Difference between revisions of "2021 Fall AMC 12A Problems/Problem 16"

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From this solution, it is clear that this claim is true: We isolate <math>A_{20}</math> and connect all cables for <math>A_1, A_2, \ldots, A_{19}</math> and <math>B_1, B_2, \ldots, B_{10}</math> to get <math>19\cdot10=190</math> cables. However, <math>A_{20}</math> cannot communicate with any of these <math>29</math> computers. <p>
 
From this solution, it is clear that this claim is true: We isolate <math>A_{20}</math> and connect all cables for <math>A_1, A_2, \ldots, A_{19}</math> and <math>B_1, B_2, \ldots, B_{10}</math> to get <math>19\cdot10=190</math> cables. However, <math>A_{20}</math> cannot communicate with any of these <math>29</math> computers. <p>
 
   <li><b>With exactly <math>\boldsymbol{191}</math> cables, all computers must be able to communicate with each other.</b></li><p>
 
   <li><b>With exactly <math>\boldsymbol{191}</math> cables, all computers must be able to communicate with each other.</b></li><p>
By the Pigeonhole Principle, we conclude that at least one brand B computer connects to all <math>20</math> brand A computers. Without loss of generality, we assume that <math>B_1</math> connects to all <math>20</math> brand A computers. On the other hand, we conclude that at least <math>11</math> brand A computer connects to all <math>10</math> brand B computers. Without loss of generality, we assume that <math>A_1, A_2, \ldots, A_{11}</math> connect to all <math>10</math> brand B computers. <p>
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By the Pigeonhole Principle, we conclude that at least one brand B computer connects to all <math>20</math> brand A computers. Without loss of generality, we assume that <math>B_1</math> connects to all <math>20</math> brand A computers. On the other hand, we conclude that at least <math>11</math> brand A computers connect to all <math>10</math> brand B computers. Without loss of generality, we assume that <math>A_1, A_2, \ldots, A_{11}</math> connect to all <math>10</math> brand B computers. <p>
 
Therefore, any two computers must be able to communicate with each other:
 
Therefore, any two computers must be able to communicate with each other:
 
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<ul style="list-style-type:square, margin-left: 3em;">

Latest revision as of 06:58, 3 November 2024

Problem

An organization has $30$ employees, $20$ of whom have a brand A computer while the other $10$ have a brand B computer. For security, the computers can only be connected to each other and only by cables. The cables can only connect a brand A computer to a brand B computer. Employees can communicate with each other if their computers are directly connected by a cable or by relaying messages through a series of connected computers. Initially, no computer is connected to any other. A technician arbitrarily selects one computer of each brand and installs a cable between them, provided there is not already a cable between that pair. The technician stops once every employee can communicate with each other. What is the maximum possible number of cables used?

$\textbf{(A)}\ 190  \qquad\textbf{(B)}\  191 \qquad\textbf{(C)}\  192 \qquad\textbf{(D)}\  195 \qquad\textbf{(E)}\ 196$

Solution

We claim that to maximize the number of cables used, we isolate one computer and connect all cables for the remaining $29$ computers, then connect one more cable for the isolated computer.

If a brand A computer is isolated, then the technician can use at most $19\cdot10+1=191$ cables. If a brand B computer is isolated instead, then the technician can use at most $20\cdot9+1=181$ cables. Therefore, the answer is $\boxed{\textbf{(B)}\  191}.$

Remark

Suppose that the computers are labeled $A_1, A_2, \ldots, A_{20}$ and $B_1, B_2, \ldots, B_{10}.$ We will prove the claim in the first paragraph of this solution:

  1. With exactly $\boldsymbol{190}$ cables, it is possible that some computers cannot communicate with each other.
  2. From this solution, it is clear that this claim is true: We isolate $A_{20}$ and connect all cables for $A_1, A_2, \ldots, A_{19}$ and $B_1, B_2, \ldots, B_{10}$ to get $19\cdot10=190$ cables. However, $A_{20}$ cannot communicate with any of these $29$ computers.

  3. With exactly $\boldsymbol{191}$ cables, all computers must be able to communicate with each other.
  4. By the Pigeonhole Principle, we conclude that at least one brand B computer connects to all $20$ brand A computers. Without loss of generality, we assume that $B_1$ connects to all $20$ brand A computers. On the other hand, we conclude that at least $11$ brand A computers connect to all $10$ brand B computers. Without loss of generality, we assume that $A_1, A_2, \ldots, A_{11}$ connect to all $10$ brand B computers.

    Therefore, any two computers must be able to communicate with each other:

    • Between $A_i$ and $A_j,$ where $i,j\in\{1,2,\ldots,20\}$ and $i\neq j$
    • The relay is $A_i\leftrightarrow B_1\leftrightarrow A_j.$

    • Between $B_i$ and $B_j,$ where $i,j\in\{1,2,\ldots,10\}$ and $i\neq j$
    • The relay is $B_i\leftrightarrow \{A_1, A_2, \ldots, A_{11}\}\leftrightarrow B_j.$

    • Between $A_i$ and $B_j,$ where $i\in\{1,2,\ldots,20\}$ and $j\in\{1,2,\ldots,10\}$
    • The relay is $A_i\leftrightarrow B_1\leftrightarrow \{A_1, A_2, \ldots, A_{11}\}\leftrightarrow B_j.$

~MRENTHUSIASM

See Also

2021 Fall AMC 12A (ProblemsAnswer KeyResources)
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 12 Problems and Solutions

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