2006 AMC 12B Problems/Problem 20
Contents
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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