Difference between revisions of "2023 AMC 10A Problems/Problem 16"
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==Solution 1 (3 min solve)== | ==Solution 1 (3 min solve)== | ||
| − | We know that the total amount of games must be the sum of games won by left and right handed players. Then, we can write <math>g = l + r</math>, and since <math>l = 1.4r</math>, <math>g = 2.4r</math>. Given that <math>r</math> and <math>g</math> are both integers, <math>g/2.4</math> also must be an integer. From here we can see that <math>g</math> must be divisible by 12, leaving only answers B and D. Now we know the formula for how many games are played in this tournament is <math>n(n-1)/2</math>, the sum of the first <math>n-1</math> triangular numbers. Now setting 36 and 48 equal to the equation will show that two consecutive numbers must equal 72 or 96. Clearly <math>72=8*9</math>, so the answer is <math>\boxed{(B)}</math>. | + | We know that the total amount of games must be the sum of games won by left and right handed players. Then, we can write <math>g = l + r</math>, and since <math>l = 1.4r</math>, <math>g = 2.4r</math>. Given that <math>r</math> and <math>g</math> are both integers, <math>g/2.4</math> also must be an integer. From here we can see that <math>g</math> must be divisible by 12, leaving only answers B and D. Now we know the formula for how many games are played in this tournament is <math>n(n-1)/2</math>, the sum of the first <math>n-1</math> triangular numbers. Now setting 36 and 48 equal to the equation will show that two consecutive numbers must equal 72 or 96. Clearly <math>72=8*9</math>, so the answer is <math>\boxed{\textbf{(B) }36}</math>. |
~~ Antifreeze5420 | ~~ Antifreeze5420 | ||
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First, every player played the other, so there's <math>n\choose2</math> games. Also, if the right-handed won <math>x</math> games, the left handed won <math>7/5x</math>, meaning that the total amount of games was <math>12/5x</math>, so the total amount of games is divisible by <math>12</math>. | First, every player played the other, so there's <math>n\choose2</math> games. Also, if the right-handed won <math>x</math> games, the left handed won <math>7/5x</math>, meaning that the total amount of games was <math>12/5x</math>, so the total amount of games is divisible by <math>12</math>. | ||
| − | Then, we do something funny and look at the answer choices. Only <math>\boxed{B}</math> satisfies our 2 findings. | + | Then, we do something funny and look at the answer choices. Only <math>\boxed{\textbf{(B) }36}</math> satisfies our 2 findings. |
== Video Solution 1 by OmegaLearn == | == Video Solution 1 by OmegaLearn == | ||
Revision as of 21:18, 9 November 2023
Contents
Problem
In a table tennis tournament every participant played every other participant exactly once. Although there were twice as many right-handed players as left-handed players, the number of games won by left-handed players was 
more than the number of games won by right-handed players. (There were no ties and no ambidextrous players.) What is the total number of games played?
Solution 1 (3 min solve)
We know that the total amount of games must be the sum of games won by left and right handed players. Then, we can write 
, and since 
, 
. Given that 
and 
are both integers, 
also must be an integer. From here we can see that must be divisible by 12, leaving only answers B and D. Now we know the formula for how many games are played in this tournament is 
, the sum of the first 
triangular numbers. Now setting 36 and 48 equal to the equation will show that two consecutive numbers must equal 72 or 96. Clearly 
, so the answer is 
.
~~ Antifreeze5420
Solution 2
First, every player played the other, so there's 
games. Also, if the right-handed won 
games, the left handed won 
, meaning that the total amount of games was 
, so the total amount of games is divisible by 
.
Then, we do something funny and look at the answer choices. Only satisfies our 2 findings.
Video Solution 1 by OmegaLearn
See Also
| 2023 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 15 |
Followed by Problem 17 | |
| 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 AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
