Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2022 AMC 10A Problems/Problem 14"
(→‎Solution 1 (Multiplication Principle))
 
(43 intermediate revisions by 16 users not shown)
Line 1: Line 1:
 +
{{duplicate|[[2022 AMC 10A Problems/Problem 14|2022 AMC 10A #14]] and [[2022 AMC 12A Problems/Problem 10|2022 AMC 12A #10]]}}
 +
 
==Problem==
 
==Problem==
  
Line 5: Line 7:
 
<math>\textbf{(A) } 108 \qquad \textbf{(B) } 120 \qquad \textbf{(C) } 126 \qquad \textbf{(D) } 132 \qquad \textbf{(E) } 144</math>
 
<math>\textbf{(A) } 108 \qquad \textbf{(B) } 120 \qquad \textbf{(C) } 126 \qquad \textbf{(D) } 132 \qquad \textbf{(E) } 144</math>
  
−
==Solution==
+
==Solution 1 (Multiplication Principle)==
  
−
Clearly, the integers from <math>8</math> through <math>14</math> must be in different pairs, and <math>7</math> must pair with <math>14.</math>
+
Clearly, the integers from <math>8</math> through <math>14</math> must be in different pairs, so are the integers from <math>1</math> through <math>7.</math> Note that <math>7</math> must pair with <math>14.</math>
  
−
Note that <math>6</math> can pair with either <math>12</math> or <math>13.</math> From here, we consider casework:
+
We pair the numbers <math>1,2,3,4,5,6</math> with the numbers <math>8,9,10,11,12,13</math> systematically:
  
−
* If <math>6</math> pairs with <math>12,</math> then <math>5</math> can pair with one of <math>10,11,13.</math> After that, each of <math>1,2,3,4</math> does not have any restrictions. This case produces <math>3\cdot4!=72</math> ways.
+
* <math>6</math> can pair with either <math>12</math> or <math>13.</math>
  
−
* If <math>6</math> pairs with <math>13,</math> then <math>5</math> can pair with one of <math>10,11,12.</math> After that, each of <math>1,2,3,4</math> does not have any restrictions. This case produces <math>3\cdot4!=72</math> ways.
+
* <math>5</math> can pair with any of the three remaining numbers from <math>10,11,12,13.</math>
  
−
Together, the answer is <math>72+72=\boxed{\textbf{(E) } 144}.</math>
+
* <math>1,2,3,4</math> can pair with the other four remaining numbers from <math>8,9,10,11,12,13</math> without restrictions.
 +
 
 +
Together, the answer is <math>2\cdot3\cdot4!=\boxed{\textbf{(E) } 144}.</math>
  
 
~MRENTHUSIASM
 
~MRENTHUSIASM
  
−
== Video Solution by OmegaLearn ==
+
==Solution 2 (Multiplication Principle)==
 +
 
 +
As said in Solution 1, clearly, the integers from <math>8</math> through <math>14</math> must be in different pairs.
 +
 
 +
We know that <math>8</math> or <math>9</math> can pair with any integer from <math>1</math> to <math>4</math>, <math>10</math> or <math>11</math> can pair with any integer from <math>1</math> to <math>5</math>, and <math>12</math> or <math>13</math> can pair with any integer from <math>1</math> to <math>6</math>. Thus, <math>8</math> will have <math>4</math> choices to pair with, <math>9</math> will then have <math>3</math> choices to pair with (<math>9</math> cannot pair with the same number as the one <math>8</math> pairs with). <math>10</math> cannot pair with the numbers <math>8</math> and <math>9</math> has paired with but can also now pair with <math>5</math>, so there are <math>3</math> choices. <math>11</math> cannot pair with <math>8</math>'s, <math>9</math>'s, or <math>10</math>'s paired numbers, so there will be <math>2</math> choices for <math>11</math>. <math>12</math> can pair with an integer from <math>1</math> to <math>5</math> that hasn't been paired with already, or it can pair with <math>6</math>. <math>13</math> will only have one choice left, and <math>7</math> must pair with <math>14</math>.
 +
 
 +
So, the answer is <math>4\cdot3\cdot3\cdot2\cdot2\cdot1\cdot1=\boxed{\textbf{(E) } 144}.</math>
 +
 
 +
~Scarletsyc
 +
 
 +
==Solution 3 (Generalization)==
 +
 
 +
The integers <math>x \in \{8, \ldots , 14 \}</math> must each be the larger elements of a distinct pair.
 +
 
 +
Assign partners in decreasing order for <math>x \in \{7, \dots, 1\}</math>:
 +
 
 +
Note that <math>7</math> must pair with <math>14</math>: <math>\mathbf{1} \textbf{ choice}</math>.
 +
 
 +
For <math>5 \leq x \leq 7</math>, the choices are <math>\{2x, \dots, 14\} - \{ \text{previous choices}\}</math>. As <math>x</math> decreases by 1, The minuend increases by 2 elements, and the subtrahend increases by 1 element, so the difference increases by 1, yielding <math>\mathbf{3!} \textbf{ combined choices}</math>.
 +
 
 +
After assigning a partner to <math>5</math>, there are no invalid pairings for yet-unpaired numbers, so there are <math>\mathbf{4!} \textbf{ ways}</math> to choose partners for <math>\{1,2,3,4\}</math>.
 +
 
