2021 AMC 12A Fall Problem 16

Attempt Problem 16 of the 2021 AMC 12A Fall below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2021 AMC 12A Fall solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

16.

An organization has 3030 employees, 2020 of whom have a brand A computer while the other 1010 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?
190190
191191
192192
195195
196196
Answer: B
Concepts:graph theoryextremal argument
Small hint:
To delay connectivity, keep the graph disconnected as long as possible, then one final cable joins everything
Big hint:
The most edges while still disconnected is a complete bipartite graph on 1919 brand A and 1010 brand B, leaving one brand A computer isolated
Solution:
The technician keeps adding cables until the graph becomes connected. To maximize the count, keep the network disconnected for as long as possible: leave a single brand A computer isolated and fully connect the remaining 1919 brand A computers to all 1010 brand B computers.
That uses 19⋅10=19019 \cdot 10 = 190 cables while still disconnected. The next cable connects the last brand A computer, joining everyone, for a total of 190+1=191.190 + 1 = 191.
Thus, the correct answer is B.
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