Given that today is yet another occurrence of Friday the 13th, it seems like a good time to apply some mathematical legerdemain in support of the true specialness of Friday. Why do I say Friday

*(and Friday the 13th) *is a special kind of day?

Consider these two little known facts:

- Friday is the most likely day to be the 13th.
- The first day of a new century can never be a Friday.

No other day can make such claims! How can such outrageous claims be true? Let us begin by considering how our calendar works.

Every fourth year is a leap year, so in 400 years there are about 300 years that have 365 days, and 100 years that have 366 days.

Except...

Every 100th year is not a leap year, so that makes 304 regular years and 96 leap years.

Except...

Every 400th year is a leap year

* (which is why 2000 was a leap year)*, so we have 303 regular years, and 97 leap years.

That's all the exceptions.

Thus the number of days in 400 years is given by

(303 x 365) + (97 x 366) = 146,097,

which is an exact multiple of 7, namely 7 x 20,871. That means the calendar repeats itself exactly, leap years and all, every 400 years.

Now all we have to consider is the 400 year cycle. In 400 years, there are 4,800 days

* (400 x 12) *labeled with the date of the 13th. But, 4,800 is not evenly divisible by 7. Thus the frequency of the day of week of the days labeled the 13th cannot be uniform. This is one of those cases where actually enumerating the days and counting them up is the simplest way to get the number of Monday the 13th ... Friday the 13th ... Sunday the 13th's. For the moment, trust me that the answer for the 400 year cycle is[1]:

**Day** | **Number of Occurrences in 400 years** |

Sat | 684 |

Sun | 687 |

Mon | 685 |

Tue | 685 |

Wed | 687 |

Thu | 684 |

Fri | 688 |

So we see that Friday is indeed the day most likely to be the 13th, edging out Sunday and Wednesday by one occurrence per 400 years..

Working out that the first day of a new century is never a Friday

*(or Sunday for that matter)* is much simpler.

There are 365 days in a common year. Dividing 7 into 365, we get 52 plus a remainder of 1. So a common year is exactly 52 weeks plus one day. Therefore, from a common year to the following year, New Year's Day advances by one day of the week

*(DOTW, since us scientists cannot live without acronyms.)*. So if New Year's Day of a common year falls on a Monday, the next year's will be a Tuesday.

For a 366-day leap year, the remainder is 2, which means that going from a leap year to the year following, New Year's Day advances by two DOTW – e.g., from a Monday to a Wednesday.

What happens to New Year's Day over a period of several centuries?

In a single century, New Year's Day advances by 100 DOTW

*(one per year)* plus an extra day for each leap year. These come every four years, of course, except that a year evenly divisible by 100 isn't a leap year. Therefore in each century there are 24 extra leap days

*(100 divided by 4, minus 1),* meaning that from one century to the next, New Year's advances by a gross total of 124 DOTW. But each complete week's worth of days doesn't do anything to advance New Year's – seven days of DOTW advancement just gets you back to the day you started on – so we divide 7 into 124 and get a remainder of 5. Thus from one century to the next the net DOTW advancement is five days.

As a second century goes by, the DOTW advances another five days, for a total of ten days from the beginning of the first century. Taking modulo 7 of 10, we get three days of net DOTW advancement over the 200 years.

A third century advances New Year's another 5 days, for a total of 15 DOTW; modulo 7 of this yields 1 day of net DOTW advancement.

For the fourth century, we have to take into account the third leap year rule: years divisible by 400 do have a leap day. So for the fourth century, we advance the DOTW another five days plus an additional leap day, plus the 15 days accumulated over the first three centuries, for a gross total of 21 days advanced. Take the modulo 7 of that, and you get a net total of 0: that is, the fifth century starts on the same day of the week that the first century did. Then the pattern starts over again.

So we have is one fixed pattern that repeats itself exactly every 400 years. Since there are only four starts-of-the-century in each four centuries, New Year's can fall on only four of the seven days of the week. The winners depend solely on where the days of the week happened to line up with the cycle when the crazy leap year system was adopted. Our system is set up so that not only Sundays get left out, but Tuesdays and Thursdays as well. The first New Year's of each 400-year cycle falls on a Friday, the next century starts on a Wednesday, the next on a Monday, and the next one – the century beginning with a year that's divisible by 400 (like, e.g., 2000) – starts on a Saturday.

But you have to remember one little pedantic factoid - the new century technically starts in the the years ending with 01. If you define your centuries as beginning with the year ending in 01, then the cycle goes Saturday-Thursday-Tuesday-Monday. So being pedantic still won't get you a century that starts on a Sunday, but it does get rid of Friday.

*{*grin*}*
[1] If you don't want to trust me about the frequency of the various days of the week over the 400 year cycle, consider this simple Excel spreadsheet calculation as elucidated by

Graeme McRae on his wonderful

MathHelp site:

... Start with the 13th of this month. Put that date in cell A1. Then in cell A2, put the following formula:

=DATE(YEAR(A1+30),MONTH(A1+30),13)

Copy the contents of cell A2 down to all the cells from A3 through A4800. Now you have a table of the dates of all the 13ths of every month for the next 400 years.

Next, in cell B1, put this formula:

=MOD(A1,7)

Copy that formula from B1 down to all the cells from B2 through B4800. Here, numbers from 0 to 6 represent the days of the week, as follows:

**Number** | **Day** |

0 | Sat |

1 | Sun |

2 | Mon |

3 | Tue |

4 | Wed |

5 | Thu |

6 | Fri |

If you like, you can have Excel format these cells to show the day of the week instead of just the number. Click Format, Cells, Custom, and type ddd as the custom format -- it's up to you.

Now, in cells C1:C7, enter the numbers 0, 1, 2, 3, 4, 5, 6 in a vertical column.

In cell D1, enter the following formula, but **don't hit enter yet!**

=SUM(IF(B$1:B$4800=C1,1,0))

After you type the formula in cell D1, hold down the Shift and Ctrl keys, and press enter. This makes the formula into an "Array Formula", so it counts the number of days in column B that match cell C1. Then copy this formula from D1 to D2 through D7.

If you format column C to make it show day-of-week (Click Format, Cells, Custom, and type ddd), then you will have a table that looks like this:

**Day** | **Frequency** |

Sat | 684 |

Sun | 687 |

Mon | 685 |

Tue | 685 |

Wed | 687 |

Thu | 684 |

Fri | 688 |