AMC12 2011 B

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 $\boxed{\textbf{(D) }\frac{34}{67}}$. Notice that for these integers $99,87,67$: $0\rightarrow 49,50,51\rightarrow 98$ $100\rightarrow 51,50,49\rightarrow 2$ $P=\frac{98}{99}$ $0\rightarrow 43,44\rightarrow 56,57\rightarrow 86$ $87\rightarrow 57,56\rightarrow 44,43\rightarrow 14$ $P=\frac{74}{87}$ $0\rightarrow 33,34\rightarrow 66$ $100\rightarrow 67,66\rightarrow 34$ $P=\frac{34}{67}$ That the probability is $\frac{2k-100}{k}=2-\frac{100}{k}$. Even for $k=13$, $P(13)=\frac{9}{13}=\frac{100}{13}-7$. And $P(11)=\frac{10}{11}=10-\frac{100}{11}$. Perhaps the probability for a given $k$ is $\left\lceil{\frac{100}{k}}\right\rceil-\frac{100}{k}$ if $\left[\frac{100}{k}\right]=\left\lfloor{\frac{100}{k}}\right\rfloor$ and $\frac{100}{k}-\left\lfloor{\frac{100}{k}}\right\rfloor$ if $\left[\frac{100}{k}\right]=\left\lceil{\frac{100}{k}}\right\rceil$. So $P>\frac{1}{2}$ and $P_\text{min}=\frac{k_\text{min}+1}{2k_\text{min}}=\frac{101}{201}$. Because $201=3\cdot 67\mid 99!$ ! Now, let's say we are not given any answer, we need to consider $k < \frac{200}{3}$. I claim that $P(k) \ge \frac{1}{2} + \frac{1}{2k}$ If $\left[\frac{100}{k}\right]$ got round down, then $1 \le n \le \frac{k}{2}$ all satisfy the condition along with $n = k$ because if $\text{fpart}\left(\frac{100}{k}\right) < \frac{1}{2}$ and $\text{fpart} \left(\frac{n}{k}\right) < \frac{1}{2}$, so must $\text{fpart} \left(\frac{100 - n}{k}\right) < \frac{1}{2}$ and for $n = k$, it is the same as $n = 0$. , which makes $P(k) \ge \frac{1}{2} + \frac{1}{2k}$. If $\left[\frac{100}{k}\right]$ got round up, then $\frac{k}{2} \le n \le k$ all satisfy the condition along with $n = 1$ because if $\text{fpart}\left(\frac{100}{k}\right) > \frac{1}{2}$ and $\text{fpart} \left(\frac{n}{k}\right) > \frac{1}{2}$ Case 1) $\text{fpart} \left(\frac{100 - n}{k}\right) < \frac{1}{2}$ Case 2)