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

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   <li><b>With exactly <math>\boldsymbol{190}</math> cables, it is possible that some computers cannot communicate with each other.</b></li><p>
 
   <li><b>With exactly <math>\boldsymbol{190}</math> cables, it is possible that some computers cannot communicate with each other.</b></li><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>
 
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>With exactly <math>191</math> cables, all computers must be able to communicate with each other.</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 the 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 the 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>
 
By the Pigeonhole Principle, we conclude that at least one brand B computer connects to all <math>20</math> brand A computers. Without the 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 the 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:

Revision as of 09:45, 27 November 2021

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, connect all cables for the remaining $29$ computers, and 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.

  2. With exactly $\boldsymbol{191}$ cables, all computers must be able to communicate with each other.

    3. By the Pigeonhole Principle, we conclude that at least one brand B computer connects to all $20$ brand A computers. Without the 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 computer connects to all $10$ brand B computers. Without the 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 $A_i\longleftrightarrow B_1\longleftrightarrow A_j.$

    • Between $B_i$ and $B_j,$ where $i,j\in\{1,2,\ldots,10\}$ and $i\neq j$

      • The relay is $B_i\longleftrightarrow \{A_1, A_2, \ldots, A_{11}\}\longleftrightarrow B_j.$

    • Between $A_i$ and $B_j,$ where $i\in\{1,2,\ldots,20\}$ and $j\in\{1,2,\ldots,10\}$

      • The relay is $A_i\longleftrightarrow B_1\longleftrightarrow \{A_1, A_2, \ldots, A_{11}\}\longleftrightarrow B_j.$

~MRENTHUSIASM

See Also

2021 Fall AMC 12A (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 12 Problems and Solutions

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