#math trivia for #April13: Second #FridayThe13th this year. Is there a third?What’s the most a year can have? The fewest?
— Burt Kaliski Jr. (@modulomathy) April 13, 2012
This one is related to the answer to #92.
The starting days of the month, as noted in that answer, have the following modular values (relative to January 1) in a leap year:
0 3 4 0 2 5 0 3 6 1 4 6
Each modular value between 0 and 6 occurs at least once, so the 13th will occur at least once on each of the seven days of the week during a leap year.
Each modular value occurs at most three times, so the 13th will occur at most three times on any given day of the week during a leap year. Since leap years can potentially start on any day of the week (this requires a little inter-year modular arithmetic), this means the maximum number of Friday the 13ths is three, as is occuring in 2012.
The pattern for non-leap-years just shifts the March to December starting positions by -1 modulo 7, due to the absence of February 29:
0 3 3 6 1 4 6 2 5 0 3 5
Again, each modular value occurs at least once and at most three times. Combined with the fact that non-leap years can also potentially start on any day of the week, the minimum and maximum number of Friday the 13ths are again one and three.
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