-1

I am wondering if there is any way to determine whether it is possible to have a year that does not contain a Friday the 13th?

robjohn
  • 345,667
llewmills
  • 117
  • 1
    There is and it's not that hard. – John Douma Sep 13 '19 at 00:25
  • 1
    It isn't; a year always has at least one Friday the 13th. Take March as the base month. June is March plus one day of the week; September is March plus two; April is March plus three; October is March plus four; May is March plus five; and August is March plus six. – Brian Tung Sep 13 '19 at 00:25
  • 2
    Of course there is a way to determine that. The exact sequence of weekdays and dates in the Gregorian calendar repeats unchanged every 400 years, so at worst you'd need to inspect 400 different years to find your answer. – hmakholm left over Monica Sep 13 '19 at 00:26
  • 1
    The calendar of a year is determined by the day of the week on which January $1$ falls, and whether or not it's a leap year, so there are only $14$ possible calendars. – saulspatz Sep 13 '19 at 00:35

2 Answers2

3

Just do it.

Tedious but easy way:

Jan 1. must fall on a monday, tues,....., sunday. And either there is or there is not a leap year. So there are only $14$ possible calendars. Check them all. They all have at least one friday the $13$ths.

Same idea but a bit more efficient.

Suppose it isn't a leap year. And assume that January first is on a day of the week we call $0$.

January has $31= 4*7 + 3$ days. So February first will fall on the day of the week $3$.

February has $28 =4*7 + 0$ days. So March 1 falls on day $3$.

March has $31=4*7+3$ days. So April 1 fall on day $6$.

April has $30=4*7+2$ days. So May 1 falls on day $8$ but the week has only $7$ days so it wraps around back to $1$.

And so on.

We get the 12 months start on the following days for the first of the month: $(0, 3,3,6,1,4,6,2,5,0,3,5)$. If it were a leap year we would get $(0, 3,4,0,2,5,0,3,6,1,4,6)$

The $13$th of every month would fall on the following days. $(6, 2,2,5,0,3,5,1,4,6,2,4)$ and $(6, 2,3,6,1,4,6,2,5,0,3,5)$.

Everday day from $0$ to $6$ is possible so xxxday the $13$th will always occur.

===

If Jan. 1st is a Sunday, and it isn't a leap year then April and July will have Friday the 13ths. If Jan. 1st is a Monday then September and December will have Friday the 13ths. Etc.

fleablood
  • 124,253
0

Whether leap year or not, May, June, July, August, September, October, and November start on different days of the week. The month that starts on a Sunday will have a Friday the $13^\text{th}$. $$ \newcommand{\Sun}{\color{#C00}{\text{Sun}}} \newcommand{\Mon}{\text{Mon}} \newcommand{\Tue}{\text{Tue}} \newcommand{\Wed}{\text{Wed}} \newcommand{\Thu}{\text{Thu}} \newcommand{\Fri}{\text{Fri}} \newcommand{\Sat}{\text{Sat}} \begin{array}{c|c} &\text{May}&\text{Jun}&\text{Jul}&\text{Aug}&\text{Sep}&\text{Oct}&\text{Nov}\\ \hline2005&\Sun&\Wed&\Fri&\Mon&\Thu&\Sat&\Tue\\ \hline2000&\Mon&\Thu&\Sat&\Tue&\Fri&\Sun&\Wed\\ \hline2001&\Tue&\Fri&\Sun&\Wed&\Sat&\Mon&\Thu\\ \hline2002&\Wed&\Sat&\Mon&\Thu&\Sun&\Tue&\Fri\\ \hline2003&\Thu&\Sun&\Tue&\Fri&\Mon&\Wed&\Sat\\ \hline2009&\Fri&\Mon&\Wed&\Sat&\Tue&\Thu&\Sun\\ \hline2004&\Sat&\Tue&\Thu&\Sun&\Wed&\Fri&\Mon\\ \hline\end{array}\\ \text{Day of the $1^\text{st}$ of Each Month} $$

robjohn
  • 345,667