# 2021 Fall AMC 12A Problems/Problem 16

## 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 computer connects 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