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Difference between revisions of "1985 AJHSME Problems/Problem 22"

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For a regular digit, there are <math>10</math> possible choices to make: <math>0</math>, <math>1</math>, <math>2</math>, <math>3</math>, <math>4</math>, <math>5</math>, <math>6</math>, <math>7</math>, <math>8</math>, or <math>9</math>. The only digit that is not regular is the first one, which prohibits <math>0</math> and <math>1</math> from taking place, resulting in <math>8</math> 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 <math>8 * 10 * 10 * 10 * 10 * 10 * 10</math>, or <math>8 * 10 ^ 6</math> possible total phone numbers (<math>b</math>).  
 
For a regular digit, there are <math>10</math> possible choices to make: <math>0</math>, <math>1</math>, <math>2</math>, <math>3</math>, <math>4</math>, <math>5</math>, <math>6</math>, <math>7</math>, <math>8</math>, or <math>9</math>. The only digit that is not regular is the first one, which prohibits <math>0</math> and <math>1</math> from taking place, resulting in <math>8</math> 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 <math>8 * 10 * 10 * 10 * 10 * 10 * 10</math>, or <math>8 * 10 ^ 6</math> possible total phone numbers (<math>b</math>).  
  
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Now that we have the denominator, the only unknown remaining is <math>b</math>. To solve for <math>a</math>, let's use the same method as we did for the denominator. For the first digit, there is only one possible value: <math>9</math>. For the last digit, there is only one possible value: <math>0</math>. However, the rest of the five digits are "free" (meaning they can be any one of <math>10</math> choices). Thus <math>a = 1 * 10 * 10 * 10 * 10 * 10 * 1</math>, or <math>10^5</math> possible phone numbers with restrictions.  
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Now that we have the denominator, the only unknown remaining is <math>a</math>. To solve for <math>a</math>, let's use the same method as we did for the denominator. For the first digit, there is only one possible value: <math>9</math>. For the last digit, there is only one possible value: <math>0</math>. However, the rest of the five digits are "free" (meaning they can be any one of <math>10</math> choices). Thus <math>a = 1 * 10 * 10 * 10 * 10 * 10 * 1</math>, or <math>10^5</math> possible phone numbers with restrictions.  
  
 
The fraction <math>\frac{a}{b}</math> is the same as <math>\frac{10^5}{8 * 10^6}</math>, which reduces to <math>\boxed{\text{B}}</math>.
 
The fraction <math>\frac{a}{b}</math> is the same as <math>\frac{10^5}{8 * 10^6}</math>, which reduces to <math>\boxed{\text{B}}</math>.
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==Solution 2==
 
==Solution 2==
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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 <math>\dfrac{1}{8}</math> possibility the first digit is 9 and a <math>\dfrac{1}{10}</math> possibility the last digit is 0. Multiplying these gives us <math>\dfrac{1}{8} \cdot \dfrac{1}{10} = \dfrac{1}{80} \Rightarrow \boxed{\text{B}}.</math>
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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 <math>\dfrac{1}{8}</math> possibility the first digit is 9 and a <math>\dfrac{1}{10}</math> possibility the last digit is 0. Multiplying these gives us <math>\dfrac{1}{8} \cdot \dfrac{1}{10} = \dfrac{1}{80} \Longrightarrow \boxed{\text{B}}.</math>
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~sanaops9
  
 
==See Also==
 
==See Also==

Latest revision as of 15:20, 6 October 2024

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 $a$. 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

See Also

1985 AJHSME (Problems • Answer Key • Resources)
Preceded by
Problem 21 		Followed by
Problem 23
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions


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