The 5th number to add is the quotient obtained from the division of the year by 12, what should I do if the year is less than 12? Suppose, 2004. Please clarify.
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.
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.
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. 💪🤙👍
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 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.
Your method works perfectly on ordinary year. But it doesn’t work on leap year but sometimes it works too. For example take 20 March 2004 it works here but it doesn’t on 16 April 2024
@@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.
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
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
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.
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.
For 1 Jan 1684 - this is not working. Calculation is Date (1) + month code (0) + year code (0) + last 2 digits of year (84) + Quotient when last 2 digits divided by 4 (21)-> Remainder (106/7) less 1 (since it is a leap year) -> 1-1 = 0, hence it should be sunday but online calendars state that this was a Tuesday. Please help
It looks like you did everything right but the problem is that this method only works for the Gregorian calendar. In 1684 very few countries used the Gregorian calendar. Most countries at that time were using the Julian calendar which was different than the Gregorian calendar by a few days. The U.S. and England didn’t switch to the Gregorian calendar until sometime around 1752. Most countries made the switch to the Gregorian calendar at different times after 1700. Italy was the first country to use the Gregorian calendar in 1582.
I believe your country was using the Julian calendar on 1 January 1684 which was a Tuesday under that calendar. I think your country went to the Gregorian calendar in 1752. This method of calculating the week day for any date only works under the Gregorian calendar. In your calculation of 1 January 1684 you used the wrong Century number. The Century number is 6. It should be 1 + 0 + 6 + 84 + 21 = 112 112 / 7 = a remainder of 0. Subtract 1 for leap year is 0 - 1 = - 1 which is a SATURDAY. Since Sunday is a 0 just count backwards ( - 1 ) to the previous day of SATURDAY. This is using the Gregorian calendar.
According to my calculations, 1st of April 2004, we get 1+6+6+04+1 = 18 Then, from 18/7 we obtain a rest of 4 as our final clue... Once 2004 was a leap year, I subtracted 1, so 4-1= 3 (wednesday) However, 1/4/2004, was on a Thursday! Where's the mistake here? Do we really need to subtract 1 on leap years??
Century chart cannot be called century chart per se as centuries do not start in 00 but 01. 21st century is not from 2000-2099 but 2001-2100. Perhaps “Hundreds Chart” is more accurate.
@@eclipticpath Ordinary year has 365 days and leap year has 366 days. *If we don't find leap years, we will get wrong answer.* Ex: In 5 years 1st year is an ordinary year. 2nd year is an ordinary year. 3rd year is an ordinary year. 4th year is a leap year. 5th year is an ordinary year. Therefore in 5 years, 4 ordinary years and 1 leap year. Total days = 4 * 365 + 1 * 366 = 1826 days.
@@manikandans7303 You only subtract 1 if the date is a leap year AND the month is either January or February. You DO NOT subtract 1 for leap years when the months are March through December.
@@manikandans7303 This method of calculating the day of the week for any date is based on 365 days. When a leap year happens, an extra day is added at the end of February making it a 366 day year. An adjustment of subtracting 1 needs to be made for all the days on and before the extra day that is added at the end of February.
0:43 WdHd (weekday hidden in date)... 0:53... 3 charts - week (sat 0 sunn 6), month(033-(aay...chhay 6maii 1 juu two 2 ). 612- 625(jul6 A had 2 legs so Aug=2 Sept S=5). 035(oct=0 looks like o.). 1:00 wkcht....1:15 month chart how to ratta see...1:35 infinite cent chat_6024_6024 ( rmbrr prant cent THISS cent== 6. 20th cent 0 all before 20c is 24"" TOO FARRRRr to b Rmbrrd TOO FARRRRR). 2:34... 6 steps see for fast, 4 quotient & 7 remainder ans @ last .
This method is checked thoroughly. Those who are claiming wrong, i will say them to check the calculation again.
The 5th number to add is the quotient obtained from the division of the year by 12, what should I do if the year is less than 12? Suppose, 2004. Please clarify.
@@subhadipmukherjee574 divide by 4
By far the best explanation and easiest to remember!!
Why did he divide 50 by 4?
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
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.
best explanation on youtube thanks for the formula bro i learned within 30 minutes 🙏🏽
but answer is tuesday 15 aug 2050
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.
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.
