Difference between revisions of "2011 AMC 12B Problems/Problem 25"
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Now the only case without rounding, <math>k = 1</math>. It must be true. | Now the only case without rounding, <math>k = 1</math>. It must be true. | ||
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== See also == | == See also == |
Revision as of 00:11, 18 February 2013
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 |