Difference between revisions of "2006 AMC 12B Problems/Problem 20"
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− | {{ | + | == Problem == |
+ | Let <math>x</math> be chosen at random from the interval <math>(0,1)</math>. What is the probability that | ||
+ | <math>\lfloor\log_{10}4x\rfloor - \lfloor\log_{10}x\rfloor = 0</math>? | ||
+ | Here <math>\lfloor x\rfloor</math> denotes the greatest integer that is less than or equal to <math>x</math>. | ||
− | + | ||
− | {{ | + | <math> |
+ | \mathrm{(A)}\ \frac 18 | ||
+ | \qquad | ||
+ | \mathrm{(B)}\ \frac 3{20} | ||
+ | \qquad | ||
+ | \mathrm{(C)}\ \frac 16 | ||
+ | \qquad | ||
+ | \mathrm{(D)}\ \frac 15 | ||
+ | \qquad | ||
+ | \mathrm{(E)}\ \frac 14 | ||
+ | </math> | ||
== Solution == | == Solution == | ||
+ | |||
+ | Let <math>k</math> be an arbitrary integer. For which <math>x</math> do we have <math>\lfloor\log_{10}4x\rfloor = \lfloor\log_{10}x\rfloor = k</math>? | ||
+ | |||
+ | The equation <math>\lfloor\log_{10}x\rfloor = k</math> can be rewritten as <math>10^k \leq x < 10^{k+1}</math>. The second one gives us <math>10^k \leq 4x < 10^{k+1}</math>. Combining these, we get that both hold at the same time if and only if <math>10^k \leq x < \frac{10^{k+1}}4</math>. | ||
+ | |||
+ | Hence for each integer <math>k</math> we get an interval of values for which <math>\lfloor\log_{10}4x\rfloor - \lfloor\log_{10}x\rfloor = 0</math>. These intervals are obviously pairwise disjoint. | ||
+ | |||
+ | For any <math>k\geq 0</math> the corresponding interval is disjoint with <math>(0,1)</math>, so it does not contribute to our answer. On the other hand, for any <math>k<0</math> the entire interval is inside <math>(0,1)</math>. Hence our answer is the sum of the lengths of the intervals for <math>k<0</math>. | ||
+ | |||
+ | For a fixed <math>k</math> the length of the interval <math>\left[ 10^k, \frac{10^{k+1}}4 \right)</math> is <math>\frac 32\cdot 10^k</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>. | ||
== 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}} |
Revision as of 18:12, 11 February 2009
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 .
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 |