Difference between revisions of "2011 AMC 12B Problems/Problem 25"
(→Proof) |
(→Proof) |
||
Line 86: | Line 86: | ||
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>. | 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}} | + | 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 /> |
Revision as of 18:40, 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
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.
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.