Difference between revisions of "2011 AMC 12B Problems/Problem 25"
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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> | ||
+ | <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[\frac{100}{k}\right]-\frac{100}{k}</math> if <math>\left[\frac{100}{k}\right]</math> is even and <math>\frac{100}{k}-\left[\frac{100}{k}\right]</math> if <math>\left[\frac{100}{k}\right]</math> is odd. | ||
+ | |||
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Revision as of 18:13, 15 February 2016
Contents
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
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 .
Proof
Notice that for these integers :
That the probability is . Even for
,
. And
.
Perhaps the probability for a given is
if
is even and
if
is odd.
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.
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.