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

POLISHING. NO EDIT PLEASE. A MILLION THANKS

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

We will prove the claim in the first paragraph:

Suppose

~MRENTHUSIASM

2021 Fall AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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2021 Fall AMC 12A Problems/Problem 16

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