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 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.
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.
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. 💪🤙👍
@@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.
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.
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. 🤙💪🇪🇸🇮🇹🇨🇭🇦🇷
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.
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
That’s because the Julian calendar was used back then. The only difference between the old Julian calendar and the current Gregorian calendar is that under the old Julian calendar every year ending in 00 was a leap year while under the current Gregorian calendar, years ending in 00 are ONLY leap years if they are equally divisible by 400. The years 1800 and 1900 were not leap years while 2000 was a leap year. October 14, 1066 ( Julian calendar ) was a Saturday.
@@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.
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
It's not working for me, because the day for jan 3rd 2007 was Wednesday, but when i follow your method it shows Saturday Also for leap year 3rd jan 2008 i got 0 as remainder what i have to do when i get zero, should i subtract zero by 1
Can someone help me with 17th July 1982, after adding all of the numbers i get 125 and that divides by 7 is 17.8, but there is no weekday that goes with number 8. I tryed doing this multiple times but got same results
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.
Ask somone
That's just impossible
U can't do it if ur not sure what day was yesterday and today,so u should just ask some1
Today is "today" 😩
Sun's position?
there are ancient structures erected to determine the position of the sun throughout the year for this purpose
By far the best explanation and easiest to remember!!
best explanation on youtube thanks for the formula bro i learned within 30 minutes 🙏🏽
extremely amazing..I understoood every thing just in one this video
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.
salute 🫡
Thank you so much, this video was very helpful 😍
Thank you👍 stay connected
the easiest method loved it bro❤
THANK YOU SO MUCH THIS VIDEO IS SO UNDERRATED
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.
how is that the easiest * cries *
Could you please send a link to a website with all the numbers that have to be memorized
?
Don't come to distrub us if you are a legend than explain it and come for advice 😏
give us the numbers to memorise
@@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.
Thank you very much for your video. Very Useful
It is kind of hard to learn the codes but when you do it is so easy this is very helpful
I'm surprised how well this worked for every-other date thank you so much for this method
Nice one. Subscribed. Waiting for more maths tricks. 👌👌
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.
yes, as you add 7 just as you can subtract 7
@@cpfigueiredonot working for 22\02\1979
@@shameermandodichalil7696 22+3+0+79+19 =123/7 remainder 4 which is Thursday
Here for subtracting purpose, 0 should be taken as 7. Then 7-1=6 i.e. Saturday.
Great- thanks!
thanks bro best video on this topic , able to solve all questions 👍👍👍
best fastest method I think all others took a lot of calculations
Very helpful!!
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 .
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.
Sir your notes are so much helpful can u plz make all reasoning topics
Yes, I have plan to do that.
Explanation is very good
great video
Thank you explorer!
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 😅
That's really work!👍👍
Hmmm, interesting that you can find the week-day for any date by just calculating it in your mind...cool
Best explanation to remember
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?
Not vedic math
@@eforexplorer8900 if i calcuate for 24 oct 2028 as it was leap year you told to subtract 1 from givwn answer got wrong
got monday instead of tuesday
@@manojk1849
Is oct not Jan or Feb , so you don't have to subtract...and if you don't subtract you get Tuesday which is correct.
@@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. 💪🤙👍
Thank you very helpful please make more of these🙏🙏🙏👍👍👍🎉🎉🎉
Thank you👍 stay connected
@@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.
@@theonexx762if the month is Jan or Feb then only you should deduct...🙏
@@eforexplorer8900 got it, thank you so much 🙏🏻
Thank You ❤
amazing video thanku so much
NICE....Like it !!!!
What do you do with let’s say 8.7 or 8.8 or 8.9?
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.
I calculated
For a leap year
Subtraction is only required for January and February
Meaning your birthday is not in those two months
Do clarify please.
You shotcut value of calculation 😊
This formula is for Gregorian dates only, I think.
where come the sentuary code 0 -2-4-6 ..
what is its explanation ?
Thank you
It's very helpful
how did he get the number 84 below 85?
Thank you,, next video plzz,,
Thank you very much
thank you so much bro😘
Sir kitni badi problem solv krdi Apne thankyu sir
Excellent 😊 thank you 🤝
The best
And also find 25 Dec 1992
Friday is crrct but leap year 1992...
-1 is the 5-1=4 Thursday?is it (leap)only works Jan/Feb?
Yes, exception is for Jan and Feb only.
How did you assign code number 2, 0, 6 etc to the respective years...?????? How one can think which code comes before which year????
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. 🤙💪🇪🇸🇮🇹🇨🇭🇦🇷
Don't you think that the quotient spelling is written wrong
How you make these hand writing video..please tell...which software you used..
Videoscribe
Thanks bro ❤
Nice this method is easy 😊😊😊
Why did you divide 50 by 4 and the total with 7
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.
so noce....❤❤❤❤❤
❤❤❤❤ good job ❤
Why to divide by 4
28 Mar 3136 ( being a leap year ) result should be Sat, but with pattern, it's Friday . What may be wrong?
Only Jan and Feb -1
28 + 3 + 0 + 31 + 7 = 69
Remainder of (69/7) = 6 = *Saturday*
thank you
Thank you boss 👍🏻, tomorrow i have a scholarship exam for CUET tution, i hope this can help me tomorrow 👍🏻
Best of luck for your exam.
