2006 AMC 12B Problems/Problem 20
Contents
[hide]Problem
Let be chosen at random from the interval
. What is the probability that
?
Here
denotes the greatest integer that is less than or equal to
.
Solution
Let be an arbitrary integer. For which
do we have
?
The equation can be rewritten as
. The second one gives us
. Combining these, we get that both hold at the same time if and only if
.
Hence for each integer we get an interval of values for which
. These intervals are obviously pairwise disjoint.
For any the corresponding interval is disjoint with
, so it does not contribute to our answer. On the other hand, for any
the entire interval is inside
. Hence our answer is the sum of the lengths of the intervals for
.
For a fixed the length of the interval
is
.
This means that our result is .
Solution 2
The largest value for is
. If
, then
doesn't fulfill the condition unless
. The same holds when you get smaller, because
for
is the lowest value such that
becomes a higher power of 10.
Recognize that this is a geometric sequence. The probability of choosing such that
and
both equal
is
, because there is a 90 percent chance of choosing
, and only values of
between
and
work in this case. Then, for
such that
and
both equal
, you have
. This is a geometric series with ratio
. Using
for the sum of an infinite geometric sequence, we get
.
Solution by Halt_CatchFire
See also
2006 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 19 |
Followed by Problem 21 |
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 |
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