#math trivia: The 20-student class gets the flu. Every day a different prime number of students is out sick. For how long can this go on?
— Burt Kaliski Jr. (@modulomathy) September 10, 2012
There are eight different prime numbers less than or equal to 20 (2, 3, 5, 7, 11, 13, 17 and 19), so the flu-related absences can go on for up to eight days if each day has a different prime number of absences.
Beyond the trivia:
- The problem is only concerned with the total number of students absent, not the particular students absent. If every day a different prime-number-size group of students were out sick, the absences could go on for a very long time. There are 20! / (2!)(18!) = (20*19)/2 = 190 different groups of size 2, (20!) / (3!)(17!) = (20*19*18)/(3*2) = 1140 different groups of size 3, and so on. The largest number of groups with a given prime-number-size happens at 11, with a total of (20!) / (9!)(11!) = 167,960 groups. With 180 school days a year, it would take nearly a millennium just to exhaust just the possible groups of size 11, by which time all the students would presumably have completed the class!
- What if the flu spread to the other four classes, potentially affecting all 100 students? There are 25 different prime numbers less than or equal to 100, so the absences could go on for an additional 17 days after the initial eight. And if it spread to the whole world, with population 7 billion? There are 323,804,352 primes less than or equal to 7,000,000,000, so our epidemic would be a very, very long one. (Check out Wolfram Alpha to see how many prime numbers there are up to a given number. The rough estimate for the number of primes up to N, according to the Prime Number Theorem, is N/(ln N) where (ln N) is the natural logarithm of N.)