Difference between revisions of "2023 AMC 12A Problems/Problem 13"
Albert1317 (talk | contribs) (→Solution 4) |
m (→Video Solution ⚡️ Under 3 min ⚡️) |
||
(25 intermediate revisions by 12 users not shown) | |||
Line 7: | Line 7: | ||
<math>\textbf{(A) }15\qquad\textbf{(B) }36\qquad\textbf{(C) }45\qquad\textbf{(D) }48\qquad\textbf{(E) }66</math> | <math>\textbf{(A) }15\qquad\textbf{(B) }36\qquad\textbf{(C) }45\qquad\textbf{(D) }48\qquad\textbf{(E) }66</math> | ||
− | ==Solution 1 | + | ==Solution 1== |
− | 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 have a product of 72 or 96. Clearly <math>72=8*9</math>, so the answer is <math>\boxed{\textbf{(B) }36}</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 have a product of 72 or 96. Clearly, <math>72=8*9</math>, so the answer is <math>\boxed{\textbf{(B) }36}</math>. |
~~ Antifreeze5420 | ~~ Antifreeze5420 | ||
Line 20: | Line 20: | ||
==Solution 4== | ==Solution 4== | ||
− | Here is the rigid way to prove that 36 is the only | + | Here is the rigid way to prove that <math>36</math> is the only answer. Let the number of left-handed players be <math>n</math>, so the number of right-handed players is <math>2n</math>. The number of games won by the left-handed players comes in two ways: |
+ | *The games played by two left-left pairs, which is <math>\frac{n(n-1)}{2}</math>, and | ||
+ | *The games played by left-right pairs, which we'll call <math>x</math>. | ||
+ | Note that <math>x\leq 2n^2,</math> which is the total number of games played by left-right pairs. Using the same logic for right-right pairs and right-left pairs, we have that <cmath>\frac{\frac{n(n-1)}{2}+x}{\frac{2n(2n-1)}{2}+2n^2-x}=1.4,</cmath> which gives <cmath>x=\frac{17n^2}{8}-\frac{3n}{8}.</cmath> | ||
+ | We know that <math>x\leq 2n^2</math>, applying that becomes | ||
+ | <cmath>\begin{align*} | ||
+ | \frac{17n^2}{8}-\frac{3n}{8} &\le 2n^2 \\ | ||
+ | 17n^2 - 3n &\le 16n^2 \\ | ||
+ | n^2 - 3n &\le 0 \\ | ||
+ | n &\le 3. | ||
+ | \end{align*}</cmath> | ||
+ | (We can safely divide by <math>n</math> because it must be positive). So the total number of players <math>p</math> can only be <math>3</math>, <math>6</math>, and <math>9</math>. | ||
+ | Since the total number of games <math>p(p-1)/2</math> is <math>(1.4+1 = 12/5)</math> times a non-negative integer number of games won by righties, <math>p(p-1)</math> must be a multiple of <math>12(2) = 24</math>. Among <math>\{3,6,9\}</math>, only <math>p = 9</math> satisfies this condition, so the total number of games is <math>(9)(8)/2 = \boxed{\textbf{(B)}\ 36}.</math> | ||
− | ~ ~ | + | ~ggao5uiuc, oinava, yingkai_0_ (minor edits) |
+ | |||
+ | ==Solution 5 (🧀Cheese🧀)== | ||
+ | If there are <math>n</math> players, the total number of games played must be <math>\binom{n}{2}</math>, so it has to be a triangular number. The ratio of games won by left-handed to right-handed players is <math>7:5</math>, so the number of games played must also be divisible by <math>12</math>. Finally, we notice that only <math>\boxed{\textbf{(B) }36}</math> satisfies both of these conditions. | ||
+ | |||
+ | ~MathFun1000 | ||
+ | |||
+ | == Video Solution by Power Solve == | ||
+ | https://youtu.be/2_ytwADrIto | ||
+ | |||
+ | ==Video Solution ⚡️ Under 2 min ⚡️ == | ||
+ | https://youtu.be/wSeSYxDCOZ8 | ||
+ | |||
+ | <i> ~Education, the Study of Everything </i> | ||
== Video Solution 1 by OmegaLearn == | == Video Solution 1 by OmegaLearn == | ||
https://youtu.be/BXgQIV2WbOA | https://youtu.be/BXgQIV2WbOA | ||
+ | |||
+ | == Video Solution by CosineMethod == | ||
+ | |||
+ | https://www.youtube.com/watch?v=N-eZMOv_ZjY | ||
== Video Solution 2 by TheBeautyofMath == | == Video Solution 2 by TheBeautyofMath == | ||
Line 36: | Line 65: | ||
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
+ | |||
+ | ==Video Solution by Math-X == | ||
+ | https://youtu.be/N2lyYRMuZuk?si=7IEmYDGBHi5i_XKt&t=1452 | ||
+ | ~Math-X | ||
+ | |||
+ | ==Video Solution by SpreadTheMathLove== | ||
+ | https://www.youtube.com/watch?v=sypOvNiR3sw | ||
==See Also== | ==See Also== |
Latest revision as of 14:31, 14 August 2024
- The following problem is from both the 2023 AMC 10A #16 and 2023 AMC 12A #13, so both problems redirect to this page.
