3. Vladik thought of a natural number $N$. He divided it by one of its prime divisors and wrote the result on the board. Then he divided this result by one of its prime divisors and wrote the new result on the board. He continued this process until he wrote 1 on the board. This 1 turned out to be the 22nd number written on the board. It turned out that the sum of all the numbers written on the board was $N / 2$. What could $N$ be?
Answer: $2 \cdot 3^{21}$. Let's write the sum starting from the beginning: $1+p_{1}+p_{1} p_{2}+p_{1} p_{2} p_{3}+\cdots=N / 2$. If $p_{1}$ were an odd divisor of $N$, the right side would be divisible by $p_{1}$, which cannot be the case since the left side is not divisible by $p_{1}$ due to the first term. Therefore, $p_{1}=2$. Substituting: $3+2 p_{2}+2 p_{2} p_{3}+\cdots=N / 2$. The left side is odd, so $N$ must have contained the single factor $p_{1}=2$, and the remaining $p_{i}$ are odd. Then the right side is divisible by $p_{2}$, so $p_{2}=3$. Substituting: $9+6 p_{3}+\cdots=N / 2$. Similarly, we sequentially find that all the other $p_{i}$ are equal to 3.