 +
The answer is <math>3! \cdot 4! = \boxed{\textbf{(E) } 144}</math>.
 +
 
 +
In general, for <math>1,\ldots,2n</math>, the same logic yields answer: <math>\left\lfloor\dfrac{n}{2}\right\rfloor! \cdot \left\lceil\dfrac{n}{2}\right\rceil!</math>
 +
 
 +
~oinava
 +
 
 +
==Video Solution by Education, the Study of Everything==
 +
https://youtu.be/k6EUl65wS9Q
 +
 
 +
== Video Solution by Sohil Rathi==
 
https://youtu.be/V1jOj8ysd_w
 
https://youtu.be/V1jOj8ysd_w
  
 
~ pi_is_3.14
 
~ pi_is_3.14
 +
 +
==Video Solution (Smart and Simple)==
 +
https://youtu.be/7yAh4MtJ8a8?si=jyIdy-jZb2raj3cM&t=1800
 +
 +
~Math-X
 +
 +
==Video Solution (RMM club)==
 +
https://youtu.be/DwCE1wu5hrA
 +
 +
== Video Solution by Lucas637 (Fast and Easy) ==
 +
https://www.youtube.com/watch?v=egQK11g54mA
 +
==Video Solution by TheBeautyofMath==
 +
https://youtu.be/0kkc4-y8TkU?t=1367
 +
 +
~IceMatrix
  
 
== See Also ==
 
== See Also ==

Latest revision as of 23:52, 23 October 2024

The following problem is from both the 2022 AMC 10A #14 and 2022 AMC 12A #10, so both problems redirect to this page.

Problem

How many ways are there to split the integers $1$ through $14$ into $7$ pairs such that in each pair, the greater number is at least $2$ times the lesser number?

$\textbf{(A) } 108 \qquad \textbf{(B) } 120 \qquad \textbf{(C) } 126 \qquad \textbf{(D) } 132 \qquad \textbf{(E) } 144$

Solution 1 (Multiplication Principle)

Clearly, the integers from $8$ through $14$ must be in different pairs, so are the integers from $1$ through $7.$ Note that $7$ must pair with $14.$

We pair the numbers $1,2,3,4,5,6$ with the numbers $8,9,10,11,12,13$ systematically:

  • $6$ can pair with either $12$ or $13.$
  • $5$ can pair with any of the three remaining numbers from $10,11,12,13.$
  • $1,2,3,4$ can pair with the other four remaining numbers from $8,9,10,11,12,13$ without restrictions.

Together, the answer is $2\cdot3\cdot4!=\boxed{\textbf{(E) } 144}.$

~MRENTHUSIASM

Solution 2 (Multiplication Principle)

As said in Solution 1, clearly, the integers from $8$ through $14$ must be in different pairs.

We know that $8$ or $9$ can pair with any integer from $1$ to $4$, $10$ or $11$ can pair with any integer from $1$ to $5$, and $12$ or $13$ can pair with any integer from $1$ to $6$. Thus, $8$ will have $4$ choices to pair with, $9$ will then have $3$ choices to pair with ($9$ cannot pair with the same number as the one $8$ pairs with). $10$ cannot pair with the numbers $8$ and $9$ has paired with but can also now pair with $5$, so there are $3$ choices. $11$ cannot pair with $8$'s, $9$'s, or $10$'s paired numbers, so there will be $2$ choices for $11$. $12$ can pair with an integer from $1$ to $5$ that hasn't been paired with already, or it can pair with $6$. $13$ will only have one choice left, and $7$ must pair with $14$.

So, the answer is $4\cdot3\cdot3\cdot2\cdot2\cdot1\cdot1=\boxed{\textbf{(E) } 144}.$

~Scarletsyc

Solution 3 (Generalization)

The integers $x \in \{8, \ldots , 14 \}$ must each be the larger elements of a distinct pair.

Assign partners in decreasing order for $x \in \{7, \dots, 1\}$:

Note that $7$ must pair with $14$: $\mathbf{1} \textbf{ choice}$.

For $5 \leq x \leq 7$, the choices are $\{2x, \dots, 14\} - \{ \text{previous choices}\}$. As $x$ decreases by 1, The minuend increases by 2 elements, and the subtrahend increases by 1 element, so the difference increases by 1, yielding $\mathbf{3!} \textbf{ combined choices}$.

After assigning a partner to $5$, there are no invalid pairings for yet-unpaired numbers, so there are $\mathbf{4!} \textbf{ ways}$ to choose partners for $\{1,2,3,4\}$.

The answer is $3! \cdot 4! = \boxed{\textbf{(E) } 144}$.

In general, for $1,\ldots,2n$, the same logic yields answer: $\left\lfloor\dfrac{n}{2}\right\rfloor! \cdot \left\lceil\dfrac{n}{2}\right\rceil!$

~oinava

Video Solution by Education, the Study of Everything

https://youtu.be/k6EUl65wS9Q

Video Solution by Sohil Rathi

https://youtu.be/V1jOj8ysd_w

~ pi_is_3.14

Video Solution (Smart and Simple)

https://youtu.be/7yAh4MtJ8a8?si=jyIdy-jZb2raj3cM&t=1800

~Math-X

Video Solution (RMM club)

https://youtu.be/DwCE1wu5hrA

Video Solution by Lucas637 (Fast and Easy)

https://www.youtube.com/watch?v=egQK11g54mA

Video Solution by TheBeautyofMath

https://youtu.be/0kkc4-y8TkU?t=1367

~IceMatrix

See Also

2022 AMC 10A (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 AMC 10 Problems and Solutions
2022 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2022_AMC_10A_Problems/Problem_14&oldid=230503"