Difference between revisions of "2011 AMC 12B Problems/Problem 25"
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− | For <math>k > \frac{200}{3}</math>, <math>\left[\frac{100}{k}\right] = 1</math>. 3 of the <math>k</math> that should consider | + | For <math>k > \frac{200}{3}</math>, <math>\left[\frac{100}{k}\right] = 1</math>. 3 of the <math>k</math> that we should consider land in here. |
For <math>n < \frac{k}{2}</math>, <math> \left[\frac{n}{k}\right] = 0</math>, then we need <math> \left[\frac{100 - n}{k}\right] = 1</math> | For <math>n < \frac{k}{2}</math>, <math> \left[\frac{n}{k}\right] = 0</math>, then we need <math> \left[\frac{100 - n}{k}\right] = 1</math> | ||
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=== Solution 2 === | === Solution 2 === | ||
− | It suffices to consider <math>0\le n | + | It suffices to consider <math>0\le n \le k-1.</math> Now for each of these <math>n,</math> let <math>f(n)=\left[\frac{n}{k}\right], g(n)=\left[\frac{100-n}{k}\right]-\left[\frac{100}{k}\right].</math> If we let <math>k=67,</math> then the following graphs result for <math>f</math> and <math>g.</math> |
<math>f:</math> | <math>f:</math> | ||
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</asy> | </asy> | ||
− | Our probability is the number of <math>0\le i | + | Our probability is the number of <math>0\le i \le k-1</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 == |
Latest revision as of 03:07, 12 June 2024
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 we should consider land 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 .
Proof:
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.