2006 AMC 12B Problems/Problem 20

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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 $x$ .

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

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

2006 AMC 12B (Problems • 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

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2006 AMC 12B Problems/Problem 20

From AoPSWiki

Problem

Solution

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