Difference between revisions of "2015 AMC 8 Problems/Problem 24"
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== Problem == | == Problem == | ||
− | A baseball league consists of two four-team divisions. Each team plays every other team in its division <math>N</math> games. Each team plays every team in the other division <math>M</math> games with <math>N>2M</math> and <math>M>4</math>. Each team plays a 76 game schedule. How many games does a team play within its own division? | + | A baseball league consists of two four-team divisions. Each team plays every other team in its division <math>N</math> games. Each team plays every team in the other division <math>M</math> games with <math>N>2M</math> and <math>M>4</math>. Each team plays a <math>76</math> game schedule. How many games does a team play within its own division? |
<math>\textbf{(A) } 36 \qquad \textbf{(B) } 48 \qquad \textbf{(C) } 54 \qquad \textbf{(D) } 60 \qquad \textbf{(E) } 72 </math> | <math>\textbf{(A) } 36 \qquad \textbf{(B) } 48 \qquad \textbf{(C) } 54 \qquad \textbf{(D) } 60 \qquad \textbf{(E) } 72 </math> | ||
− | == | + | ==Solution 1== |
+ | On one team they play <math>3N</math> games in their division and <math>4M</math> games in the other. This gives <math>3N+4M=76</math>. | ||
− | + | Since <math>M>4</math> we start by trying M=5. This doesn't work because <math>56</math> is not divisible by <math>3</math>. | |
− | |||
− | + | Next, <math>M=6</math> does not work because <math>52</math> is not divisible by <math>3</math>. | |
− | + | We try <math>M=7</math> does work by giving <math>N=16</math> ,<math>~M=7</math> and thus <math>3\times 16=\boxed{\textbf{(B)}~48}</math> games in their division. | |
− | |||
− | |||
<math>M=10</math> seems to work, until we realize this gives <math>N=12</math>, but <math>N>2M</math> so this will not work. | <math>M=10</math> seems to work, until we realize this gives <math>N=12</math>, but <math>N>2M</math> so this will not work. | ||
− | + | ==Solution 2== | |
<math>76=3N+4M > 10M</math>, giving <math>M \le 7</math>. | <math>76=3N+4M > 10M</math>, giving <math>M \le 7</math>. | ||
Since <math>M>4</math>, we have <math>M=5,6,7</math>. | Since <math>M>4</math>, we have <math>M=5,6,7</math>. | ||
Since <math>4M</math> is <math>1</math> <math>\pmod{3}</math>, we must have <math>M</math> equal to <math>1</math> <math>\pmod{3}</math>, so <math>M=7</math>. | Since <math>4M</math> is <math>1</math> <math>\pmod{3}</math>, we must have <math>M</math> equal to <math>1</math> <math>\pmod{3}</math>, so <math>M=7</math>. | ||
− | This gives <math>3N=48</math>, as desired. The answer is <math>\boxed{\textbf{(B)}~48}</math> | + | This gives <math>3N=48</math>, as desired. The answer is <math>\boxed{\textbf{(B)}~48}</math>. |
+ | |||
+ | ==Solution 3== | ||
+ | Notice that each team plays <math>N</math> games against each of the three other teams in its division. So that's <math>3N</math>. | ||
+ | |||
+ | Since each team plays <math>M</math> games against each of the four other teams in the other division, that's <math>4M</math>. | ||
+ | |||
+ | So <math>3N+4M=76</math>, with <math>M>4, N>2M</math>. | ||
+ | |||
+ | Let's start by solving this Diophantine equation. In other words, <math>N=\frac{76-4M}{3}</math>. | ||
+ | |||
+ | So <math>76-4M\equiv0 \pmod{3}</math> (remember: <math>M</math> must be divisible by 3 for <math>N</math> to be an integer!). Therefore, after reducing <math>76</math> to <math>1</math> and <math>-4M</math> to <math>2M</math> (we are doing things in <math>\pmod{3}</math>), we find that <math>M\equiv1 \pmod{3}</math>. | ||
+ | |||
+ | Since <math>M>4</math>, so the minimum possible value of <math>M</math> is <math>7</math>. However, remember that <math>N>2M</math>! To find the greatest possible value of M, we assume that <math>N=2M</math> and that is the upper limit of <math>M</math> (excluding that value because <math>N>2M</math>). Plugging <math>N=2M</math> in, <math>10M=76</math>. So <math>M<7.6</math>. Since you can't have <math>7.6</math> games, we know that we can only check <math>M=7</math> since we know that since <math>M>4, M<7.6, M\equiv1 \pmod{3}</math>. After checking <math>M=7</math>, we find that it works. | ||
