2011 AMC 12B Problems/Problem 25

Problem

For every $m$ and $k$ integers with $k$ odd, denote by $\left[\frac{m}{k}\right]$ the integer closest to $\frac{m}{k}$ . For every odd integer $k$ , let $P(k)$ be the probability that

$\left[\frac{n}{k}\right] + \left[\frac{100 - n}{k}\right] = \left[\frac{100}{k}\right]$

for an integer $n$ randomly chosen from the interval $1 \leq n \leq 99!$ . What is the minimum possible value of $P(k)$ over the odd integers $k$ in the interval $1 \leq k \leq 99$ ?

$\textbf{(A)}\ \frac{1}{2} \qquad \textbf{(B)}\ \frac{50}{99} \qquad \textbf{(C)}\ \frac{44}{87} \qquad \textbf{(D)}\ \frac{34}{67} \qquad \textbf{(E)}\ \frac{7}{13}$

Solution

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 should consider lands 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 (D).

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 $n = k$ , 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 $n = 1$ , which makes

$P(k) \ge \frac{1}{2} + \frac{1}{2k}$ .

2011 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 24	Followed by Last Problem
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

Page

Toolbox

Search

2011 AMC 12B Problems/Problem 25

Problem

Solution

See also