This is a slight re-hashing of an old puzzle, to make it more topical. Do you recognise it? Can you remember the solution? Or can you solve it anyway?
Every person who catches this particular virus becomes infectious one day later, and then remain infectious for two weeks. Every person who has the virus and is infectious will infect one other person every day until they become noninfectious.
In a particular population, one person catches the virus, and then the infection pattern described above goes into action. By the time the original person becomes noninfectious, how many people have the virus, assuming that the population is not yet completely infected?
This puzzle is based on a puzzle about breeding rabbits, first set by Fibonacci himself. Which might give a clue to the answer.
Person A catches the virus on day one, and there is one person with the virus.
On day two person A becomes infectious, and there is still one person with the virus.
On day three person A infects person B, so there are two people with the virus, one who's infectious and one who is not.
On day four person A infects person C and person B becomes infectious. Now there are three people with the virus, two of whom are infectious and one who is not.
On day five person A and person B infect person D and person E, while person C becomes infectious. So now there are five people with the virus, three of whom are infectious and two how are not.
This pattern continues following Fibonacci's famous sequence 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610. And so on day 15, the day person A becomes noninfectious, 610 people have the virus.