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Logic Puzzles by The Logic Fiend
Good day my lamebrained friend. Be addled by my riddles. You may be rewarded or punished for your answers. So beware!

Friday the 13th
Let's get back to basics. In a year, how many Friday the 13th's can there be? Yes, I am asking for a range (e.g. minimum of 0, maximum of 12 is possible).

Explain your answer. A random guess without reason is worthless.

25 Comments.

...
I say about 3 or 4. Why? because I looked at my calendar! =) well, last year there was three, and from the looks of the calendar, the days around them [12,14] land around that day quite a bit. lol. errr. i tried.
karendipity on 2003-07-09 05:24:20


1-3 because i checked my computer's calendar from 2003-2010..most of em were 1 or 2...one of em was 3
aznkuo on 2003-07-09 06:01:28

please, please.
I am looking for somewhat of a more mathematical explanation.
theLogicFiend on 2003-07-09 08:33:53


1-3. Mathematically, there is a huge formula that a calendar follows in regards to the rotation of the days of the week and how they line up in number according to the day. In order to have a Friday the 13th, the first of the month needs to be on a Sunday. There is a finite number of combinations, due to the inconsistancy of the numbering of the months, (30 days has September...), as well as the addition of a day every 4 years, that can land the first of a month on a Sunday. I'm not sure of all the mathematics that are involved, but I know that they are there. There is a thing called a perpetual calendar that has exhausted all possiblities in the combinations of the alignment of numbers, and if you reference one of those, you might be able to figure out the math involved. But the answer you seek is 1-3 Firday the 13ths.
AmberlyMeleKatherine on 2003-07-13 10:57:22

re: AmberlyMeleKatherine
Hmm. You have slightly eased my anger, so I will give you half credit. 25 points.

For full credit, provide more insight into this formula. It's quite straightforward, and there are only 2 cases to consider.
theLogicFiend on 2003-07-15 12:04:56

eased anger
Gald to have achieved at least that much. I've been thinking about the exact math necessary, and the best I can come up with is that a month has to begin on a Sunday in order for the 13th to be a Friday. The way that 3 Friday the 13ths are achieved is that February begins on a Sunday and it isn't a leap year. On non-leap years, February have 28 days. 28/7=3. So that's 3 equal weeks in that month, making the first of March a Sunday as well. March 13th would be the second Firday the 13th. March has 31 days, so the next time the first of a month is on a Sunday won't be until the November of that month. That is the ONLY way that this can work. If february first is a Sunday, but it's a leap year, it throws the whole thing off.
I hope this makes things a bit more clear.
AmberlyMeleKatherine on 2003-07-18 11:40:12

re amberlymelekatherine
uhm..28/7 is 4 ....haha..im workin on the formula i dont have it just yet
swis on 2003-07-23 01:48:21


whoa freaky. what's with all the lil pop up evil smurf guys
idcwutam on 2003-07-23 02:40:27

28/7
okay, so i'm dumb. sue me.
AmberlyMeleKatherine on 2003-07-24 01:31:56

Bad Luck?
Any year there must be at least one month and at most three months for which the 13th of the month falls on Friday. First: The average number of Friday the 13th's is really quick to compute: There are 12 months (therefore 12 '13th's') and 7 weekdays for them to fall on, so the average number of Friday the 13th's in a year is 12/7, or about 1.714. Second: The trick here is to figure out which months start on the same day of the week. In a standard (non-leap) year, if January starts on a Monday, then February starts on a Thursday, March also starts on a Thursday, etc. Since January has 31 days, it has 4 full weeks (28 days) plus 3 days. Hence, February must start 3 weekdays later. Similarly, since February has 28 days (4 weeks + 0 days,) March starts on the same weekday as February. Continuing this pattern, you can see which months start on the same days, and you can count how many months start on each day (the counts for both a standard year and a leap year). If months (like February and March) start on the same weekday, logically the 13th of those months will also fall on the same weekday- which will, in certain years be a Friday. So whatever was the highest count of months starting on the same weekday, that number is the most Friday the 13th's there could possibly be.
Mari on 2003-07-24 04:49:47

so whats your answer
haha
swis on 2003-07-24 11:58:59

using the good theory of mari
i got 1 minimum and 3 maximum..how?well here..[1=1st possible day2=2nd possible 3 = 3rd possible...] year starts on 1 [+3 after jan]4 [+0 after feb]4 [+3 after Mar]7 [+2 after Apr]2[it cycles like a week] [+3 after May]5 [+2 after Jun]7 [+ 3 after Jul]3 [+3 after Aug]6 [+2 after Sep]1 [+3 after Oct]4 [+2 after Nov]6 december doesnt matter after that i got two 1 two 2 one 3 three 4 one 5 two 6 two 7 so max is 3 min is 1!
swis on 2003-07-25 12:07:52

mathematicly
t-28=x, d+x=[S,N] t=total days in month x=extra days after the 28th d=day month starts S=sunday N=Non-sunday if t-28=x and d+x=S then the month has a friday the 13th if t-28=x and d+x=N then the month has no friday the 13Th
swis on 2003-07-25 12:16:18

How about an average?
1.714 12 Friday the 13th's divided by 7 days of a week that they can be on.
chris on 2003-08-07 08:41:03

explain
why is it impossible to have no friday '13s' in a year, again?
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