Calendrical Conundrum
In any calendar year how many Friday the thirteenths can there be? If is the maximum number and is the minimum number, find .
The answer is 4.
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1 solution
Consider 2018: Jan 1 was on a Mon and the 13th's are on: Jan - Sat; Feb - Tues; Mar - Tues, Apr - Fri, May - Sun; June - Wed; July - Fri, Aug - Mon, Sept - Thurs, Oct - Sat, Nov - Tue, Dec - Thur
If we shift the "year" to start on any given day of the week, the respective days of the 13th's will shift equally. Thus, 13th's that are on a Monday this year WOULD HAVE been a Friday in a year where Jan 1 was on a different weekday.
So: we have the 13ths on: 1 Monday, 3 Tuesdays, 1 Wednesday, 2 Thursday, 2 Fridays, 2 Saturdays, 1 Sunday
So the most a 13th can fall on any weekday is 3 times; the least is 1 time; and thus the answer of least + most = 1 + 3 = 4.
(If you account for leap years, you have to add a weekday to all the months following Feb - OR - subtract a day from January - in which case there would be one less Saturday (brining total to 1), and one additional Sunday (bringing its total to 2) - in other words, not changing our answer