Difference between revisions of "2011 AMC 12B Problems/Problem 25"
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− | If <math> \left[\frac{100}{k}\right] </math> got round down, then <math>1 \le n \le \frac{k}{2}</math> all satisfy the condition along with <math>n = k</math>, which makes | + | If <math> \left[\frac{100}{k}\right] </math> got round down, then <math>1 \le n \le \frac{k}{2}</math> all satisfy the condition along with <math>n = k</math> |
+ | |||
+ | because if <math>\text{fpart}\left(\frac{100}{k}\right) < \frac{1}{2}</math> and <math>\text{fpart} \left(\frac{n}{k}\right) < \frac{1}{2}</math>, so must <math>\text{fpart} \left(\frac{100 - n}{k}\right) < \frac{1}{2}</math> | ||
+ | |||
+ | and for <math>n = k</math>, it is the same as <math>n = 0</math>. | ||
+ | |||
+ | , which makes | ||
<math>P(k) \ge \frac{1}{2} + \frac{1}{2k}</math>. | <math>P(k) \ge \frac{1}{2} + \frac{1}{2k}</math>. | ||
<br /> | <br /> | ||
− | If <math> \left[\frac{100}{k}\right] </math> got round up, then <math>\frac{k}{2} \le n \le k</math> all satisfy the condition along with <math>n = 1</math>, which makes | + | If <math> \left[\frac{100}{k}\right] </math> got round up, then <math>\frac{k}{2} \le n \le k</math> all satisfy the condition along with <math>n = 1</math> |
+ | |||
+ | |||
+ | because if <math>\text{fpart}\left(\frac{100}{k}\right) > \frac{1}{2}</math> and <math>\text{fpart} \left(\frac{n}{k}\right) > \frac{1}{2}</math> | ||
+ | |||
+ | Case 1) | ||
+ | <math>\text{fpart} \left(\frac{100 - n}{k}\right) < \frac{1}{2}</math> | ||
+ | |||
+ | -> <math>\text{fpart}\left(\frac{100}{k}\right) = \text{fpart} \left(\frac{n}{k}\right) +\text{fpart} \left(\frac{100 - n}{k}\right)</math> | ||
+ | |||
+ | Case 2) | ||
+ | |||
+ | <math>\text{fpart} \left(\frac{100 - n}{k}\right) > \frac{1}{2}</math> | ||
+ | |||
+ | -> <math>\text{fpart}\left(\frac{100}{k}\right) + 1 = \text{fpart} \left(\frac{n}{k}\right) +\text{fpart} \left(\frac{100 - n}{k}\right)</math> | ||
+ | |||
+ | |||
+ | and for <math>n = 1</math>, since <math>k</math> is odd, <math>\left[\frac{99}{k}\right] \neq \left[\frac{100}{k}\right]</math> | ||
+ | |||
+ | -> <math>99.5 = k (p + .5)</math> -> <math>199 = k (2p + 1)</math>, and <math>199</math> is prime so <math>k = 1</math> or <math>k =199</math>, which is not in this set | ||
+ | |||
+ | , which makes | ||
<math>P(k) \ge \frac{1}{2} + \frac{1}{2k}</math>. | <math>P(k) \ge \frac{1}{2} + \frac{1}{2k}</math>. | ||
+ | |||
+ | <br /> | ||
+ | Now the only case without rounding, <math>k = 1</math>. It must be true. | ||
== See also == | == See also == |
Revision as of 10:30, 11 March 2011
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 |
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 |