Difference between revisions of "2016 AMC 10B Problems/Problem 22"
m (→Solution 2 (Cheap Solution)) |
m |
||
| Line 24: | Line 24: | ||
Speaking of cheap solutions, you can let the teams be <math>1,2,3,..21</math> and have <math>i</math> beat <math>i+1,i+2, ... i+10 \pmod{21}</math> and lose to <math>i+11,i+12, ... i+20 \pmod{21}.</math> You can choose any random team as <math>A</math> for <math>21</math> ways but overcount since set, so divide by <math>3</math> so multiply by <math>7.</math> Now we can proceed w/casework (which isn't too hard). Eventually you realize the sum is <math>7(1+2+\cdots+10) = \boxed{\textbf{(A)}}</math>. | Speaking of cheap solutions, you can let the teams be <math>1,2,3,..21</math> and have <math>i</math> beat <math>i+1,i+2, ... i+10 \pmod{21}</math> and lose to <math>i+11,i+12, ... i+20 \pmod{21}.</math> You can choose any random team as <math>A</math> for <math>21</math> ways but overcount since set, so divide by <math>3</math> so multiply by <math>7.</math> Now we can proceed w/casework (which isn't too hard). Eventually you realize the sum is <math>7(1+2+\cdots+10) = \boxed{\textbf{(A)}}</math>. | ||
| − | + | [[Category:Introductory_Combinatorics_Problems]] | |
==See Also== | ==See Also== | ||
{{AMC10 box|year=2016|ab=B|num-b=21|num-a=23}} | {{AMC10 box|year=2016|ab=B|num-b=21|num-a=23}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 16:27, 6 April 2021
Problem
A set of teams held a round-robin tournament in which every team played every other team exactly once. Every team won 
games and lost games; there were no ties. How many sets of three teams 
were there in which 
beat 
, beat 
, and beat 
Solution 1
There are 
teams. Any of the 
sets of three teams must either be a fork (in which one team beat both the others) or a cycle:
![[asy]size(7cm);label("X",(5,5));label("Z",(10,0));label("Y",(0,0));draw((4,4)--(1,1),EndArrow);draw((6,4)--(9,1),EndArrow); label("X",(20,5));label("Z",(25,0));label("Y",(15,0));draw((19,4)--(16,1),EndArrow);draw((16,0)--(24,0),EndArrow);draw((24,1)--(21,4),EndArrow); [/asy]](http://latex.artofproblemsolving.com/c/f/b/cfb866b5865b62be078af1b1915ffc5fed69bab6.png)
But we know that every team beat exactly other teams, so for each possible 
at the head of a fork, there are always exactly 
choices for 
and 
as A beat exactly 10 teams and we are choosing 2 of them. Therefore there are 
forks, and all the rest must be cycles.
Thus the answer is 
which is 
.
Solution 2 (Cheap Solution)
Since there are teams and for each set of three teams there is a cycle, there are a total of cycles of three teams. Because about 
of the cycles satisfy the conditions of the problems, our answer is close to 
. Looking at the answer choices, we find that 
is closer to 
than any other answer choices, and that the next closest is 
which is twice of , so our answer is which is .
Speaking of cheap solutions, you can let the teams be 
and have 
beat 
and lose to 
You can choose any random team as for ways but overcount since set, so divide by 
so multiply by 
Now we can proceed w/casework (which isn't too hard). Eventually you realize the sum is 
.
See Also
| 2016 AMC 10B (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 AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
