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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So for AMC purpose, answer is <math>\boxed{\textbf{(D) }\frac{34}{67}}</math>. | So for AMC purpose, answer is <math>\boxed{\textbf{(D) }\frac{34}{67}}</math>. | ||
− | + | [b]Proof:[/b] | |
Notice that for these integers <math>99,87,67</math>: | Notice that for these integers <math>99,87,67</math>: | ||
Line 135: | Line 136: | ||
<br /> | <br /> | ||
Now the only case without rounding, <math>k = 1</math>. It must be true. | 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> | ||
+ | <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> | ||
+ | |||
+ | <math>g:</math> | ||
+ | <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 <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{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}} | {{MAA Notice}} |
Revision as of 17:05, 25 June 2016
Problem
For every and integers with odd, denote by the integer closest to . For every odd integer , let be the probability that
for an integer randomly chosen from the interval . What is the minimum possible value of over the odd integers in the interval ?
Solution
Solution 1
Answer:
First of all, you have to realize that
if
then
So, we can consider what happen in and it will repeat. Also since range of is to , it is always a multiple of . So we can just consider for .
Let 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 , , , .
For , . 3 of the that should consider lands in here.
For , , then we need
else for , , then we need
For ,
So, for the condition to be true, . ( , no worry for the rounding to be )
, so this is always true.
For , , so we want , or
For k = 67,
For k = 69,
etc.
We can clearly see that for this case, has the minimum , which is . Also, .
So for AMC purpose, answer is .
[b]Proof:[/b]
Notice that for these integers :
That the probability is . Even for , . And .
Perhaps the probability for a given is if and if .
So and . Because !
Now, let's say we are not given any answer, we need to consider .
I claim that
If got round down, then all satisfy the condition along with
because if and , so must
and for , it is the same as .
, which makes
.
If got round up, then all satisfy the condition along with
because if and
Case 1)
->
Case 2)
->
and for , since is odd,
-> -> , and is prime so or , which is not in this set
, which makes
.
Now the only case without rounding, . It must be true.
Solution 2
It suffices to consider Now for each of these let If we let then the following graphs result for and
Our probability is the number of such that over Of course, this always holds for If we let vary, then the graph of is always very similar to what it looks like above (groups of dots). However, the graph of can vary greatly. In the above diagram, for all while it is possible for for all In order to minimize the number of which satisfy we either want for or for This way, we see that at least half of the numbers from to satisfy the given equation. So, our desired probability is at least As shown by the diagram above, the probability is for Clearly no better solutions can exist when On the other hand, for and do not yield better probabilities. Therefore, our answer is
See also
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.