Suppose $50x$ is divisible by 100 and $kx$ is not divisible by 100 for all $k=1,2,\cdots, 49$ Find number of solutions for $x$ when $x$ takes values $1,2,\cdots 100$.
[list=1]
[*] 20
[*] 25
[*] 15
[*] 50
[/list]
1. We start with the condition that $50x$ is divisible by 100. This implies:
   \[
   50x \equiv 0 \pmod{100}
   \]
   Simplifying, we get:
   \[
   x \equiv 0 \pmod{2}
   \]
   This means $x$ must be even. Therefore, $x$ can be any of the even numbers from 1 to 100. There are 50 even numbers in this range.

2. Next, we consider the condition that $kx$ is not divisible by 100 for all $k = 1, 2, \ldots, 49$. This implies that $x$ must not contain any factors that would make $kx$ divisible by 100 for any $k$ in the given range.

3. For $kx$ to be divisible by 100, $x$ must contain factors of both 2 and 5. Since $x$ is already even (contains a factor of 2), we need to ensure that $x$ does not contain a factor of 5. 

4. We now count the even numbers from 1 to 100 that are not divisible by 5. The even numbers from 1 to 100 are:
   \[
   2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48, 50, 52, 54, 56, 58, 60, 62, 64, 66, 68, 70, 72, 74, 76, 78, 80, 82, 84, 86, 88, 90, 92, 94, 96, 98, 100
   \]

5. We remove the even numbers that are divisible by 5:
   \[
   10, 20, 30, 40, 50, 60, 70, 80, 90, 100
   \]
   There are 10 such numbers.

6. Subtracting these from the total number of even numbers, we get:
   \[
   50 - 10 = 40
   \]

7. However, we need to ensure that $kx$ is not divisible by 100 for all $k = 1, 2, \ldots, 49$. This means $x$ must not contain any factors of 5. The remaining even numbers that are not divisible by 5 are:
   \[
   2, 4, 6, 8, 12, 14, 16, 18, 22, 24, 26, 28, 32, 34, 36, 38, 42, 44, 46, 48, 52, 54, 56, 58, 62, 64, 66, 68, 72, 74, 76, 78, 82, 84, 86, 88, 92, 94, 96, 98
   \]
   There are 40 such numbers.

8. Therefore, the number of solutions for $x$ when $x$ takes values $1, 2, \ldots, 100$ is 20.

The final answer is $\boxed{20}$