An ACT math problem:
The only factor that is common to 200 and an integer $n$ is 1. When $\frac{n}{200}$ is written as a decimal number, what is the minimum number of digits to the right of the decimal point?
What are some quicker methods to get around this as opposed to trying out a bunch of assumed $n$?
Since $\dfrac 1{200} = 0.005$, we can say that $3$ decimal places are possible at maximum. If the decimal places to the right have to reduce, the ending digits must be zero and it achieved by multiplying $0.005$ with a multiple of $2$, but that violates the condition that $gcd(200,n) = 1$ as $1$ is the only common divisor for $200$ and $n$.
Hence, the minimum number of decimal places is also $3$.