Difference between revisions of "2012 AMC 8 Problems/Problem 14"

Latest revision as of 12:39, 26 August 2025

Problem

In the BIG N, a middle school football conference, each team plays every other team exactly once. If a total of 21 conference games were played during the 2012 season, how many teams were members of the BIG N conference?

$\textbf{(A)}\hspace{.05in}6\qquad\textbf{(B)}\hspace{.05in}7\qquad\textbf{(C)}\hspace{.05in}8\qquad\textbf{(D)}\hspace{.05in}9\qquad\textbf{(E)}\hspace{.05in}10$

Solution 1

This problem is very similar to a handshake problem. We use the formula $\frac{n(n-1)}{2}$ to usually find the number of games played (or handshakes). Now we have to use the formula in reverse.

So we have the equation $\frac{n(n-1)}{2} = 21$ . Solving, we find that the number of teams in the BIG N conference is $\boxed{\textbf{(B)}\ 7}$ .

Solution 2

(If someone understands what I'm trying to do here and can explain it better, please edit it)We know that every team has to play a game with every other team, so we just need to find out how many consecutive numbers, $1$ to $x$ , can fit into 21. We know that $6+5+4+3+2+1=21$ , and since this doesn't count to $7th$ team that shook hands with the other $6$ , we know that there are $\boxed{\textbf{(B)}\ 7}$ teams in the BIG N conference.

EDIT: So basically reiterating this, we are using a factorial and adding one because the team that they are talking about is not shaking the hands. Also as the other person said, we need to see how many consecutive numbers can add up to 21.

Video Solution

https://youtu.be/zzU98Bk1TrE ~savannahsolver

2012 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.

@@ Line 5: / Line 5: @@
 ==Solution 1==
-This problem is very similar to a handshake problem. We use the formula <math> \frac{n(n+1)}{2} </math> to usually find the number of games played (or handshakes). Now we have to use the formula in reverse.
+This problem is very similar to a handshake problem. We use the formula <math> \frac{n(n-1)}{2} </math> to usually find the number of games played (or handshakes). Now we have to use the formula in reverse.
 So we have the equation  <math> \frac{n(n-1)}{2} = 21 </math>. Solving, we find that the number of teams in the BIG N conference is  <math> \boxed{\textbf{(B)}\ 7} </math>.
 ==Solution 2==
-(if someone understands what I'm trying to do here and can explain it better, please edit it)We know that every team has to shake hands with every other team, so we just need to find out how many consecutive numbers, <math>1 to x</math>, can fit into 21. We know that <math>6+5+4+3+2+1=21</math>, and since this doesn't count to <math>7th</math> team that shook hands with the other <math>6</math>,we know that there are <math> \boxed{\textbf{(B)}\ 7} </math> teams in the BIG N conference.
+(If someone understands what I'm trying to do here and can explain it better, please edit it)We know that every team has to play a game with every other team, so we just need to find out how many consecutive numbers, <math>1</math> to <math>x</math>, can fit into 21. We know that <math>6+5+4+3+2+1=21</math>, and since this doesn't count to <math>7th</math> team that shook hands with the other <math>6</math>, we know that there are <math> \boxed{\textbf{(B)}\ 7} </math> teams in the BIG N conference.
+EDIT: So basically reiterating this, we are using a factorial and adding one because the team that they are talking about is not shaking the hands. Also as the other person said, we need to see how many consecutive numbers can add up to 21.
+==Video Solution==
+https://youtu.be/zzU98Bk1TrE ~savannahsolver
 ==See Also==
 {{AMC8 box|year=2012|num-b=13|num-a=15}}
 {{MAA Notice}}

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