+ | |||
+ | So <math>M=7, N=16</math>. So each team plays 16 games against each team in its division. Since they are asking for games in it division, which equals <math>3n = 48</math>. Select <math>\boxed{\textbf{(B)}~48}</math>. | ||
+ | |||
+ | This might be too complicated. But you should know what's happening if you read the ''The Art of Problem Solving: Introduction to Number Theory'' by Mathew Crawford. Notice how I used chapter 12's ideas of basic modular arithmetic operations and chapter 14's ideas of solving linear congruences. Remember: the Introduction Series books by AoPS are for 6th-10th graders! So make sure to read the curriculum books! | ||
+ | |||
+ | This goes same for high school students as well, and especially for those who want to continue on their path of AIME/USA(J)MO. | ||
+ | |||
+ | ~hastapasta | ||
+ | |||
+ | ===Video Solution by OmegaLearn=== | ||
+ | https://youtu.be/HISL2-N5NVg?t=4968 | ||
===Video Solutions=== | ===Video Solutions=== | ||
Line 29: | Line 51: | ||
https://www.youtube.com/watch?v=bJSWtw91SLs - Oliver Jiang | https://www.youtube.com/watch?v=bJSWtw91SLs - Oliver Jiang | ||
+ | |||
+ | https://youtu.be/0ZMDsuOYGqE | ||
+ | |||
+ | ~savannahsolver | ||
==See Also== | ==See Also== |
Revision as of 20:41, 17 January 2024
Contents
Problem
A baseball league consists of two four-team divisions. Each team plays every other team in its division games. Each team plays every team in the other division
games with
and
. Each team plays a
game schedule. How many games does a team play within its own division?
Solution 1
On one team they play games in their division and
games in the other. This gives
.
Since we start by trying M=5. This doesn't work because
is not divisible by
.
Next, does not work because
is not divisible by
.
We try does work by giving
,
and thus
games in their division.
seems to work, until we realize this gives
, but
so this will not work.
Solution 2
, giving
.
Since
, we have
.
Since
is
, we must have
equal to
, so
.
This gives , as desired. The answer is
.
Solution 3
Notice that each team plays games against each of the three other teams in its division. So that's
.
Since each team plays games against each of the four other teams in the other division, that's
.
So , with
.
Let's start by solving this Diophantine equation. In other words, .
So (remember:
must be divisible by 3 for
to be an integer!). Therefore, after reducing
to
and
to
(we are doing things in
), we find that
.
Since , so the minimum possible value of
is
. However, remember that
! To find the greatest possible value of M, we assume that
and that is the upper limit of
(excluding that value because
). Plugging
in,
. So
. Since you can't have
games, we know that we can only check
since we know that since
. After checking
, we find that it works.
So . So each team plays 16 games against each team in its division. Since they are asking for games in it division, which equals
. Select
.
This might be too complicated. But you should know what's happening if you read the The Art of Problem Solving: Introduction to Number Theory by Mathew Crawford. Notice how I used chapter 12's ideas of basic modular arithmetic operations and chapter 14's ideas of solving linear congruences. Remember: the Introduction Series books by AoPS are for 6th-10th graders! So make sure to read the curriculum books!
This goes same for high school students as well, and especially for those who want to continue on their path of AIME/USA(J)MO.
~hastapasta
Video Solution by OmegaLearn
https://youtu.be/HISL2-N5NVg?t=4968
Video Solutions
https://youtu.be/LiAupwDF0EY - Happytwin
https://www.youtube.com/watch?v=bJSWtw91SLs - Oliver Jiang
~savannahsolver
See Also
2015 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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.