In the step -divide by four if the sum is less than 4 then? Ans will come in decimal..how to deal with that?
Take quotient as 0.
Is the exception only for leap years with date in january or february?
Yes
What if it is not jan or feb in a leap year.. how do we calculate then for march - dec in a leap year?
Then normal calculation
Very helpful sir
Thank u so much
Nice 👍👍
Thanks for this video. Just saying you missed the letter p in the word explorer at the starting of the video
Yes that's right.
U r super awesome 👍
How do we know Whether It was a Leap Year Or Not ? Is There any Hypothesis For This ?
Divide the year by 4, if no remainder comes it's leap year. But for century like 2000, 1900, 2100 etc should be divided by 400.
In the century chart u told a basic thing 2064,2064 but here u take 99 years difference so if suppose 1300-1399 is 2 or 0 or 4...?
This method is valid for any date after 1582 AD
How can we find the day if the year won't be divisible by 4??
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
Thnx so much
Thankyou 2 marks is sure for my 2023 csat paper.
I’ve watched this over 5 times and have never gotten it to work what am I doing wrong
Maybe you could try solving the questions first than look at the video then see where you went wrong?
i can’t get this to work for at least the battle of hastings ie 14 october 1066 if somebody could help would be appreciated
15th August 1956 is leap year and this formula does not give right answer which is Wednesday.
That’s because the Julian calendar was used back then. The only difference between the old Julian calendar and the current Gregorian calendar is that under the old Julian calendar every year ending in 00 was a leap year while under the current Gregorian calendar, years ending in 00 are ONLY leap years if they are equally divisible by 400. The years 1800 and 1900 were not leap years while 2000 was a leap year.
October 14, 1066 ( Julian calendar ) was a Saturday.
Sir can u give an example of 1 Jan 1901 and 3 mar 2000
Please calculate it for 7 march 2016 . It gives sunday but it has to be monday
Do properly
It's just Monday 7+3+6+16+4=36divide by 7 we got balance 1 and as per chart it's Monday got it.
2016 is a leap year so subtract answer with 01 then u will get final answer as 0 so it's sunday
@@inayathinayath4506 Subtraction is applicable only for Jan and Feb of leap year. For March, no subtraction and the day is Monday.
Thank you sir
Regarding the exception, just to clarify, you only subtract 1 when there is a leap year AND the month is Jan/Feb? The two have to coincide?
Also, I tried 2 January 1903, which gives you 2 + 0 + 0 + 3 at first. 4 can't fit into 3 though, so how does that work?
@@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.
@@eclipticpath many thanks ❤
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
Wrong calculation
Tnx love from Bharat. ❤
I need help, what am i doing wrong: May 10th 1990: 10+1+0+90+11=112/7=16... What am i doing wrong here? Thanks in advance!
Check calculation
It's not working for me, because the day for jan 3rd 2007 was Wednesday, but when i follow your method it shows Saturday
Also for leap year 3rd jan 2008 i got 0 as remainder what i have to do when i get zero, should i subtract zero by 1
Follow the same method as shown with proper calculation
You didn't tell what to do when it comes zero remainder for leap year in the video
@@shekinajoseph 0-1=6
@@eforexplorer8900 OK but the correct answer is Thursday = 4, anyways thank you
@@shekinajoseph yes if you apply this method you will always get right answer, thank you.
What if its a leap year and you have 0 left what do you do
If month is Jan/Feb then substract... 0-1=6
How would you know if it's a leap year?
Divide by 400 if last two digits are zero otherwise divide by 4
what about a date like 2003 you have to divide that by 4?
Yes
Thank you..
It's works
Thanks
Thank uhh so much it is very easiest way to calculate 😋
Пенис
@@williamaydelotte6420 same
@@williamaydelotte6420 That insight is on a whole other level man
@@notturkeysalt1208 lol
Tqq...
13 June 2730 can you tell me about it I had tried to calculate according to the method described by you but it fails
Check calculation
Thnk u so much ❤
thanks
Sir,why you can take 2000 - 2009 as 6
Pls reply for me sir
Sir if I use this method in cbse paper then wouldn’t it violate cbse marking scheme???
Plz reply sir
This problem is for competitive exams not for CBSE.
do you have some sources on this or you created this? what's your name pls and the name of this equation?
Sources are youtube videos
Woaaaaaaaaaaaa 🔥🔥🔥🔥🔥
Can someone help me with 17th July 1982, after adding all of the numbers i get 125 and that divides by 7 is 17.8, but there is no weekday that goes with number 8. I tryed doing this multiple times but got same results
the weekday is 6 (saturday), cause 125 devided by 7 is 17,8 and 17 multiplied by 7 is 119. 125 minus 119 is then 6, witch indicates the weekday
7) 125 (17 This is quotient
- 7
------------
55
- 49
-------------
6 This is remainder
-------------
*Remainder of (125 / 7 )= 6*
Thankyou sir
can you provide pdf of this presentation , this is very interesting and understandable in a very easy method , hats off to you sir
Sorry I don't have.
How I divide 2002 last digit 02, with 4?
Here quotient 0.