2011 AMC 12B Problems/Problem 25
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 (D).
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 |
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All AMC 12 Problems and Solutions |
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