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

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 $\boxed{\text{B}}$.

(~thelinguist46295)

Solution 2

There are 8 possibilities for the first digit (all digits except 0 or 1) and 10 possibilities for the last digit (any digit). There is a $\dfrac{1}{8}$ possibility the first digit is 9 and a $\dfrac{1}{10}$ possibility the last digit is 0. Multiplying these gives us $\dfrac{1}{8} \cdot \dfrac{1}{10} = \dfrac{1}{80} \Longrightarrow \boxed{\text{B}}.$

~sanaops9