Difference between revisions of "2023 AMC 8 Problems/Problem 8"

Revision as of 19:45, 20 February 2023

Problem

Lola, Lolo, Tiya, and Tiyo participated in a ping pong tournament. Each player competed against each of the other three players exactly twice. Shown below are the win-loss records for the players. The numbers $1$ and $0$ represent a win or loss, respectively. For example, Lola won five matches and lost the fourth match. What was Tiyo’s win-loss record?

$\begin{tabular}{c | c} Player & Result \\ \hline Lola & \texttt{111011}\\ Lolo & \texttt{101010}\\ Tiya & \texttt{010100}\\ Tiyo & \texttt{??????} \end{tabular}$ $\textbf{(A)}\ \texttt{000101} \qquad \textbf{(B)}\ \texttt{001001} \qquad \textbf{(C)}\ \texttt{010000} \qquad \textbf{(D)}\ \texttt{010101} \qquad \textbf{(E)}\ \texttt{011000}$

Solution 1

We can calculate the total number of wins ( $1$ 's) by seeing how many matches were players, which is $12$ matches played. Then, we can calculate the # of wins already on the table, which is $5 + 3 + 2 = 10$ , so there are $12 - 10 = 2$ wins left in the mystery player. Now, we will make the key observation that there is only $2$ wins ( $1$ 's) per column as there are $2$ winners and $2$ losers in each round. Strategically looking through the columns counting the $1$ 's and putting our own $2$ $1$ 's when the column isn't already full yields $\boxed{\textbf{(A)}\ \texttt{000101}}$ .

~SohumUttamchandani -edits apex304, lpieleanu

Solution 2 (Similar to #1 but More Detailed)

In total, there will be $\binom{4}{2} \cdot 2 = 6 \cdot 2 = 12$ games because there are $\binom{4}{2}$ ways to choose a pair of people from the four players. And, each player will play each other player exactly twice. Each of these $12$ games will have $1$ winner and $1$ loser, so there will be a total of $12$ $1$ 's and $12$ $0$ 's in the win-loss table. Therefore, Tiyo will have $12-10=2$ $1$ 's and $12-8=4$ $0$ 's in his record.

Now, all we have to do is figure out the order of these $1$ 's and $0$ 's. In every round, there are two games; the players are split into two pairs and the people in those pairs play each other. Thus, every round should have $2$ winners and $2$ losers which means that every column of the win-loss table should have $2$ $1$ 's and $2$ $0$ 's. Looking at the filled-in table so far, we see that columns $4$ and $6$ need one more $1$ , so Tiyo must have $1$ 's in those columns and $0$ 's gone in the others. Therefore, our answer is $\boxed{\textbf{(A)}\ \texttt{000101}}.$

~lpieleanu

Remark: Note that we can skip straight to the reasoning used in the second paragraph (which is Solution 3 below).

Solution 3 (Faster)

We can look one by one. We see that Lola and Lolo won the first game and Tiya lost. This shows that Tiyo must have lost as well because the results must be $2$ wins and $2$ loses. We use the same logic for games $2$ and $3$ , giving us $0$ 's again. We look at the choices, and we see A is the only one that has $3$ $0$ 's

This shows our answer is $\boxed{\textbf{(A)}\ \texttt{000101}}.$

~stevens0209

Video Solution by Magic Square

https://youtu.be/-N46BeEKaCQ?t=5016

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=EcrktBc8zrM

Video Solution by Interstigation

https://youtu.be/1bA7fD7Lg54?t=497

(Creative Thinking) Video Solution

https://youtu.be/p8NciAa6_d8

~Education the Study of everything

@@ Line 36: / Line 36: @@
 ==Video Solution by Interstigation==
 https://youtu.be/1bA7fD7Lg54?t=497
+==(Creative Thinking) Video Solution==
+https://youtu.be/p8NciAa6_d8
+~Education the Study of everything
 ==See Also==
 {{AMC8 box|year=2023|num-b=7|num-a=9}}
 {{MAA Notice}}

2023 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
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

Page

Toolbox

Search