products in each subsequent decade also end in 6. Therefore, it is sufficient to remove one more factor, for example, 8. After this, the product of the remaining factors will end in 2.