Difference between revisions of "2011 AMC 12B Problems/Problem 25"

Revision as of 17:13, 15 February 2016

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 $\boxed{\textbf{(D) }\frac{34}{67}}$ .

Proof

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 $\left[\frac{100}{k}\right]-\frac{100}{k}$ if $\left[\frac{100}{k}\right]$ is even and $\frac{100}{k}-\left[\frac{100}{k}\right]$ if $\left[\frac{100}{k}\right]$ is odd.

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) $\text{fpart} \left(\frac{100 - n}{k}\right) < \frac{1}{2}$

-> $\text{fpart}\left(\frac{100}{k}\right) = \text{fpart} \left(\frac{n}{k}\right) +\text{fpart} \left(\frac{100 - n}{k}\right)$

Case 2)

$\text{fpart} \left(\frac{100 - n}{k}\right) > \frac{1}{2}$

-> $\text{fpart}\left(\frac{100}{k}\right) + 1 = \text{fpart} \left(\frac{n}{k}\right) +\text{fpart} \left(\frac{100 - n}{k}\right)$

and for $n = 1$ , since $k$ is odd, $\left[\frac{99}{k}\right] \neq \left[\frac{100}{k}\right]$

-> $99.5 = k (p + .5)$ -> $199 = k (2p + 1)$ , and $199$ is prime so $k = 1$ or $k =199$ , which is not in this set

, which makes

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

Now the only case without rounding, $k = 1$ . It must be true.

@@ Line 62: / Line 62: @@
 <math>0\rightarrow 49,50,51\rightarrow 98</math>
 <math>100\rightarrow 51,50,49\rightarrow 2</math>
 <math>P=\frac{98}{99}</math>
 <math>0\rightarrow 43,44\rightarrow 56,57\rightarrow 86</math>
 <math>87\rightarrow 57,56\rightarrow 44,43\rightarrow 14</math>
 <math>P=\frac{74}{87}</math>
 <math>0\rightarrow 33,34\rightarrow 66</math>
 <math>100\rightarrow 67,66\rightarrow 34</math>
 <math>P=\frac{34}{67}</math>

2011 AMC 12B (Problems • Answer Key • Resources)
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Difference between revisions of "2011 AMC 12B Problems/Problem 25"

Revision as of 17:13, 15 February 2016

Contents

Problem

Solution

Proof

See also