Difference between revisions of "2006 AMC 12B Problems/Problem 20"
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This means that our result is <math>\frac 32 \left( 10^{-1} + 10^{-2} + \cdots \right) = \frac 32 \cdot \frac 19 = \boxed{\frac 16}</math>. | This means that our result is <math>\frac 32 \left( 10^{-1} + 10^{-2} + \cdots \right) = \frac 32 \cdot \frac 19 = \boxed{\frac 16}</math>. | ||
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+ | == Solution 2 == | ||
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+ | The largest value for <math>x</math> is <math>10^{0}</math>. If <math>x > 10^{-1}</math>, then <math>\lfloor\log_{10}4x\rfloor</math> doesn't fulfill the condition unless <math>10^{-2} \leq x < 0.25 * 10^{-1}</math>. The same holds when you get smaller, because <math>x = 0.25 * 10^{n}</math> for <math>n \leq 0</math> is the lowest value such that <math>4x</math> becomes a higher power of 10. | ||
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+ | Recognize that this is a geometric sequence. The probability of choosing <math>x</math> such that <math>\lfloor\log_{10}4x\rfloor</math> and <math>\lfloor\log_{10}x\rfloor</math> both equal <math>-1</math> is <math>(9/10)* (15/90) =15/100</math>, because there is a 90 percent chance of choosing <math>x > 10^{-1}</math>, and only values of <math>x</math> between <math>10^{-1}</math> and <math>0.25*10^{0}</math> work in this case. Then, for <math>x</math> such that <math>\lfloor\log_{10}4x\rfloor</math> and <math>\lfloor\log_{10}x\rfloor</math> both equal <math>-2</math>, you have <math>(1/10) * ((9/10) *(15/90))</math>. This is a geometric series with ratio <math>1/10</math>. Using <math>a/(1-r)</math> for the sum of an infinite geometric sequence, we get <math>(15/100)/(1-(1/10)) = \boxed{\frac 16}</math>. | ||
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+ | Solution by Halt_CatchFire | ||
== See also == | == See also == | ||
{{AMC12 box|year=2006|ab=B|num-b=19|num-a=21}} | {{AMC12 box|year=2006|ab=B|num-b=19|num-a=21}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 14:54, 26 December 2017
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.