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

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(→‎Solution 1: Handshaking problem formula should be n(n-1)/2, rather than n(n+1)/2.)
 
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==Solution 1==
 
==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>.
 
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==
 
==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</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.
+
(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?

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

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

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

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