Number of Friday 13ths in a Year
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Theorem
In any given year, there are between $1$ and $3$ (inclusive) months in which the $13$th falls on a Friday.
Proof
The day of the week on which the $13$th falls is directy dependent upon the day of the week that the $1$st of the month falls.
From Months that Start on the Same Day of the Week, the months can be grouped into equivalence classes according to which day the month starts:
For a non-leap year, the set of equivalence classes is:
- $\set {\set {\text {January}, \text {October} }, \set {\text {February}, \text {March}, \text {November} }, \set {\text {April}, \text {July} }, \set {\text {May} }, \set {\text {June} }, \set {\text {August} }, \set {\text {September}, \text {December} } }$
For a leap year, the set of equivalence classes is:
- $\set {\set {\text {January}, \text {April}, \text {July} }, \set {\text {February}, \text {August} }, \set {\text {March}, \text {November} }, \set {\text {May} }, \set {\text {June} }, \set {\text {September}, \text {December} }, \set {\text {October} } }$
It can be seen that none of these equivalence classes has fewer than $1$ day, and none has more than $3$.
Hence the result.
$\blacksquare$
Sources
- 1975: T.S. Blyth: Set Theory and Abstract Algebra ... (previous) ... (next): $\S 6$. Indexed families; partitions; equivalence relations: Exercise $7$