2006 AMC 12B Problem 20

Question

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.

\dfrac{1}{8}

\dfrac{3}{20}

\dfrac{1}{6}

\dfrac{1}{5}

\dfrac{1}{4}

The condition says

x

and

4x

have the same power-of-ten magnitude

On each interval

[10^n, 10^{n+1}),

the condition holds for

x \lt \dfrac{10^{n+1}}{4}

The equation says

\lfloor \log_{10} x \rfloor = \lfloor \log_{10} 4x \rfloor,

i.e.

x

and

4x

lie in the same interval

[10^n, 10^{n+1}).

This holds exactly when

10^n \le x

and

4x \lt 10^{n+1},

that is

10^n \le x \lt \dfrac{10^{n+1}}{4}.

Within

[10^n, 10^{n+1}),

the favorable fraction is

\begin{aligned} &\frac{\frac{10^{n+1}}{4} - 10^n}{10^{n+1} - 10^n} \\ &= \frac{\frac{10}{4} - 1}{10 - 1} = \frac{1}{6}. \end{aligned}

Since this fraction is the same on every such interval, the overall probability is

\dfrac{1}{6}.

Thus, the correct answer is C.

Problem 20 in other years