Problem 1930

For each odd integer $n$ from $1$ to $49$, define the group of $n$ as the set of integers from $1$ to $50$ which can be expressed as $n \cdot 2^k$ for a non-negative integer $k$. Then each integer from $1$ to $50$ belongs to only one group.
For any two integers in the same group, one is a divisor of the other. Therefore we can choose at most one integer from the same group when we choose $25$ integers satisfying the condition. The number of groups is $25$, hence we need to choose one integer from each group.
Firstly, there is only one way to choose from each group of $27$, $29$, $31$, $33$, $35$, $37$, $39$, $41$, $43$, $45$, $47$, $49$. Then $27$ is chosen, hence the integer chosen from the group of $9$ is a multiple of $18$. Since $18$ is a multiple of $3$ and $6$, the integer chosen from the group of $3$ is a multiple of $12$. Since $12$ is a multiple of $1$, $2$ and $4$, the integer chosen from the group of $1$ is a multiple of $8$. Integers $39$ and $45$ are also chosen, thus we need to choose $26$ and $30$ from each group of $13$ and $15$. In the following, we consider how to choose integers from other groups.

Since $33$ is chosen, $11$ is not chosen. Integers $22$ and $44$ can not be a divisor of integers in other groups and they can not be a multiple of integers in other groups since $1$, $2$, and $4$ are not chosen. Therefore the number of ways to choose one integer from the group of $11$ is $2$, which is independent of the choice from other groups.
Integers $25$ and $50$ can not be a divisor of integers in other groups. Since $30$ is chosen, $1$, $2$, $5$, and $10$ are not chosen, thus $25$ and $50$ can not be a multiple of integers in other groups. Therefore the number of ways to choose one integer from the group of $25$ is $2$, which is independent of the choice from other groups.
Integers $17$ and $34$ can not be a divisor of integers in other groups and they can not be a multiple of integers in other groups since $1$ and $2$ are not chosen. Therefore the number of ways to choose one integer from the group of $17$ is $2$, which is independent of the choice from other groups. This is the same for the group of $19$ and $23$.
The choice of one integer each from the group of $7$ and $21$ must be $(14, 21)$, $(28, 21)$, or $(28, 42)$. Integers $14$, $28$, $21$, and $42$ can not be a divisor of integers in other groups and they can not be a multiple of integers in other groups since $1$, $2$, $3$, $4$, and $6$ are not chosen. Therefore the number of ways to choose one integer each from the group of $7$ and $21$ is $3$, which is independent of the choice from other groups. We need to determine the way to choose from the group of $1$, $3$, $5$, and $9$.

From the group of $1$ we need to choose a multiple of $8$.
* If $8$ is chosen, the choice from each group of $3$, $5$, and $9$ must be $12$, $20$, and $18$, thus we have only one way.
* If $16$ is chosen, the choice from the group of $5$ must be $20$ or $40$, and the choice from the group of $3$ and $9$ must be $(12, 18)$, $(24, 18)$, or $(24, 36)$.
* If $32$ is chosen, the choice from the group of $5$ must be $20$ or $40$, and the choice from the group of $3$ and $9$ must be $(12, 18)$, $(24, 18)$, $(24, 36)$, $(48, 18)$, or $(48, 36)$.
Hence the number of ways to choose from the group of $1$, $3$, $5$, and $9$ is $1 + 2 \cdot 3 + 2 \cdot 5 = 17$.

Therefore the answer is $2^5 \cdot 3 \cdot 17 = 1632$.