Difference between revisions of "2003 AMC 8 Problems/Problem 18"
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==Solution== | ==Solution== | ||
There are <math>3</math> people who are friends with only each other who won't be invited, plus <math>1</math> person who has no friends, and <math>2</math> people who are friends of friends of friends who won't be invited. So the answer is <math>\boxed{\textbf{(D)}\ 6}</math>. | There are <math>3</math> people who are friends with only each other who won't be invited, plus <math>1</math> person who has no friends, and <math>2</math> people who are friends of friends of friends who won't be invited. So the answer is <math>\boxed{\textbf{(D)}\ 6}</math>. | ||
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+ | ==Video Solution== | ||
+ | https://www.youtube.com/watch?v=TBncumM5bFQ | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2003|num-b=17|num-a=19}} | {{AMC8 box|year=2003|num-b=17|num-a=19}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 21:20, 4 November 2022
Contents
Problem
Each of the twenty dots on the graph below represents one of Sarah's classmates. Classmates who are friends are connected with a line segment. For her birthday party, Sarah is inviting only the following: all of her friends and all of those classmates who are friends with at least one of her friends. How many classmates will not be invited to Sarah's party?
Solution
There are people who are friends with only each other who won't be invited, plus person who has no friends, and people who are friends of friends of friends who won't be invited. So the answer is .
Video Solution
https://www.youtube.com/watch?v=TBncumM5bFQ
See Also
2003 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
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.