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

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== Solution ==
 
== Solution ==
  
 +
=== Solution 1 ===
 
Answer: <math>(D)  \frac{34}{67}</math>
 
Answer: <math>(D)  \frac{34}{67}</math>
  
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<br />
 
<br />
LET <math>\text{fpart}(x)</math> be the fractional part function
+
Let <math>\text{fpart}(x)</math> 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 <math>k = 99</math>, <math>87</math>, <math>67</math>, <math>13</math>. <math>1\le n \le k</math>
 
This is an AMC exam, so use the given choices wisely. With the given choices, and the previous explanation, we only need to consider <math>k = 99</math>, <math>87</math>, <math>67</math>, <math>13</math>. <math>1\le n \le k</math>
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We can clearly see that for this case, <math>k = 67</math> has the minimum <math>P(k)</math>, which is <math>\frac{34}{67}</math>. Also, <math>\frac{7}{13} > \frac{34}{67}</math> .
 
We can clearly see that for this case, <math>k = 67</math> has the minimum <math>P(k)</math>, which is <math>\frac{34}{67}</math>. Also, <math>\frac{7}{13} > \frac{34}{67}</math> .
  
So for AMC purpose, answer is (D).
+
So for AMC purpose, answer is <math>\boxed{\textbf{(D) }\frac{34}{67}}</math>.
 +
 
 +
===Proof:===
 +
 
 +
Notice that for these integers <math>99,87,67</math>:
 +
 
 +
 
 +
<math>0\rightarrow 49,50,51\rightarrow 98</math>
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 +
<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>
 +
 
 +
 
 +
That the probability is <math>\frac{2k-100}{k}=2-\frac{100}{k}</math>. Even for <math>k=13</math>, <math>P(13)=\frac{9}{13}=\frac{100}{13}-7</math>. And <math>P(11)=\frac{10}{11}=10-\frac{100}{11}</math>.
 +
 
 +
Perhaps the probability for a given <math>k</math> is <math>\left\lceil{\frac{100}{k}}\right\rceil-\frac{100}{k}</math> if <math>\left[\frac{100}{k}\right]=\left\lfloor{\frac{100}{k}}\right\rfloor</math> and <math>\frac{100}{k}-\left\lfloor{\frac{100}{k}}\right\rfloor</math> if <math>\left[\frac{100}{k}\right]=\left\lceil{\frac{100}{k}}\right\rceil</math>.
 +
 
 +
So <math>P>\frac{1}{2}</math> and <math>P_\text{min}=\frac{k_\text{min}+1}{2k_\text{min}}=\frac{101}{201}</math>. Because <math>201=3\cdot 67\mid 99!</math>  !
  
 
<br />
 
<br />
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<br />
 
<br />
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)
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<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.
 +
 +
=== Solution 2 ===
 +
 +
It suffices to consider <math>0\le n < k-1.</math> Now for each of these <math>n,</math> let <math>f(n)=[\frac{n}{k}], g(n)=[\frac{100}{k}]-[\frac{100-n}{k}].</math> If we let <math>k=67,</math> then the following graphs result for <math>f</math> and <math>g.</math>
 +
 +
<math>f:</math>
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<asy>
 +
size(10cm);
 +
for (int i = 0; i < 67; ++i) {
 +
if (i<=33) dot((i,0));
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    else dot((i,1));
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}
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label("(0,0)",(0,0),SW);
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label("(66,1)",(66,1),NE);
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</asy>
 +
 +
<math>g:</math>
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<asy>
 +
size(10cm);
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for (int i = 0; i < 67; ++i) {
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dot((i,0));
 +
}
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label("(0,0)",(0,0),NW);
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label("(66,0)",(66,0),SE);
 +
</asy>
 +
 +
Our probability is the number of <math>0\le i<k</math> such that <math>f(i)+g(i)=0</math> over <math>k.</math> Of course, this always holds for <math>i=0.</math> If we let <math>k</math> vary, then  the graph of <math>f</math> is always very similar to what it looks like above (groups of <math>\frac{k+1}{2},\frac{k-1}{2}</math> dots). However, the graph of <math>g</math> can vary greatly. In the above diagram, <math>g(i)=0</math> for all <math>i,</math> while it is possible for <math>g(i)=-1</math> for all <math>i\neq 0.</math> In order to minimize the number of <math>i</math> which satisfy <math>f(i)+g(i)=0,</math> we either want <math>g(i)=0</math> for <math>0\le i<k,</math> or <math>g(i)=-1</math> for <math>1\le i<k.</math> This way, we see that at least half of the numbers from <math>1</math> to <math>k-1</math> satisfy the given equation. So, our desired probability is at least <math>\frac{k+1}{2k}.</math> As shown by the diagram above, the probability is <math>\frac{34}{67}</math> for <math>k=67.</math> Clearly no better solutions can exist when <math>k<67.</math> On the other hand, for <math>k>67</math> <math>87</math> and <math>99</math> do not yield better probabilities. Therefore, our answer is <math>\boxed{\frac{34}{67}}.</math>
  
 
== See also ==
 
== See also ==
 
  
 
{{AMC12 box|year=2011|num-b=24|after=Last Problem|ab=B}}
 
{{AMC12 box|year=2011|num-b=24|after=Last Problem|ab=B}}
 +
{{MAA Notice}}

Revision as of 20:40, 6 February 2018

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

Solution 1

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

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

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.

Solution 2

It suffices to consider $0\le n < k-1.$ Now for each of these $n,$ let $f(n)=[\frac{n}{k}], g(n)=[\frac{100}{k}]-[\frac{100-n}{k}].$ If we let $k=67,$ then the following graphs result for $f$ and $g.$

$f:$ [asy] size(10cm); for (int i = 0; i < 67; ++i) { 	if (i<=33) dot((i,0));     else dot((i,1)); } label("(0,0)",(0,0),SW); label("(66,1)",(66,1),NE); [/asy]

$g:$ [asy] size(10cm); for (int i = 0; i < 67; ++i) { 	dot((i,0)); } label("(0,0)",(0,0),NW); label("(66,0)",(66,0),SE); [/asy]

Our probability is the number of $0\le i<k$ such that $f(i)+g(i)=0$ over $k.$ Of course, this always holds for $i=0.$ If we let $k$ vary, then the graph of $f$ is always very similar to what it looks like above (groups of $\frac{k+1}{2},\frac{k-1}{2}$ dots). However, the graph of $g$ can vary greatly. In the above diagram, $g(i)=0$ for all $i,$ while it is possible for $g(i)=-1$ for all $i\neq 0.$ In order to minimize the number of $i$ which satisfy $f(i)+g(i)=0,$ we either want $g(i)=0$ for $0\le i<k,$ or $g(i)=-1$ for $1\le i<k.$ This way, we see that at least half of the numbers from $1$ to $k-1$ satisfy the given equation. So, our desired probability is at least $\frac{k+1}{2k}.$ As shown by the diagram above, the probability is $\frac{34}{67}$ for $k=67.$ Clearly no better solutions can exist when $k<67.$ On the other hand, for $k>67$ $87$ and $99$ do not yield better probabilities. Therefore, our answer is $\boxed{\frac{34}{67}}.$

See also

2011 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 24
Followed by
Last Problem
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All AMC 12 Problems and Solutions

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