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 quantilenumOD <- 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…

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