Well, apparently it is! At least this year. Next year it will be Sunday.

Maybe some of you had already heard of the “Doomsday Rule” of taking any date on the calendar (accounting for the 10-day Julian-Gregorian change in the 1500s, you can go back pretty far) and mentally determining what day of the week that date fell on, but I had never stumbled across it. I am a fan of some of the work Mr. John Conway, the mathematician that came up with the method, but, still, I had never heard of it until yesterday.

I was familiar with such algorithms before, but this one seems much easier to me than those others, requiring one to merely memorize a few “century days,” a few patterns of “doomsdays,” and a rather short and straightforward set of arithmetic steps. Leap Year adjustments seem fairly simple. (Why they are called “doomsdays,” I have no idea.)

The key recognition is that in any given year, the following dates (given here in the form M/D, as in: today is 3/17) are always on the same day of the week, called the “doomsday” of that year: 4/4, 6/6, 8/8, 10/10, 12/12. For instance, this year, 2009, all of those dates are on Saturday. (Really, just go look at your calendar — weird, huh?) For other months, it isn’t so bad — just not as easy as for the previous five. For instance, four of them occur in pairs: 5/9 & 9/5 and 7/11 & 11/7 (with the suggested mnemonic: “I work from 9-to-5 at the 7-11”). That leaves only the first three months, which are all handled with the fact that the last day of February — leap year or not — is always on this same day. (Again, check the calendar!)

Anyone interested can pursue the matter himself (or herself!) by checking out the Wikipedia page. For now, let me just demonstrate with an easily verifiable example. What day of the week will July 4 be this year? Here’s the method in action (it’s going to seem a little confusing, but check out the Wiki page for the details of why the steps are done)…

This year is 2009, so I take the last two digits, “09” or just 9, and divide them by 12, getting 0 (zero) with a remainder of 9. Then take the previous remainder, 9, and divide by 4 to get 2 with a remainder of 1. Ignore the last remainder (1) and add the other three numbers that we have generated: 0 + 9 + 2 = 11. From this, I am free to subtract as many 7s as I please to reduce the size of the number (not a necessary step but a helpful one). I’ll subtract one 7 to bring the number to 11 – 7 = 4.

Now, I need to know the “anchor day” for this century, which is Tuesday. Using the number 4 I just generated, I add this number of days to Tuesday to get Saturday. That means that Saturday is the “doomsday” for 2009 and therefore 7/11 — that is July 11 — is on a Saturday in 2009. July 4 — or 7/4 — is exactly 7 days before 7/11, so July 4 will be on a Saturday this year. (He looks at the calendar to check, and — huzzah! — it is, indeed, correct.)

OK, so I could have looked at the calendar for that one to begin with. But for days of long ago — or days still ahead of us — it is a fun way to spend a little time. Or even for days not so far away: we went through each of the kids’ birthdays last night, for instance, figuring out what day of the week they were born on.

July 4, 1776? Thursday. July 14, 1789? Tuesday. October 21, 1492 (Julian Calendar: October 12, 1492)? Monday — and day 7 of the Feast of Tabernacles, by the way.

Of course, fun is in the eye of the beholder, I suppose. 🙂

Man…you’ve really come a long way downhill recently. From Euler’s number to THIS! You should fast and repent next “Saturday”! Phi on you, Mr. Smith.