Difference between revisions of "1985 AJHSME Problems/Problem 22"
5849206328x (talk | contribs) (New page: ==Problem== Assume every 7-digit whole number is a possible telephone number except those that begin with <math>0</math> or <math>1</math>. What fraction of telephone numbers begin with ...) |
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==Problem== | ==Problem== | ||
− | Assume every 7-digit whole number is a possible telephone number except those that begin with <math>0</math> or <math>1</math>. What fraction of telephone numbers begin with <math>9</math> and end with <math>0</math>? | + | Assume every 7-digit [[whole number]] is a possible telephone number except those that begin with <math>0</math> or <math>1</math>. What [[fraction]] of telephone numbers begin with <math>9</math> and end with <math>0</math>? |
<math>\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}</math> | <math>\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}</math> | ||
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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. | ||
− | + | 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>). | |
− | + | 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. | |
− | + | 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>. | |
− | + | (~thelinguist46295) | |
− | <math>\boxed{\text{B}}</math> | + | ==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>\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> | ||
+ | |||
+ | ~sanaops9 | ||
==See Also== | ==See Also== | ||
− | [[ | + | {{AJHSME box|year=1985|num-b=21|num-a=23}} |
+ | [[Category:Introductory Combinatorics Problems]] | ||
+ | [[Category:Introductory Probability Problems]] | ||
+ | |||
+ | |||
+ | {{MAA Notice}} |
Latest revision as of 13:47, 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
~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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