2012 AMC 8 Problems/Problem 14

Revision as of 23:45, 12 January 2023 by Craftyowl73 1 (talk | contribs) (Solution 1: Handshaking problem formula should be n(n-1)/2, rather than n(n+1)/2.)
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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?


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

See Also

2012 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
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All AJHSME/AMC 8 Problems and Solutions

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