Allan Wechsler <acwacw@gmail.com> wrote:

I'm guessing that anybody that has dived into calendar math for any time at all can figure this one out.

I don't want to give spoilers, so I will digress a bit, with apologies to Keith. A vaguely related curiosity is that given dates of the year fall on the seven days of the week with *almost* equal probability -- almost but not exactly equal.

Right. Most notoriously, the 13th is slightly more likely to land on a Friday than to be any one of the other six days of the week.

This is because the central postulate of the Gregorian calendar is that 400 years equals exactly 146,097 days. Unfortunately, this period in days is exactly divisible by seven, so the day-of-the-week pattern in our calendar repeats exactly every four centuries.

Right. Isaac Asimov got this wrong, saying that the calendar repeats every 2800 years. As you said, it repeats every 400. (Except for the dates of Easter, which repeat every 5,700,000 years.) But the calendar repetition period has nothing to do with the solution to my problem, which would be the same even if the calendar never repeated. (An example of a non-repeating calendar is one in which a day is added to the year for every odd number the year number is divisible by and subtracted from the year for every even number the year number is divisible by. The average year length would then be a transcendental number, hence the calendar would never repeat. (There would also be zero-length and negative-length years, but those would be extremely rare.))

Calendar math is full of weirdness like this. I *still* haven't wrapped my head around the Hebrew calendar. The Islamic calendar is far simpler in principle, and (intentionally) full of mystery in practice.

I thought the Islamic calendar was unpredictable, as it depends on when, after each new moon, the moon is first observed by someone in authority, which in turn depends on the weather. The solution to a related but more difficult calendar puzzle *does* depend on the details of our calendar. What about power-of-ten dates of *all* months? Is there any day which is *not* such a day? (Using the proleptic Gregorian calendar.) The 1st and the 10th of each month are of course the 0th and 1st powers of 10 of that month. But what about the other 26, 27, 28, or 29 days of that month? About once a month is the 100th day of some previous month, and about once a month is the 1000th day of some other previous month, and about once a month is the 10,000th day of some other previous month, etc. So most likely yes. But with an average of about 30.4 days per month, we'd have to go to at least 30th powers of 10, meaning most such previous months would be long before the Big Bang. Probably a lot more than 30th powers, since (presumably) nothing keeps such days from landing on the same dates, leaving other dates empty. Since there's no shortage of negative-numbered years (given that we've abandoned the constraint that the Gregorian calendar, the Earth, or even the universe, needs to have existed at the time) is it inevitable that every day has this property? Not necessarily, since the calendar does repeat. Successive powers of 10 number of days are 2.7379070069885... 27.379070069885..., 273.79070069885... years ago, etc. Since that number (400/146097) is rational, the decimal expansion repeats, so there are only a finite number of distinct fractional numbers of years ago, no matter how many powers of 10 you try. I don't know the period of repetition of 400/146097, but since 146097 is 3^3 * 7 * 773, I think it's at most 6*772=4632 decimal digits. Also, since the calendar does repeat, if we find solutions for 146097 consecutive days, we'll have found a solution for all days. A simpler version is what if every month had the same length? If so, I'm pretty sure that there would be no solutions for some days unless that month length was relatively prime to 10.