1985 AJHSME Problems/Problem 22
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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
An equivalent problem is finding the probability that a randomly selected telephone number begins with and ends with .
There are possibilities for the first digit in total, and only that works, so the probability the number begins with is
Out of the possibilities for the last digit, only works: , so the probability the number ends with is
Since these are independent events, the probability both happens is
Solution 2
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)
See Also
1985 AJHSME (Problems • Answer Key • Resources) | ||
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Followed by Problem 23 | |
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