Re: [math-fun] September 10000, 1993
"Keith F. Lynch" <kfl@KeithLynch.net> wrote:
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? . . .
Also, since the calendar does repeat, if we find solutions for 146097 consecutive days, we'll have found a solution for all days.
I have done exactly that, and confirmed that *every* day can indeed be expressed as day 10^N of some prior month in infinitely many ways, and that the most recent such N never exceeds 306. Here are the last few and next few days: W 1/13/2021 4/100000/1747 R 1/14/2021 12/10^55/1795 F 1/15/2021 6/10^64/1971 S 1/16/2021 9/10000/1993 N 1/17/2021 6/10^18/1913 M 1/18/2021 4/10^35/1954 T 1/19/2021 8/10^183/1833 W 1/20/2021 9/10^28/1963 R 1/21/2021 11/10^52/1790 F 1/22/2021 6/10^176/1713 S 1/23/2021 4/10^23/2087 N 1/24/2021 5/1000/2018 M 1/25/2021 3/1000000/2083 T 1/26/2021 2/10^12/1814 W 1/27/2021 1/10^16/1736 Note that the year on the right is usually wrong, sometimes by an immense number of years. But always an exact multple of 400 years. For instance the 24th of this month is of course not actually day million of March 2083, it's day million of March 718 BCE, exactly 2800 years earlier.
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.
It is exactly 1/4 of that. 1158 digits. Hence January 16 wasn't just September 10^4, 1993, it was also September 10^1162 of an enormously earlier year, September 10^2320 of an even earlier year, ad infinitum. Phrased another way, 10^1159 = 1 (mod 146097). My full table, which shows just the most recent solution for each day of a full Gregorian period (400 years = 146097 days), can be downloaded from http://KeithLynch.net/146097.txt. The code that generated it can be downloaded from http://KeithLynch.net/146097.c.
Except for the fact that there was certainly no month more that 10^20 (or so) years ago, because the universe didn't exist yet and/or more specifically, time as we define it doesn't reach that far into the past. But yes, the answer to the original question relies only on residues (mod 7) of powers of 10 , so it is essentially the same for any calender that has weeks of seven days and no exceptions to that. It is quite obvious that one particular value will never appear as such a residue, slightly less obvious that all others will (and do so for N < 7 — incidentally corresponding to roughly the age of mankind). - Maximilian On Tue, 19 Jan 2021, 22:56 Keith F. Lynch, <kfl@keithlynch.net> wrote:
I have done exactly that, and confirmed that *every* day can indeed be expressed as day 10^N of some prior month in infinitely many ways, and that the most recent such N never exceeds 306.
(...)
September 10^4, 1993, it was also September 10^1162 of an enormously earlier year, September 10^2320 of an even earlier year, ad infinitum. Phrased another way, 10^1159 = 1 (mod 146097).
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Keith F. Lynch -
M F Hasler