#math trivia #104 solution

This one is related to the answer to #92.

The starting days of the month, as noted in that answer, have the following modular values (relative to January 1) in a leap year:

0 3 4 0 2 5 0 3 6 1 4 6

Each modular value between 0 and 6 occurs at least once, so the 13th will occur at least once on each of the seven days of the week during a leap year. 

Each modular value occurs at most three times, so the 13th will occur at most three times on any given day of the week  during a leap year.  Since leap years can potentially start on any day of the week (this requires a little inter-year modular arithmetic), this means the maximum number of Friday the 13ths is three, as is occuring in 2012.

The pattern for non-leap-years just shifts the March to December starting positions by -1 modulo 7, due to the absence of February 29:

0 3 3 6 1 4 6 2 5 0 3 5

Again, each modular value occurs at least once and at most three times.  Combined with the fact that non-leap years can also potentially start on any day of the week, the minimum and maximum number of Friday the 13ths are again one and three.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s