[Number Theory]
Problem: What is the smallest positive integer $x$ for which $\frac{1}{32}=\frac{x}{10^{y}}$ for some positive integer $y$ ?

Solution: Since $10^{y} \neq 0$, the equation $\frac{1}{32}=\frac{x}{10^{y}}$ is equivalent to $10^{y}=32 x$.

So the given question is equivalent to asking for the smallest positive integer $x$ for which $32 x$ equals a positive integer power of 10 .

Now $32=2^{5}$ and so $32 x=2^{5} x$.

For $32 x$ to equal a power of 10, each factor of 2 must be matched with a factor of 5 .

Therefore, $x$ must be divisible by $5^{5}$ (that is, $x$ must include at least 5 powers of 5 ), and so $x \geq 5^{5}=3125$.

But $32\left(5^{5}\right)=2^{5} 5^{5}=10^{5}$, and so if $x=5^{5}=3125$, then $32 x$ is indeed a power of 10 , namely $10^{5}$.

This tells us that the smallest positive integer $x$ for which $\frac{1}{32}=\frac{x}{10^{y}}$ for some positive integer $y$ is $x=5^{5}=3125$.

Final answer: 3125