Difference between revisions of "2012 AMC 8 Problems/Problem 14"
(→Solution) |
(→Solution 2) |
||
(5 intermediate revisions by 2 users not shown) | |||
Line 4: | Line 4: | ||
<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+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== | ||
+ | (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 to 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. | ||
==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 19:55, 13 July 2021
Contents
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 shake hands with every other team, so we just need to find out how many consecutive numbers, , 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.
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.