2012 AMC 8 Problems/Problem 14

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.

Video Solution

https://youtu.be/zzU98Bk1TrE ~savannahsolver

2012 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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2012 AMC 8 Problems/Problem 14

Contents

Problem

Solution 1

Solution 2

Video Solution

See Also