Answer: 64 and 96.

## Solution:

In the next step, after erasing the ones, all prime numbers will be erased (they have no divisors other than 1 and themselves). Next, the numbers that have exactly two prime factors (not necessarily distinct) in their factorization will be erased, followed by the numbers that have exactly three prime factors, and so on. Since $2^{7}>100$, there are no numbers among those written on the board that have more than six prime factors in their factorization. Therefore, on the last step, the numbers that have exactly six prime factors in their factorization will be erased. Since $3^{2} \cdot 2^{4}>100$, there are only two such numbers: $2^{6}=64$ and $3 \cdot 2^{5}=96$.

## Criteria:

For the presence in the solution of the idea that on the $k-th$ step, the numbers having $k$ prime factors in their factorization are erased, 2 points should already be given.
5. On the board are all natural numbers from 1 to 100: 1, 2, 3, 4, ..., 99, 100. Each student who enters the classroom circles the numbers that are not divisible by any of the other written numbers, and then erases all the circled numbers (obviously, the first student to enter will erase only the number 1). When Karl erased some numbers, no numbers were left on the board. Which numbers did he erase?