1985 AJHSME Problems/Problem 22

Problem

Assume every 7-digit whole number is a possible telephone number except those that begin with $0$ or $1$ . What fraction of telephone numbers begin with $9$ and end with $0$ ?

$\text{(A)}\ \frac{1}{63} \qquad \text{(B)}\ \frac{1}{80} \qquad \text{(C)}\ \frac{1}{81} \qquad \text{(D)}\ \frac{1}{90} \qquad \text{(E)}\ \frac{1}{100}$

Note: All telephone numbers are 7-digit whole numbers.

Solution 1

An equivalent problem is finding the probability that a randomly selected telephone number begins with $9$ and ends with $0$ .

There are $10-2=8$ possibilities for the first digit in total, and only $1$ that works, so the probability the number begins with $9$ is $\frac{1}{8}$

There are $10$ possibilities for the last digit, and only $1$ that works $(0)$ , so the probability the number ends with $0$ is $\frac{1}{10}$

Since these are independent events, the probability both happens is $\frac{1}{8}\cdot \frac{1}{10}=\frac{1}{80}$

$\boxed{\text{B}}$

Solution 2

The fraction is simply the number of $7$ -digit phone numbers with the restrictions applied divided by the total number of phone numbers. Let $a$ denote the numerator, and $b$ denote the denominator. Let's first work on finding $b$ , the total number.

For a regular digit, there are $10$ possible choices to make: $0$ , $1$ , $2$ , $3$ , $4$ , $5$ , $6$ , $7$ , $8$ , or $9$ . The only digit that is not regular is the first one, which prohibits $0$ and $1$ from taking place, resulting in $8$ possible choices to make for that first digit. Since each digit is independent of one another, we multiply the number of choices for each digit, resulting in $8 * 10 * 10 * 10 * 10 * 10 * 10$ , or $8 * 10 ^ (6)$ possible total phone numbers ( $b$ ).

Now that we have the denominator, the only unknown remaining is $b$ . To solve for $a$ , let's use the same method as we did for the denominator. For the first digit, there is only one possible value: $9$ . For the last digit, there is only one possible value: $0$ . However, the rest of the five digits are "free" (meaning they can be any one of $10$ choices). Thus $a = 1 * 10 * 10 * 10 * 10 * 10 * 1$ , or $10^5$ possible phone numbers with restrictions.

The fraction $\frac{a}{b}$ is the same as $\frac{10^5}{8 * 10^(6)}$ , which reduces to $\boxed{\text{B}}$ .

1985 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
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1985 AJHSME Problems/Problem 22

Contents

Problem

Solution 1

Solution 2

See Also