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2
- 4 Solution 3
- 5 Solution 4
- 6 Solution 5 (🧀Cheese🧀)
- 7 Video Solution by Power Solve
- 8 Video Solution ⚡️ Under 2 min ⚡️
- 9 Video Solution 1 by OmegaLearn
- 10 Video Solution by CosineMethod
- 11 Video Solution 2 by TheBeautyofMath
- 12 Video Solution
- 13 Video Solution by Math-X
- 14 Video Solution by SpreadTheMathLove
- 15 See Also
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
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 have a product of 72 or 96. Clearly,
, so the answer is
.
~~ Antifreeze5420
Solution 2
First, we know that every player played every other player, so there's a total of games since each pair of players forms a bijection to a game. Therefore, that rules out D. Also, if we assume the right-handed players won a total of
games, the left-handed players must have won a total of
games, meaning that the total number of games played was
. Thus, the total number of games must be divisible by
. Therefore leaving only answer choices B and D. Since answer choice D doesn't satisfy the first condition, the only answer that satisfies both conditions is
Solution 3
Let be the amount of games the right-handed won. Since the left-handed won
games, the total number of games played can be expressed as
, or
, meaning that the answer is divisible by 12. This brings us down to two answer choices,
and
.
We note that the answer is some number
choose
. This means the answer is in the form
. Since answer choice D gives
, and
has no integer solutions, we know that
is the only possible choice.
Solution 4
Here is the rigid way to prove that is the only answer. Let the number of left-handed players be
, so the number of right-handed players is
. The number of games won by the left-handed players comes in two ways:
- The games played by two left-left pairs, which is
, and
- The games played by left-right pairs, which we'll call
.
Note that which is the total number of games played by left-right pairs. Using the same logic for right-right pairs and right-left pairs, we have that
which gives
We know that
, applying that becomes
(We can safely divide by
because it must be positive). So the total number of players
can only be
,
, and
.
Since the total number of games is
times a non-negative integer number of games won by righties,
must be a multiple of
. Among
, only
satisfies this condition, so the total number of games is
~ggao5uiuc, oinava, yingkai_0_ (minor edits)
Solution 5 (🧀Cheese🧀)
If there are players, the total number of games played must be
, so it has to be a triangular number. The ratio of games won by left-handed to right-handed players is
, so the number of games played must also be divisible by
. Finally, we notice that only
satisfies both of these conditions.
~MathFun1000
Video Solution by Power Solve
Video Solution ⚡️ Under 2 min ⚡️
~Education, the Study of Everything
Video Solution 1 by OmegaLearn
Video Solution by CosineMethod
https://www.youtube.com/watch?v=N-eZMOv_ZjY
Video Solution 2 by TheBeautyofMath
https://www.youtube.com/watch?v=sLtsF1k9Fx8&t=227s
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution by Math-X
https://youtu.be/N2lyYRMuZuk?si=7IEmYDGBHi5i_XKt&t=1452 ~Math-X
Video Solution by SpreadTheMathLove
https://www.youtube.com/watch?v=sypOvNiR3sw
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 |
2023 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 12 |
Followed by Problem 14 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.