Complete method: how to find the week-day for any date?

[eforexplorer]

This method is checked thoroughly. Those who are claiming wrong, i will say them to check the calculation again.

[mcmerry]

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.

[MrPathakhemant]

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.

[thecalendarninja]

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.

[there.are-no.handlesAVAILABLE.]

By far the best explanation and easiest to remember!!

[cpfigueiredo]

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!

[JustJoe-yhdw]

Excellent presentation. One thing to add concerning leap years which need to be determined when the month is either January or February. Leap years are those years that are equally divisible by 4 such a 2000, 2004, 2008, 20012 … and so on. There is 1 exception when determining a leap year which is: years that end in 00 are ONLY leap years if the year is equally divisible by 400. Therefore years 1700, 1800, and 1900 were NOT leap years because they are NOT equally divisible by 400. The year 2000 WAS a leap year because it is equally divisible by 400. Going forward, years 2100, 2200, and 2300 will NOT be leap years because they are NOT equally divisible by 400 while 2400 WILL BE a leap year. Remember, all of the above only apply to years ending in 00.

[mrpowerup]

best explanation on youtube thanks for the formula bro i learned within 30 minutes 🙏🏽

[waqasmansoor]

extremely amazing..I understoood every thing just in one this video

[rexcavalier]

This formula is for Gregorian dates only, I think.

[joepiacente]

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?

[LakshpreetKaur.Rollno]

in 1582 when the gragorian callender was started .
10 days was emitted .
but in 1582 years; surplus days
become 12 days.
plz explain this gap .

[Sahil-sjo]

guys believe me....of all the videos in youtube....this is the only video which is accurate....i have calculated many times and the answer that i get is always right

[simranmittal]

Thank you so much, this video was very helpful 😍

[vijayakumari]

Ok now 1947 August is friday but if we calculate according to this formula 15+2+0+47+16=80 and 80÷7 we get remainder as 3 but it is showing it as Wednesday so it is wrong

[nipunkarthikeyan]

Thanks for this video. Just saying you missed the letter p in the word explorer at the starting of the video

[archishmanbera]

Don't you think that the quotient spelling is written wrong

[slly_sunny]

My grandpa somehow knows how to do this so im learning this to be able to bond with him lol since we dont really have much to talk about together 😅

[Mhike]

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.

[AbdulMannan-mzwd]

It is kind of hard to learn the codes but when you do it is so easy this is very helpful