# Five Fridays and Sundays on October

How to prove that if you take any 400 consecutive Octobers then exactly 14 % of those years have five Fridays and Sundays?

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To have five Fridays and Sundays, October must start on Friday. A span of 400 years has 146097 days in it-remember the years that are multiples of 100 but not 400 are not leap years. This is a multiple of 7. So the pattern repeats. Count how many of the 400 start October with Thursday or Friday. You should find 56.

Corrected-used to say start on Thursday or Friday.

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Is there any nice way to count that 56 Octobers starts from Thursday or Friday? – Jaska Oct 29 '10 at 14:03
I would just go to Excel make a column of 400 years starting with, say, 10/1/2000. The fill command will increment by year. Then in the next column use the weekday function to get the day of the week. Alternately, find what October started on some (this?) year and think about the pattern how it moves from one year to the next. Usually it advances one day, but there are leap years. – Ross Millikan Oct 29 '10 at 14:17
I see. I had the same idea on my mind although I replaced Excel by PHP. Probably there is no trick to do it easier. – Jaska Oct 29 '10 at 14:38
Another way to do it is to invert the Doomsday algorithm: en.wikipedia.org/wiki/Doomsday_rule If an October starts on Thursday or Friday, the Doomsday (10/10) must be Saturday or Sunday (day 6 or day 0). for each of the four centuries you can solve $\lfloor 5y/4 \rfloor\mod 7 +$ offset $= 0$ or $6$. Or you can just count the number of Saturday/Sunday Doomsdays on the table in that article near the bottom. – Willie Wong Oct 29 '10 at 15:19

As people have said, the best way to do this is brute force, once you know that the calendar repeats with period 400 years.

I'd note that to have five Fridays and five Sundays, October must start on a Friday (like it does this year), and therefore run from Friday, October 1 to Sunday, October 31.

If it feels like cheating to use date rules that somebody has already computed, you could (for example) note that:

• this year, October 1 was a Friday, the sixth day of the week;

• if October 1, year $n$ is the $k$th day of the week, then October 1, year $n+1$ is either the $k+1$st or $k+2$nd day of the week, depending on whether $n+1$ is an ordinary year or a leap year. (Of course the $8$th and $9$th days of the week can be converted to the $1$st and $2$nd days.

Finally, it's easy to see that the answer should be "approximately 1/7" because the calendar rules don't make reference to any particular days of the week, although of course this is not a proof. This is why the claim that went around the Internet earlier this month that an October like the current one only occurs every 823 years was so laughable; see here for a debunking.

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I've seen the same 823-year claim made for July, 2011. – Gerry Myerson Jul 5 '11 at 7:04
It actually showed up in my e-mail. (From someone who seems to think I'm someone they know and has the wrong e-mail address for me. Everyone I know knows better than to send me things like this.) – Michael Lugo Jul 5 '11 at 14:55