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 $x.$

$\dfrac{1}{8}$

$\dfrac{3}{20}$

$\dfrac{1}{6}$

$\dfrac{1}{5}$

$\dfrac{1}{4}$

Solution:

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 $\frac{10^{n+1}/4 - 10^n}{10^{n+1} - 10^n} = \frac{10/4 - 1}{10 - 1} = \frac{1}{6}.$

Since this fraction is the same on every such interval, the overall probability is $\dfrac{1}{6}.$

Thus, the correct answer is C.

2006 AMC 12B Problem 20

Solution:

Problem 20 in Other Years