smc

Answer: $(D)\frac{34}{67}$ First of all, you have to realize that if $\left[\frac{n}{k}\right]+\left[\frac{100-n}{k}\right]=\left[\frac{100}{k}\right]$ then $\left[\frac{n-k}{k}\right]+\left[\frac{100-(n-k)}{k}\right]=\left[\frac{100}{k}\right]$ So, we can consider what happen in $1\le n\le k$ and it will repeat. Also since range of $n$ is $1$ to $99!$, it is always a multiple of $k$. So we can just consider $P(k)$ for $1\le n\le k$. Let $\text{fpart}(x)$ be the fractional part function This is an AMC exam, so use the given choices wisely. With the given choices, and the previous explanation, we only need to consider $k=99$, $87$, $67$, $13$. $1\le n\le k$ For $k>\frac{200}{3}$, $\left[\frac{100}{k}\right]=1$. 3 of the $k$ that we should consider land in here. For $n<\frac{k}{2}$, $\left[\frac{n}{k}\right]=0$, then we need $\left[\frac{100-n}{k}\right]=1$ else for $\frac{k}{2}<n<k$, $\left[\frac{n}{k}\right]=1$, then we need $\left[\frac{100-n}{k}\right]=0$ For $n<\frac{k}{2}$, $\left[\frac{100-n}{k}\right]=\left[\frac{100}{k}-\frac{n}{k}\right]=1$ So, for the condition to be true, $100-n>\frac{k}{2}$ . ( $k>\frac{200}{3}$, no worry for the rounding to be $>1$) $100>k>\frac{k}{2}+n$, so this is always true. For $\frac{k}{2}<n<k$, $\left[\frac{100-n}{k}\right]=0$, so we want $100-n<\frac{k}{2}$, or $100<\frac{k}{2}+n$ $100<\frac{k}{2}+n<\frac{3k}{2}$ For k = 67, $67>n>100-\frac{67}{2}=66.5$ For k = 69, $69>n>100-\frac{69}{2}=67.5$ etc. We can clearly see that for this case, $k=67$ has the minimum $P(k)$, which is $\frac{34}{67}$. Also, $\frac{7}{13}>\frac{34}{67}$ . So for AMC purpose, answer is $\boxed{\frac{34}{67}}$.