11.2. What is the minimum number of factors that need to be crossed out from the number $99!=1 \cdot 2 \cdot \ldots \cdot 99$ so that the product of the remaining factors ends in $2?$
Answer: 20 factors.

Solution. From the number 99! it is necessary to remove all factors that are multiples of 5, otherwise the product will end in 0. There are a total of 19 such factors (ending in 0 or 5).

The product of the remaining factors ends in 6. Indeed, since the product $1 \times 2 \cdot 3 \cdot 4 \cdot 6 \cdot 7 \cdot 8 \cdot 9$ ends in 6, then similar products in each subsequent decade also end in 6. Therefore, removing 19 factors is insufficient. But 20 is sufficient. If you remove another number ending in 3 or 8, the product of the numbers in the corresponding decade will end in 2. The product of this number and a number ending in 6 also ends in 2.

## Grading Criteria

+ correct answer and a fully justified solution

$\pm$ correct answer and a generally correct solution, but there are minor gaps in the justification

干 correct answer and a correct example, but it is not explained why the specified number of removed factors is the smallest

$\mp$ correct answer, but some of the numbers to be removed are incorrectly identified in the reasoning

干 correct approach to the solution, but an error was made that affected the answer

干 an estimate is provided without an example

- only the answer is provided
- the problem is not solved or is solved incorrectly