Difference between revisions of "1998 AHSME Problems/Problem 24"
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\qquad\mathrm{(D)}\ 20,000 | \qquad\mathrm{(D)}\ 20,000 | ||
\qquad\mathrm{(E)}\ 20,100</math> | \qquad\mathrm{(E)}\ 20,100</math> | ||
| − | == Solution == | + | == Solution A == |
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| + | In this problem, we only need to consider the digits <math>\overline{d_4d_5d_6d_7}</math>. Each possibility of <math>\overline{d_4d_5d_6d_7}</math> gives <math>2</math> possibilities for <math>\overline{d_1d_2d_3}</math>, which are <math>\overline{d_1d_2d_3}=\overline{d_4d_5d_6}</math> and <math>\overline{d_1d_2d_3}=\overline{d_5d_6d_7}</math> with the exception of the case of <math>d_4=d_5=d_6=d_7</math>, which only gives one sequence. After accounting for overcounting, the answer is <math>(10 \times 10 \times 10 \times 10) \times 2 - 10=19990 \Rightarrow \boxed{\mathrm{(C)}}</math> | ||
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| + | == Solution B == | ||
Let <math>A</math> represent the set of telephone numbers with <math>\overline{d_1d_2d_3} = \overline{d_4d_5d_6}</math> (of which there are <math>1000</math> possibilities for <math>\overline{d_1d_2d_3}</math> and <math>10</math> for <math>d_7</math>), and <math>B</math> those such that <math>\overline{d_1d_2d_3} = \overline{d_5d_6d_7}</math>. Then <math>A \cap B</math> (the telephone numbers that belong to both <math>A</math> and <math>B</math>) is the set of telephone numbers such that <math>d_1 = d_2 = d_3 = d_4 = d_5 = d_6 = d_7</math>, of which there are <math>10</math> possibilities. By the [[Principle of Inclusion-Exclusion]], | Let <math>A</math> represent the set of telephone numbers with <math>\overline{d_1d_2d_3} = \overline{d_4d_5d_6}</math> (of which there are <math>1000</math> possibilities for <math>\overline{d_1d_2d_3}</math> and <math>10</math> for <math>d_7</math>), and <math>B</math> those such that <math>\overline{d_1d_2d_3} = \overline{d_5d_6d_7}</math>. Then <math>A \cap B</math> (the telephone numbers that belong to both <math>A</math> and <math>B</math>) is the set of telephone numbers such that <math>d_1 = d_2 = d_3 = d_4 = d_5 = d_6 = d_7</math>, of which there are <math>10</math> possibilities. By the [[Principle of Inclusion-Exclusion]], | ||
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[[Category:Introductory Combinatorics Problems]] | [[Category:Introductory Combinatorics Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 10:14, 10 August 2016
Contents
Problem
Call a 
-digit telephone number 
memorable if the prefix sequence 
is exactly the same as either of the sequences 
or 
(possibly both). Assuming that each 
can be any of the ten decimal digits 
, the number of different memorable telephone numbers is
Solution A
In this problem, we only need to consider the digits 
. Each possibility of gives 
possibilities for 
, which are 
and 
with the exception of the case of 
, which only gives one sequence. After accounting for overcounting, the answer is 
Solution B
Let 
represent the set of telephone numbers with 
(of which there are 
possibilities for and 
for 
), and 
those such that 
. Then 
(the telephone numbers that belong to both and ) is the set of telephone numbers such that 
, of which there are possibilities. By the Principle of Inclusion-Exclusion,
![\[|A \cup B| = |A| + |B| - |A \cap B| = 1000 \times 10 + 1000 \times 10 - 10 = 19990 \Rightarrow \mathrm{(C)}\]](http://latex.artofproblemsolving.com/3/f/e/3febd99dc1853249b3c76b1f8546190e0d6a2b5f.png)
See also
| 1998 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 23 |
Followed by Problem 25 | |
| 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 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
