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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}}

Latest revision as of 13:09, 21 May 2021

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 $\frac 32 \left( 10^{-1} + 10^{-2} + \cdots \right) = \frac 32 \cdot \frac 19 = \boxed{\frac 16}$.

Solution 2

The largest value for $x$ is $10^{0}$. If $x > 10^{-1}$, then $\lfloor\log_{10}4x\rfloor$ doesn't fulfill the condition unless $10^{-2} \leq x < 0.25 * 10^{-1}$. The same holds when you get smaller, because $x = 0.25 * 10^{n}$ for $n \leq 0$ is the lowest value such that $4x$ becomes a higher power of 10.

Recognize that this is a geometric sequence. The probability of choosing $x$ such that $\lfloor\log_{10}4x\rfloor$ and $\lfloor\log_{10}x\rfloor$ both equal $-1$ is $(9/10)* (15/90) =15/100$, because there is a 90 percent chance of choosing $x > 10^{-1}$, and only values of $x$ between $10^{-1}$ and $0.25*10^{0}$ work in this case. Then, for $x$ such that $\lfloor\log_{10}4x\rfloor$ and $\lfloor\log_{10}x\rfloor$ both equal $-2$, you have $(1/10) * ((9/10) *(15/90))$. This is a geometric series with ratio $1/10$. Using $a/(1-r)$ for the sum of an infinite geometric sequence, we get $(15/100)/(1-(1/10)) = \boxed{\frac 16}$.

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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