Difference between revisions of "1985 AJHSME Problems/Problem 22"
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''Note: All telephone numbers are 7-digit whole numbers.'' | ''Note: All telephone numbers are 7-digit whole numbers.'' | ||
| − | ==Solution | + | ==Solution 1== |
The fraction is simply the number of <math>7</math>-digit phone numbers with the restrictions applied divided by the total number of phone numbers. Let <math>a</math> denote the numerator, and <math>b</math> denote the denominator. Let's first work on finding <math>b</math>, the total number. | The fraction is simply the number of <math>7</math>-digit phone numbers with the restrictions applied divided by the total number of phone numbers. Let <math>a</math> denote the numerator, and <math>b</math> denote the denominator. Let's first work on finding <math>b</math>, the total number. | ||
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(~thelinguist46295) | (~thelinguist46295) | ||
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| + | ==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 <math>\dfrac18</math> possibility the first digit is 9 and a <math>\dfrac1{10}</math> possibility the last digit is 0. Multiplying these gives us <math>\dfrac18 \cdot \dfrac1{10} = \dfrac{1}{80} \longRightarrow \boxed{\text{B}}.</math> | ||
==See Also== | ==See Also== | ||
Revision as of 13:45, 1 August 2024
Contents
Problem
Assume every 7-digit whole number is a possible telephone number except those that begin with 
or 
. What fraction of telephone numbers begin with 
and end with ?
Note: All telephone numbers are 7-digit whole numbers.
Solution 1
The fraction is simply the number of 
-digit phone numbers with the restrictions applied divided by the total number of phone numbers. Let 
denote the numerator, and 
denote the denominator. Let's first work on finding , the total number.
For a regular digit, there are 
possible choices to make: , , 
, 
, 
, 
, 
, , 
, or . The only digit that is not regular is the first one, which prohibits and from taking place, resulting in 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 
, or 
possible total phone numbers ().
Now that we have the denominator, the only unknown remaining is . To solve for , let's use the same method as we did for the denominator. For the first digit, there is only one possible value: . For the last digit, there is only one possible value: . However, the rest of the five digits are "free" (meaning they can be any one of choices). Thus 
, or 
possible phone numbers with restrictions.
The fraction 
is the same as 
, which reduces to 
.
(~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 
possibility the first digit is 9 and a 
possibility the last digit is 0. Multiplying these gives us $\dfrac18 \cdot \dfrac1{10} = \dfrac{1}{80} \longRightarrow \boxed{\text{B}}.$ (Error compiling LaTeX. Unknown error_msg)
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