Thank you so much, this video was very helpful 😍
Thank you👍 stay connected
extremely amazing..I understoood every thing just in one this video
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 😅
It is kind of hard to learn the codes but when you do it is so easy this is very helpful
the easiest method loved it bro❤
I'm surprised how well this worked for every-other date thank you so much for this method
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 boss 👍🏻, tomorrow i have a scholarship exam for CUET tution, i hope this can help me tomorrow 👍🏻
Best of luck for your exam.
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.
easy way to learn 😊😊thnku brdr ❤
thanks bro best video on this topic , able to solve all questions 👍👍👍
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.
Amazing, loved the method.
Thank you so much sir ❤ it's very helpful for me 🎉😢
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 .
Sir kitni badi problem solv krdi Apne thankyu sir
Hmmm, interesting that you can find the week-day for any date by just calculating it in your mind...cool
Sir your notes are so much helpful can u plz make all reasoning topics
Yes, I have plan to do that.
Nice one. Subscribed. Waiting for more maths tricks. 👌👌
This formula is for Gregorian dates only, I think.
best fastest method I think all others took a lot of calculations
Your method works perfectly on ordinary year. But it doesn’t work on leap year but sometimes it works too. For example take 20 March 2004 it works here but it doesn’t on 16 April 2024
for me... any year perfectly divided by 4 is a leap year.
not always, end of century years have to be divisible by 400, so 1800 isnt a leap year
You shotcut value of calculation 😊
THANK YOU SO MUCH THIS VIDEO IS SO UNDERRATED
Why is it not working for 31st OCT 1957?????
No bro it's work i can explain you
=31+0+0+57+14=102
So 102/7=4 and according to week chart the 4 is Thursday ✅
Don't you think that the quotient spelling is written wrong
NICE....Like it !!!!
It is not well as I am trying to me
this is good but will take a long time to do it mentally
tq u sir
Thank you sir
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.
Best explanation to remember
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 🙏🏻
Explanation is very good
great video
Thank you very much for your video. Very Useful
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
It should be 15 + 2 + 0 + 47 +11 = 75 so remainder is 5 which is Friday correct ans
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
Here is a challenge
2020 march 31st is Tuesday
But as per this method we will get Monday
A
2020 is a leap year
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.
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.
For every leap year or only for your example we have to minus 1
@@shivatejavangala7023 😄
In leap year it is only for Jan feb or any month. Abe jaldi bta
Are re re train miss hoi gawa ka, it's for jan and feb only
So my birthday is on Monday in the year 2050!! He used my birthday date btw!! ;)
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.
Thankyou very much sir you have made 10 marks for me before my cuet exam
Thank You ❤
And if you have the first 4 years of a century, you don't Devise by 4, you just add them EG.
19.8.2001
19+2+6+1+1=29÷7=4(leftover 0)=sunday
Thankyou sir
The 15th century starts from 1401 to 1500 and not 1400 to 1499.
The 1400 is the last year of 14th century.
@@kironggg yes you are right, that was a mistake.
For 1 Jan 1684 - this is not working. Calculation is Date (1) + month code (0) + year code (0) + last 2 digits of year (84) + Quotient when last 2 digits divided by 4 (21)-> Remainder (106/7) less 1 (since it is a leap year) -> 1-1 = 0, hence it should be sunday but online calendars state that this was a Tuesday. Please help
It looks like you did everything right but the problem is that this method only works for the Gregorian calendar. In 1684 very few countries used the Gregorian calendar. Most countries at that time were using the Julian calendar which was different than the Gregorian calendar by a few days. The U.S. and England didn’t switch to the Gregorian calendar until sometime around 1752. Most countries made the switch to the Gregorian calendar at different times after 1700. Italy was the first country to use the Gregorian calendar in 1582.
I believe your country was using the Julian calendar on 1 January 1684 which was a Tuesday under that calendar. I think your country went to the Gregorian calendar in 1752. This method of calculating the week day for any date only works under the Gregorian calendar.
In your calculation of 1 January 1684 you used the wrong Century number. The Century number is 6. It should be 1 + 0 + 6 + 84 + 21 = 112 112 / 7 = a remainder of 0. Subtract 1 for leap year is 0 - 1 = - 1 which is a SATURDAY. Since Sunday is a 0 just count backwards ( - 1 ) to the previous day of SATURDAY. This is using the Gregorian calendar.
Bro, you added the year code incorrect it is 6..
And for your kind information, it was a Saturday when I searched it up
Thankyou 2 marks is sure for my 2023 csat paper.
