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 shake hands 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