## a [Gregorian] calendar riddle

**A** simple riddle express this week on The Riddler, about finding the years between 2001 and 2099 with the most cases when day x month = year [all entries with two digits]. For instance, this works for 1 January, 2001 since 01=01 x 01. The only difficulty in writing an R code for this question is to figure out the number of days in a given month of a given year (in order to include leap years).

The solution was however quickly found on Stack Overflow and the resulting code is

#safer beta quantile numOD <- function(date) { m <- format(date, format="%m") while (format(date, format="%m") == m) date <- date + 1 return(as.integer(format(date - 1, format="%d"))) } dayz=matrix(31,12,99) for (i in 2001:2099) for (j in 2:11) dayz[j,i-2000]=numOD(as.Date( paste(i,"-",j,"-1",sep=""),"%Y-%m-%d")) monz=rep(0,99) for (i in 1:99){ for (j in 1:12) if ((i==(i%/%j)*j)&((i%/%j)<=dayz[j,i])) monz[i]=monz[i]+1}

The best year in this respect being 2024, with 7 occurrences of the year being the product of a month and a day…

April 17, 2018 at 8:18 am

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