## how many Friday 13th?

**A** short Riddler’s riddle on the maximum number of Fridays 13th over a calendar year, of which I found 9 by a dumb exploration :

bi=c(1:31,1:29,1:31,1:30,1:31,1:30,1:31,1:31,1:30,1:31,1:30,1:31)

oy=bi[-60]

for(j in 0:(length(cy<-c(bi,oy,oy,oy))-1)){#any day in quartade

dy=c(cy[(j+1):length(cy)],cy[1:j])

for(i in 0:6){

dz=(i+(1:length(cy)))%%7

if((k<-sum((dz==5)*(cy==13)))>9)print(c(i,j,k))}}

with no change whatsoever when starting another day of the year, including a Friday 13.(since this only gains 13 days!). An example of a quartade (!) with nine such days is the sequence 2012-2015 with 3+2+1+3 occurences….

December 21, 2020 at 4:51 pm

Just for fun, did you verify that a 28-day Feb doesn’t help (e.g. if such a Feb has Fri-13, then so does March) ?

December 21, 2020 at 5:03 pm

Yes, to account for this I started with every possible weekday over five years in a row. A Friday the 13th happening on a 28 Feb year sees one extra Friday the 13th, but, while beneficial, this only happens at most once over the four years. The solution from The Riddler was also having a maximum of 9 as the maximum, so I am afraid there is not so much variability as one could hope for!

December 18, 2020 at 8:20 am

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