2012 AMC 10A Problems/Problem 25
Problem
Real numbers ,
, and
are chosen independently and at random from the interval
for some positive integer
. The probability that no two of
,
, and
are within 1 unit of each other is greater than
. What is the smallest possible value of
?
Solution
Without loss of generality, assume that .
Then, the possible choices for ,
, and
are represented by the expression
.
There are two restrictions: and
.
Now, let . Then,
and
.
Then, let . Combining the two inequalities gives us
.
Since , then
. Thus,
.
There are ways to choose each number; the successful choices are represented by
.
The probability then, is which must be greater than
.
Plug in values. Trying gives us
, which is less than
. Try the next integer,
, which gives us
which is greater than
.
Thus, our answer is .
See Also
2012 AMC 10A (Problems • Answer Key • Resources) | ||
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