2006 AMC 12B Problems/Problem 20

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

2006 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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2006 AMC 12B Problems/Problem 20

Contents

Problem

Solution

Solution 2

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