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

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m (→‎Solution)
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<br />
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If <math> \left[\frac{100}{k}\right] </math> got round down, then <math>1 \le n \le \frac{k}{2}</math> all satisfy the condition along with <math>n = k</math>, which makes  
+
If <math> \left[\frac{100}{k}\right] </math> got round down, then <math>1 \le n \le \frac{k}{2}</math> all satisfy the condition along with <math>n = k</math>
 +
 
 +
because if <math>\text{fpart}\left(\frac{100}{k}\right) < \frac{1}{2}</math> and <math>\text{fpart} \left(\frac{n}{k}\right) < \frac{1}{2}</math>, so must <math>\text{fpart} \left(\frac{100 - n}{k}\right) < \frac{1}{2}</math>
 +
 
 +
and for <math>n = k</math>, it is the same as <math>n = 0</math>.
 +
 
 +
, which makes  
  
 
<math>P(k) \ge \frac{1}{2} + \frac{1}{2k}</math>.
 
<math>P(k) \ge \frac{1}{2} + \frac{1}{2k}</math>.
  
 
<br />
 
<br />
−
If <math> \left[\frac{100}{k}\right] </math> got round up, then <math>\frac{k}{2} \le n \le k</math> all satisfy the condition along with <math>n = 1</math>, which makes  
+
If <math> \left[\frac{100}{k}\right] </math> got round up, then <math>\frac{k}{2} \le n \le k</math> all satisfy the condition along with <math>n = 1</math>
 +
 
 +
 
 +
because if <math>\text{fpart}\left(\frac{100}{k}\right) > \frac{1}{2}</math> and <math>\text{fpart} \left(\frac{n}{k}\right) > \frac{1}{2}</math>
 +
 
 +
Case 1)
 +
<math>\text{fpart} \left(\frac{100 - n}{k}\right) < \frac{1}{2}</math>
 +
 
 +
-> <math>\text{fpart}\left(\frac{100}{k}\right) = \text{fpart} \left(\frac{n}{k}\right) +\text{fpart} \left(\frac{100 - n}{k}\right)</math>
 +
 
 +
Case 2)
 +
 
 +
<math>\text{fpart} \left(\frac{100 - n}{k}\right) > \frac{1}{2}</math>
 +
 
 +
-> <math>\text{fpart}\left(\frac{100}{k}\right) + 1 = \text{fpart} \left(\frac{n}{k}\right) +\text{fpart} \left(\frac{100 - n}{k}\right)</math>
 +
 
 +
 
 +
and for <math>n = 1</math>, since <math>k</math> is odd, <math>\left[\frac{99}{k}\right] \neq \left[\frac{100}{k}\right]</math>
 +
 
 +
-> <math>99.5 = k (p + .5)</math>  -> <math>199 = k (2p + 1)</math>, and <math>199</math> is prime so <math>k = 1</math> or <math>k =199</math>, which is not in this set
 +
 
 +
, which makes  
  
 
<math>P(k) \ge \frac{1}{2} + \frac{1}{2k}</math>.
 
<math>P(k) \ge \frac{1}{2} + \frac{1}{2k}</math>.
 +
 +
<br />
 +
Now the only case without rounding, <math>k = 1</math>. It must be true.
  
 
== See also ==
 
== See also ==

Revision as of 09:30, 11 March 2011

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 $n = k$

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.

See also

2011 AMC 12B (Problems • Answer Key • Resources)
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