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Difference between revisions of "2006 AMC 12B Problems/Problem 20"

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{{empty}}
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== Problem ==
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Let <math>x</math> be chosen at random from the interval <math>(0,1)</math>. What is the probability that
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<math>\lfloor\log_{10}4x\rfloor - \lfloor\log_{10}x\rfloor = 0</math>?
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Here <math>\lfloor x\rfloor</math> denotes the greatest integer that is less than or equal to <math>x</math>.
  
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== Problem ==
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{{problem}}
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<math>
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\mathrm{(A)}\ \frac 18
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\qquad
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\mathrm{(B)}\ \frac 3{20}
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\qquad
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\mathrm{(C)}\ \frac 16
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\qquad
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\mathrm{(D)}\ \frac 15
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\qquad
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\mathrm{(E)}\ \frac 14
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</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>.
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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>.
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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>.
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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>.
  
 
== 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 17:12, 11 February 2009

Problem

Let $x$ be chosen at random from the interval $(0,1)$. What is the probability that $\lfloor\log_{10}4x\rfloor - \lfloor\log_{10}x\rfloor = 0$? Here $\lfloor x\rfloor$ denotes the greatest integer that is less than or equal to $x$.


$\mathrm{(A)}\ \frac 18 \qquad \mathrm{(B)}\ \frac 3{20} \qquad \mathrm{(C)}\ \frac 16 \qquad \mathrm{(D)}\ \frac 15 \qquad \mathrm{(E)}\ \frac 14$

Solution

Let $k$ be an arbitrary integer. For which $x$ do we have $\lfloor\log_{10}4x\rfloor = \lfloor\log_{10}x\rfloor = k$?

The equation $\lfloor\log_{10}x\rfloor = k$ can be rewritten as $10^k \leq x < 10^{k+1}$. The second one gives us $10^k \leq 4x < 10^{k+1}$. Combining these, we get that both hold at the same time if and only if $10^k \leq x < \frac{10^{k+1}}4$.

Hence for each integer $k$ we get an interval of values for which $\lfloor\log_{10}4x\rfloor - \lfloor\log_{10}x\rfloor = 0$. These intervals are obviously pairwise disjoint.

For any $k\geq 0$ the corresponding interval is disjoint with $(0,1)$, so it does not contribute to our answer. On the other hand, for any $k<0$ the entire interval is inside $(0,1)$. Hence our answer is the sum of the lengths of the intervals for $k<0$.

For a fixed $k$ the length of the interval $\left[ 10^k, \frac{10^{k+1}}4 \right)$ is $\frac 32\cdot 10^k$.

This means that our result is $\frac 32 \left( 10^{-1} + 10^{-2} + \cdots \right) = \frac 32 \cdot \frac 19 = \boxed{\frac 16}$.

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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