1.) I simply pre-memorized all 366 possible dates as a value from -3 to 3. 2.) Then I just add that to one of the 4 possible century codes of 2,0,-2, or 3 (7 possible century codes for Julian from -3 to 3) and finally, 3.) Add that to a pre-memorized value for each of the 100 years from 00 to 99. For example, my birthdate of March 19, 1982 goes: -2+3-3=-2 so Friday! Or shortcutting, I just cancel the 3 and -3, and -2 is Friday! The fastest way to do this is to pre-memorize all 100 year values from 00 to 99. Then I just add that to the century code and finally add that to a pre-memorized value for each of the possible 366 dates (these 3 values can be added together in any order you like). For example, June 15, 2012 would simply be: 2+2+1=5, so Friday! June 15 always has a value of 2 (pre-memorized) 2000's has a century code of 2 and, year 12 always has a value of 1 (pre-memorized) 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.
@@mirkosabato8205 Just practice, it’s actually a lot easier than you think, since there’s a pattern to it all. For example if i forget a code or a number, I can just reference it to one nearby to get the answer. I memorized 365 different values, one for each day on a common year, plus all 100 year codes. If I forget for example that 99 is code 4 or -3, i can just think of 98 as code 3 and just add one to it. Or if I forget March 18 is code -3, then I just remember March 19 is code -2 and go from there. None of the codes or things necessary to memorize are random, they all follow an easy pattern that just repeats over and over without changing.
There is a simpler method instead of calculating leap years and then subtracting to figure out normal years. For every year add 1. For every leap year add extra 1 (this is equivalent to leap year having 2 extra days and normal year having 1 extra day ).
I simply use a version of Conway’s Doomsday Algorithm that i adjusted a bit. I can calculate any day of the week in just a second or so infinitely into the future or the past for both Julian and Gregorian AD and BC as well.
I'm also a user of the Doomsday Algorithm, I can calculate the day of the week from 1582 to the distant future. I do calculations for the Julian calendar as well. However, I don't know how to calculate days from the BC era. Can you tell me how it is done as well?
Sir meri attendance lagaiye👍🏻 Amazing Video to amaze friends 😍🔥 This is a time of celebration of Diwali and other festivals. May the lights of Diwali brighten each day of your life🪔❤
Great Paryag ❤️ keep it up Brother dear.... Can you some Maths to describe space and the it's smallest Possible Value. And is space makes meaning at the smallest level and what doe it comprises of ?
Blessed solar return, my sincerest wish is that your life will continue to be filled with many more years of elation. May this day be an occasion for you to realize all of your desires for elation and manifest marvelous things. I wish you tremendous success in all endeavors you undertake soon, and may God continue to bless you abundantly. Have a blessed day
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.
I suggest learning the Doomsday Algorithm method. Based on my experience, it is the simplest method for calculating the day of the week for any date. So simple that it can be done mentally.
I am still confused as to how 1800 is not a leap year because if 1600 was a leap year, 200 years would pass to be 1800 and 200 is divisable by 4. So, 1800 should be a leap year. May I kindly get an explanation as to how 1800 is not a leap year or was this an error in the video?
It is commonly known to have leap year in every four years but there's also a rule that for every hundred century (e.g. 1600, 1700, 1800,...2000) a leap year must be skipped if it is not divisible by 400. This is to adjust the calendar of the extra time those leap year every four years accumulates. So they need to skip the years 1700, 1800, 1900 since it is not divisible by 400. Also, in the next year 2100, 2200, 2300 will also skip leap years.
Hello matho genius I am big fan you and always waiting for your new video! I have gain a lot of inspiration of maths from you! I have maths competition next week and i don't know what to practice
My method, which I discovered myself, for the day in the year, if the year is between 1900 and 2100 (which means no weird or annoying dates). I wrote it c++ style, which should be easy to follow, because I don't know c++ either. x=months y=years-1900 If x>2: x=x-3 Else: x=x+9 y=y-1 The date is: z+⌊z/4⌋+2y+⌊(6y+4)/10⌋+days (mod 7) I can't remember the day it should be on, but using my algorithm, today (Wednesday) is 5+16+122+30+2≡0 mod 7, so Wednesday is at 0 modulo 7. This means that you need to add 3 to get to the intuitive date, where Wednesday is 3 modulo 7. My algorithm is not onky much shorter, it is also a lot faster. Try it. I won't spoil the date :)
I like it, but mine is even faster! Mine works in a second or less. The problem with my method is that it requires a lot of memorization. The payoff is super fast results that are actually very easy after the memorization and practice has been done.
