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 == |
+ | |||
+ | 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> | ||
+ | |||
+ | == 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]], | ||
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,
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 |
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