Thank you,, next video plzz,,
According to my calculations, 1st of April 2004, we get 1+6+6+04+1 = 18
Then, from 18/7 we obtain a rest of 4 as our final clue...
Once 2004 was a leap year, I subtracted 1, so 4-1= 3 (wednesday)
However, 1/4/2004, was on a Thursday!
Where's the mistake here? Do we really need to subtract 1 on leap years??
For Jan / Feb month only
Better answer this question without doing problem you can use the time for other sums
Nice this method is easy 😊😊😊
amazing video thanku so much
Thank you
Thnx so much
Nice 👍👍
U Have Explained One Exception At Last For Leap Year, It Is Applicable For Only Jan/Feb Months Aaa Bro...🤔
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*
Very helpful sir
Tnx love from Bharat. ❤
Tqq...
Whats the logic behind this algorithm , how does it work ?
Great- thanks!
I born 2003 .. 03 not devided by 4 .. then what i do ?
Century chart cannot be called century chart per se as centuries do not start in 00 but 01. 21st century is not from 2000-2099 but 2001-2100. Perhaps “Hundreds Chart” is more accurate.
Yes you are right, i just casually gave the name -' century chart'.
Thank you very much
❤❤❤❤ good job ❤
Tell me 2 things
Why did you divided 50 by 4
And by did you divided 85 by 7
1). To find how many leap years in a given year.
2). Week has 7 days.
@@duggirambabu7792 Why do you have to find out the amount of leap years?
@@eclipticpath
Ordinary year has 365 days and leap year has 366 days.
*If we don't find leap years, we will get wrong answer.*
Ex: In 5 years
1st year is an ordinary year.
2nd year is an ordinary year.
3rd year is an ordinary year.
4th year is a leap year.
5th year is an ordinary year.
Therefore in 5 years, 4 ordinary years and 1 leap year.
Total days = 4 * 365 + 1 * 366 = 1826 days.
This is not true for all...just consider, 14 May 2004, according to this method it should be a Wednesday, but it is actually a Friday.
. 14+1+6+4+1=26 26/7 = remainder of 5 which is a Friday. I think you used the quotient of 3 instead of the remainder.
didn't work for 28 feb 2016, by this method answer comes 1 >> monday but the actual week day is SUNDAY
It was the leap year so U need to minus 1 as per taught 😊
It's very helpful
Thank you explorer!
Daut achhi please au gote video banai dia please
For 5th Oct 2016 I got Tuesday but the day was really Wednesday
Plz solve this
5+0+6+16+4=31 31/7 = 4 with a remainder of 3 which is a Wednesday
@@JustJoe-yh5dw but it is leap yr ,so we have to subtract 1 from 3 and we get 2 which is Tuesday
But the answer is Wednesday how?
@@manikandans7303 You only subtract 1 if the date is a leap year AND the month is either January or February. You DO NOT subtract 1 for leap years when the months are March through December.
@@JustJoe-yh5dw why sir or it is rule ah sir ?
@@manikandans7303 This method of calculating the day of the week for any date is based on 365 days. When a leap year happens, an extra day is added at the end of February making it a 366 day year. An adjustment of subtracting 1 needs to be made for all the days on and before the extra day that is added at the end of February.
U should make more videos..
I'll
In this if it is jan or feb but not a leap year?
I did January 1 0001 and got Saturday
What is logic behind divide by 4 and 7?
The derivation needs a separate video
Why to divide by 4
Why you multiplied by 4 and 7 only why not other number
How would you know if it's a leap year?
Divide by 400 if last two digits are zero otherwise divide by 4
Since century code for 1900 is 0. Should I add the zero in the equation?
adding zero doesn’t do anything
Yeah 😂
@@Orion225 so it wouldn’t matter if you add it or not
Very helpful!!
0:43 WdHd (weekday hidden in date)... 0:53... 3 charts - week (sat 0 sunn 6), month(033-(aay...chhay 6maii 1 juu two 2 ). 612- 625(jul6 A had 2 legs so Aug=2 Sept S=5). 035(oct=0 looks like o.). 1:00 wkcht....1:15 month chart how to ratta see...1:35 infinite cent chat_6024_6024 ( rmbrr prant cent THISS cent== 6. 20th cent 0 all before 20c is 24"" TOO FARRRRr to b Rmbrrd TOO FARRRRR). 2:34... 6 steps see for fast, 4 quotient & 7 remainder ans @ last .
That's really work!👍👍
i think week chart is wrong
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?