1.) I simply pre-memorized all 366 possible dates as a value from -3 to 3. 2.) Then I just add that to one of the 4 possible century codes of 2,0,-2, or 3 (7 possible century codes for Julian from -3 to 3) and finally, 3.) Add that to a pre-memorized value for each of the 100 years from 00 to 99. For example, my birthdate of March 19, 1982 goes: -2+3-3=-2 so Friday! Or shortcutting, I just cancel the 3 and -3, and -2 is Friday! The fastest way to do this is to pre-memorize all 100 year values from 00 to 99. Then I just add that to the century code and finally add that to a pre-memorized value for each of the possible 366 dates (these 3 values can be added together in any order you like). For example, June 15, 2012 would simply be: 2+2+1=5, so Friday! June 15 always has a value of 2 (pre-memorized) 2000's has a century code of 2 and, year 12 always has a value of 1 (pre-memorized) 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 believe my method is the shortest possible, however most probably most people won’t bother to memorize all the items that I did in order to keep everything so short and sweet. After a lot of practice I realized that instead of calculating year codes and month codes with dates, that it can just be memorized to the point of almost no calculation. I got to the point that I can calculate the day of the week for any date much faster than I can even figure out a person’s age based upon their birthdate.
@@thecalendarninja Your method is definitely the fastest, it goes straight to the point. My method is just a compromise between learning time and calculating time. I shift January and February into the previous year, and the amount of days in the months has a recursive representation, implying that the sum of the months can be written as [ax+b], a linear equation rounded down, or to the nearest integer. This famous sequence of zeros and ones 010110101101101011... can be expressed recursively by f(0)=01, f(1)=011, where the ratio between the zeros and the ones approaches the golden ratio. How did you get your rare hobby?
I hope you enjoyed !
Can you do a video on the doomsday algorithm
okay will look at it
Love from Bangladesh 🇧🇩🇧🇩🇧🇩
Really very well ...
its 400 too
thank you so much sir , you've teached me so much more than my teachers
1.) I simply pre-memorized all 366 possible dates as a value from -3 to 3.
2.) Then I just add that to one of the 4 possible century codes of 2,0,-2, or 3 (7 possible century codes for Julian from -3 to 3) and finally,
3.) Add that to a pre-memorized value for each of the 100 years from 00 to 99.
For example, my birthdate of March 19, 1982 goes:
-2+3-3=-2 so Friday! Or shortcutting, I just cancel the 3 and -3, and -2 is Friday! The fastest way to do this is to pre-memorize all 100 year values from 00 to 99. Then I just add that to the century code and finally add that to a pre-memorized value for each of the possible 366 dates (these 3 values can be added together in any order you like). For example, June 15, 2012 would simply be:
2+2+1=5, so Friday!
June 15 always has a value of 2 (pre-memorized)
2000's has a century code of 2
and,
year 12 always has a value of 1 (pre-memorized)
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.
How did you memorized like 500 different numbers?
@@mirkosabato8205 Just practice, it’s actually a lot easier than you think, since there’s a pattern to it all. For example if i forget a code or a number, I can just reference it to one nearby to get the answer. I memorized 365 different values, one for each day on a common year, plus all 100 year codes. If I forget for example that 99 is code 4 or -3, i can just think of 98 as code 3 and just add one to it. Or if I forget March 18 is code -3, then I just remember March 19 is code -2 and go from there. None of the codes or things necessary to memorize are random, they all follow an easy pattern that just repeats over and over without changing.
@@thecalendarninja it is for every date ever even in julian calendar? wow! i think you need to make a tutorial about this. it is interesting
@@krukrok5218 yes sir
So where can I find the 365 date values so I can memorize it ?
