Here goes a much more difficult question...how to know what day is today if you don't have a calendar, a date a phone, etc...you just have your compass and a clear night.
Year part should always be previous year( completed), rest codes need to be revised accordingly ; date folling in Mar.-Dec. of Leap year:Total to be deducted by 1 to get correct day. I have programmed in QB64 and I get correct result for any year 0000 -4000, checked with Mobile & website.
The fastest way (not easiest until after doing the required memorization) to do this is to prememorize all 100 year values from 00 to 99. Then I just add that to the century code and reduce mod7 and finally add that to a prememorized value for each of the possible 366 dates. For example, June 15, 2012 would simply be: 2+2+1=5, so Friday! June 15 always has a value of 2 (prememorized) 2000's has a century code of 2 and, year 12 always has a value of 1 (prememorized) 2+2+1=5, Friday! Anytime leapyear happens just subtract one from the answer for January or February only, all other months stay the same! I calculate any day of the week using my method in less than a second, almost instantly to 2 seconds tops if I'm thinking a bit. I do both the Julian and Gregorian calendars in both AD and BC infinitely into the past or the future.I do other cool tricks that aren't heard of very often, for example since this calendar math can be viewed as just a simple addition problem, we can also solve for a missing value using subtraction! For example, if we already know the day of the week to be Friday for example, we can solve for a possible month/months or even a date from 1 to 31 or even a possible year or century. Example: Which month or months did Friday the 13th fall on in 1899? 1899 was a Tuesday year (-2+4=2) or (5-3=2). So, Tuesday was the 10th since 3 days later was Friday the 13th. So, 2+13+x=5 (Friday) x=4 because 15+(-10)=5 -10+14=4 Mod7 arithmetic Only January and October had a month code of 4 😁. So it only happened twice that year, in January and in October. Also we can say more intuitively that since 1899 was a Tuesday year, and Friday being 3 days later, which month/months had the Doomsday of 3,10,17,24,31? and obviously that is only January and October. If the question changed to what were all of the Fridays in January and October for example in 1899? We can say "the 6th, 13th, 20th, and 27th" If someone were to instead ask for example on Friday the 13th from 1883 to 1900 what year/years did that happen? We can say "only in 1893 and 1899" In this situation we just needed to find the missing year code of (4 or -3) and only 93 and 99 have that year code in that date range. Likewise, we could have instead solved for the century 😉 also if already given a year.
@@navlikesdesign I apologize, it's not necessarily easy until after all the necessary memorization is done. It's the fastest way though for sure, which is the payoff for doing the memorization required to use my method.
The original goal of the Gregorian calendar was to change the date of Easter. In 1582, when Pope Gregory XIII introduced his Gregorian calendar, Easter, traditionally observed on March 21, fell further away from the spring equinox by 10 days, not 12. Since it was the Council of Nicaea in 325 that decreed that "Easter should be observed on the first Sunday following the first full moon after the spring equinox on March 21", and by that point the Julian calendar was already off by two days, it stayed that way. So Gregory fixed the calendar only up to 325 skipping 10 days. And the two days error from 53 B.C to 325 A.D stayed with us to this day.
That depends of century years. Still. Our century from 2000 till 2099 have the *CODE 6* , the last CENTURY from 1900 to 1999 had *CODE 0* , beetwen 1800-1899 had the *CODE 2* and beetwen 1700-1799 had the *CODE 4* . So every 4 century will have the same code, for example 1700-1799 = 2100-2199 = *code 4* !! About all of those CODES have a reason. 🤙💪🇪🇸🇮🇹🇨🇭🇦🇷
There are so many explaining the formula, This is the only one that no questions have to be answered Thank you very much! You made it so simple to understand. Do you also teach Vedic math?
@@manojk1849 yet not!! You need to subtract *-1* only if you want to try to find the day of week from JANUARY and FEBRUARY about every LEAP YEAR. Example for your date *24 Oct 2028* even if is a leap year, you don't need to subtract - 1. If the date were 16 February 2028 then you need to SUBTRACT -1. So the day of week from 24 Oct 2028 will be : 7+0+6+24+0= 37 = 37:7= *2 remainder* = Tuesday and NOT how you thought before. 💪🤙👍
Their is hidden the formulae ofcourse.7 means it's one week yes I think it's because of we have calculated it for the year having 28 days for February month it is my guess.for leap year calculation will differ.
@@eforexplorer8900 I have a question, why does 6th of July 1944 (Thursday) not work. Only when I don't count the -1 from the total of 4. But that doesn't make sense, because its a leap year and I should technically deduct it.
@@dunlop9161 That's correct, 2.1.1903 is Friday. 4 goes into 3 zero times so you add 0. And yes, only when both conditions (leap year&jan/feb) are met, you subtract one.
The date 5 Feb 2000 results in a final remainder of zero: (5 + 3 + 6 + 0 + 0) / 7 = 2 (with remainder 0). This date fits in the exception of Jan or Feb in a leap year. Hence I should subtract 1 from the remainder, but in this case I would get -1. I suppose I should "wrap" this result such as -1 + 7 = 6 ? I confirmed that this date was a Saturday!
@Anaya Ch It has worked for me. I even coded this method in a microcontroller. The only issue I had with the method was the one I mentioned in the previous comment. If it is not working for you, maybe you are missing some little detail.
First, Get the day that falls under the code 0... That is sunday. And when you subtract 1 day from sunday, it will give you SATURDAY. That is the answer.
Every year is divisible by 4. If the year is less than 4 the quotient is 0 and remainder is the year. If the year is 00 00/4 = quotient of 0 and remainder of 0 If the year is 01 01/4 = quotient of 0 and remainder of 1 If the year is 02 02/4 = quotient of 0 and remainder of 2 If the year is 03 03/4 = quotient of 0 and remainder of 3
think of it as 0802 for example (802) but use 0 for a placeholder. and if the last 2 numbers are less than a 4 like in this case, then the quotient will be 0
The name of that mathematician you’re talking about is John H. Conway. He figured out the algorithm to finding out the day for a past or future date by just using one hand. Btw, I checked my birthdays with 3 leap years (2024, 2028, 2044) and no subtractions were necessary. They were all correct with just regular calculation. Perhaps subtraction isn’t necessary.
