Difference between revisions of "2012 AMC 8 Problems/Problem 14"
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<math> \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 </math> | <math> \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 </math> | ||
| − | ==Solution== | + | ==Solution 1== |
| − | This problem is very similar to a handshake problem. We use the formula <math> \frac{n(n | + | 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>. | 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 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. | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/zzU98Bk1TrE ~savannahsolver | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2012|num-b=13|num-a=15}} | {{AMC8 box|year=2012|num-b=13|num-a=15}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 23:45, 12 January 2023
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?
Solution 1
This problem is very similar to a handshake problem. We use the formula 
to usually find the number of games played (or handshakes). Now we have to use the formula in reverse.
So we have the equation 
. Solving, we find that the number of teams in the BIG N conference is 
.
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, 
to 
, can fit into 21. We know that 
, and since this doesn't count to 
team that shook hands with the other 
, we know that there are teams in the BIG N conference.
Video Solution
https://youtu.be/zzU98Bk1TrE ~savannahsolver
See Also
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
