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?
Must each new year always contain a Friday the 13th?
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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 $13^\text{th}$. $$ \newcommand{\Sun}{\color{#C00}{\text{Sun}}} \newcommand{\Mon}{\text{Mon}} \newcommand{\Tue}{\text{Tue}} \newcommand{\Wed}{\text{Wed}} \newcommand{\Thu}{\text{Thu}} \newcommand{\Fri}{\text{Fri}} \newcommand{\Sat}{\text{Sat}} \begin{array}{c|c} &\text{May}&\text{Jun}&\text{Jul}&\text{Aug}&\text{Sep}&\text{Oct}&\text{Nov}\\ \hline2005&\Sun&\Wed&\Fri&\Mon&\Thu&\Sat&\Tue\\ \hline2000&\Mon&\Thu&\Sat&\Tue&\Fri&\Sun&\Wed\\ \hline2001&\Tue&\Fri&\Sun&\Wed&\Sat&\Mon&\Thu\\ \hline2002&\Wed&\Sat&\Mon&\Thu&\Sun&\Tue&\Fri\\ \hline2003&\Thu&\Sun&\Tue&\Fri&\Mon&\Wed&\Sat\\ \hline2009&\Fri&\Mon&\Wed&\Sat&\Tue&\Thu&\Sun\\ \hline2004&\Sat&\Tue&\Thu&\Sun&\Wed&\Fri&\Mon\\ \hline\end{array}\\ \text{Day of the $1^\text{st}$ of Each Month} $$
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.
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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.