2007/08 AHSMC Part 1 Problem 15

Problem 15 of 16

An integer $n$ is randomly chosen from $10^{99}$ to $10^{100} - 1$ inclusive. The real number $m$ is defined by $m = \dfrac{9n}{5}$ . Of the following five numbers, the one closest to the probability that $10^{99} \le m \le 10^{100} - 1$ is

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Solution

We have $10^{99} \times \dfrac{5}{9} \le n \le (10^{100}-1)\dfrac{5}{9}$ . Since $n$ is an integer, $5 \times 10^{98} < n < 5 \times 10^{99}$ . However, we must eliminate those values of $n$ where $5 \times 10^{98} < n < 10^{99}$ . Thus the number of acceptable values of $n$ is about $4.5 \times 10^{99}$ . Since $10^{99} \le n \le 10^{100}-1$ , the desired probability is very close to $\dfrac{1}{2}$ . The answer is (C).

University of Alberta

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