How many Friday the 13ths can there be in a year? (What is the maximum? What is the minimum?) And if the year is a leap year?
We only consider the case of a non-leap year (the case of a leap year is treated in exactly the same way). Having a Friday the 13th is equivalent to having a Monday the 1st, so we will count the Mondays the 1st. Furthermore, $31 \equiv 3(\bmod 7)$, thus we obtain the shift in the week between January 1st and February 1st (it is 3 days: if January 1st is a Monday, then February 1st will be a Thursday). We then do this for all the months:

$$
3+28 \equiv 3(\bmod 7) \text { for March }
$$

$$
\begin{gathered}
3+31 \equiv 6 \quad(\bmod 7) \text { for April } \\
6+30 \equiv 1 \quad(\bmod 7) \text { for May } \\
1+31 \equiv 4 \quad(\bmod 7) \text { for June } \\
4+30 \equiv 6 \quad(\bmod 7) \text { for July } \\
6+31 \equiv 2 \quad(\bmod 7) \text { for August } \\
2+31 \equiv 5 \quad(\bmod 7) \text { for September } \\
5+30 \equiv 0 \quad(\bmod 7) \text { for October } \\
0+31 \equiv 3 \quad(\bmod 7) \text { for November } \\
3+30 \equiv 5 \quad(\bmod 7) \text { for December }
\end{gathered}
$$

We thus get 2 times 0 modulo 7, 1 time 1, 1 time 3, 3 times 3, 1 time 4, 2 times 5, and 2 times 6. Therefore, there will be only one Friday the 13th if the year starts on a Sunday (day -1 of the week which results in only one 1 in the residues modulo 7), a Saturday, or a Thursday. There will be 2 if the year starts on a Monday, a Tuesday, or a Wednesday, and finally, there will be 3 if the year starts on a Friday. Note that if the year is a leap year, there can be 4 Fridays the 13th.