# Must every year have at least one Friday which is on the 13th?

1. Aug 5, 2006

### Harmony

Question:
Must every year have at least one Friday which is on the 13th?

By listing out all the dates, I found that for any year, day 1st-7th must appear at least once for the the first Friday of the months. Hence, there must be at least a month which the first Friday is on the 6th. Thus, I deduce that every year must have at least one Friday which is on the 13th.

But how do I prove it mathematically? Does the dates form sequence? Is calculus needed for this question?

2. Aug 5, 2006

### HallsofIvy

What you did sounds like a perfectly good mathematics proof!

(Did you do this for both leap years and non-leap years?)

3. Aug 6, 2006

### Harmony

I do that for leap years and non-leap years. But how do I prove it through calculus or anything mathematical?

4. Aug 6, 2006

### Office_Shredder

Staff Emeritus
You just did I thought

5. Aug 6, 2006

### StatusX

Math is ultimately just logic. Not all kinds of math involve numbers or equations.

6. Aug 6, 2006

### pallidin

Here's something I found. The entire article is even more interesting.
Source:http://www.aug.edu/dvskel/Beveridge2003.htm [Broken]

There are fourteen possible calendars in the Gregorian system. One for January 1st falling on each of the seven days of the week in a standard 365 day year and one for January 1st falling on each of the seven days during a leap year. However, because of the interaction between the seven-day week and the four-year leap year cycle, the calendar must go through a 28-year period before returning to its starting place.

The number 365 is congruent to 1 modulo 7. In other words, there is a remainder of 1 when 365 is divided by 7. This causes the calendar to shift forward by one day each year. For instance, if January 1st is on Sunday during a non-leap year, it will fall on a Monday the following year. Without leap years, each date on the calendar would move forward one day per year and so would fall on a particular day every seven years. However, every four years, a leap year of 366 days causes each date to be pushed ahead two days, as 366 is congruent to 2 modulo 7 [see Table 1]. This irregularity is what causes the 28-year cycle of calendars, 28 being the least common multiple of 4 and 7.

Now the occurrences of Friday the 13th begin to make more sense. The examination of several years time will produce no pattern as the sequence of calendars repeats only once every 28 years. However, this does not take into account the Gregorian adjustment to the leap years, for three out of every four century years are not leap years. If we go back to the last century year before 2000 that was a leap year, an interesting pattern will emerge. The year 1600 was a leap year and occurred 18 years after the first countries began to adopt the Gregorian calendar, which makes it a convenient reference point. So, beginning in 1600, the calendar would progress through three 28-year periods before coming upon an irregularity.

The year 1700 would normally be a leap year and would keep the 28-year period running smoothly. Instead, what happens is that in the middle of the 28-year period there occurs the year 1700, which should be a leap year, but by the Gregorian adjustment is not. This causes the previous 12 calendars to repeat and so the pattern is not resumed until 12 years later, which causes the 28-year period to stretch to 40 years[see Table 2].

Last edited by a moderator: May 2, 2017
7. Aug 6, 2006

### GeoMike

I'm a novice when it comes to proofs, but if you want to put your reasoning into more mathematical terms, then how about the following:

If you add 13 to the number of days in each previous month (starting with 0 for January) you get the number of days after the beginning of the year that the thirteenth of each month falls on:

13 (jan)
44 (feb)
72 (mar)
103 (apr)
133 (may)
164 (jun)
194 (jul)
225 (aug)
256 (sep)
286 (oct)
317 (nov)
347 (dec)

Now take the modulus of each number with 7 (for the weekly cycles):

6 (jan)
2 (feb)
2 (mar)
5 (apr)
0 (may)
3 (jun)
5 (jul)
1 (aug)
4 (sep)
6 (oct)
2 (nov)
4 (dec)

The list above gives the day of the week (based on the first day of the year = 0) that the 13th of each month falls on.
For example, if the year starts on a Wednesday then Wednesday = 0, Tuesday = 1, Thursday = 2, etc.
So in a non-leap year beginning on Wednesday, July 13 is a Sunday.
From the list above you can see that every day of the week (0-6) is represented at least once (regardless of which day of the week is "day zero").

For leap years the first list and the second list are both offset by +1 after February:
13 (jan)
44 (feb)
73 (mar)
104 (apr)
134 (may)
165 (jun)
195 (jul)
226 (aug)
257 (sep)
287 (oct)
318 (nov)
348 (dec)

And:
6 (jan)
2 (feb)
3 (mar)
6 (apr)
1 (may)
4 (jun)
6 (jul)
2 (aug)
5 (sep)
0 (oct)
3 (nov)
5 (dec)

Again, each day of the week is represented at least once.
So, during any year, leap or non-leap, the thirteenth will fall on a Friday at least once (and every other day of the week at least once).

Assuming my numbers and logic are correct of course...
-GeoMike-

Last edited: Aug 6, 2006