Wow I learn something new every time I watch your video. Keep it up👍
You're a Briliant guy sir
I realy apreciate your work,love your calming vibe! You explain very well im just stupid!
There is a simpler method instead of calculating leap years and then subtracting to figure out normal years.
For every year add 1. For every leap year add extra 1 (this is equivalent to leap year having 2 extra days and normal year having 1 extra day ).
I simply use a version of Conway’s Doomsday Algorithm that i adjusted a bit. I can calculate any day of the week in just a second or so infinitely into the future or the past for both Julian and Gregorian AD and BC as well.
Can U teach us that one
@@raahimkhan391 absolutely 👍 yes
I'm also a user of the Doomsday Algorithm, I can calculate the day of the week from 1582 to the distant future. I do calculations for the Julian calendar as well. However, I don't know how to calculate days from the BC era. Can you tell me how it is done as well?
absolutely
You make understand everything too briefly.❤️❤️
老師教的很好。換位思考是根本。多謝。多謝。
Thank You ❤, It helps to impress my friends. I really enjoyed it.
It excellent and elated method brother. Keep it up bro.🥰
I understood the concept sir thank you so much sir.
From your channel i learnt so many mental math sir.
Thank you sir
THANK YOU
Great playlist! Enjoyed watching it.
Sir meri attendance lagaiye👍🏻
Amazing Video to amaze friends 😍🔥
This is a time of celebration of Diwali and other festivals.
May the lights of Diwali brighten each day of your life🪔❤
Can you try July 4, 2024, please. I was trying to compute it and it was Wedenesday based on computation. However, it is actually Thursday.
Thanks bro
I am having great difficulty figuring out what you are describing because I just don't get it. Somebody help.
Great Paryag ❤️ keep it up Brother dear.... Can you some Maths to describe space and the it's smallest Possible Value. And is space makes meaning at the smallest level and what doe it comprises of ?
Thank you, Sir, 🙏🌺😇
How do you do the same for century years for example 1 january 2000? pls answer
Blessed solar return, my sincerest wish is that your life will continue to be filled with many more years of elation.
May this day be an occasion for you to realize all of your desires for elation and manifest marvelous things.
I wish you tremendous success in all endeavors you undertake soon, and may God continue to bless you abundantly.
Have a blessed day
Sir I got 80/80 because of you in 1hour in my class thank you sir we will support you but you upload Daily video please
Sir can you please teach aptitudes??
you were born thursday
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.
amazing!
Best video
I can't tell if I'm dumb, but I'm still confused lol 😓
I suggest learning the Doomsday Algorithm method. Based on my experience, it is the simplest method for calculating the day of the week for any date. So simple that it can be done mentally.
@@polygongaming8859 ok il try tysm 🥲🤚
@@marcyroblox_edits6657 Cool, let me know how it goes for you.
@@polygongaming8859 I will (when I have the time to check it out lol)
While calculating the day for srinivasan ramanunjan why did we add 3odd days for the year 1800 shouldn't it be 4 odd days
200 years have 3 odd days . that's why
Oh sorry I was making a silly mistake
And also love your content 🙂
Imagine him as a teacher 😭
Amazing .. u super... so much good
I simply do not understand all of the explaining here. Please help.
So that means march 22 2007 was a thursday
May i know from where extra 4 odd days came?
Big fan you 🙇
Tuesday
Will the Next renewal be 2000 and if its is how to do for single digits last number eg 2001,2002 ,2003 , .......
yes 2000 is a renewal one
Thursday
Happy diwali
I am still confused as to how 1800 is not a leap year because if 1600 was a leap year, 200 years would pass to be 1800 and 200 is divisable by 4. So, 1800 should be a leap year. May I kindly get an explanation as to how 1800 is not a leap year or was this an error in the video?
It is commonly known to have leap year in every four years but there's also a rule that for every hundred century (e.g. 1600, 1700, 1800,...2000) a leap year must be skipped if it is not divisible by 400. This is to adjust the calendar of the extra time those leap year every four years accumulates. So they need to skip the years 1700, 1800, 1900 since it is not divisible by 400. Also, in the next year 2100, 2200, 2300 will also skip leap years.
@@jijoayop Thanks for the explanation
I am present.
❤️
Which country do you live ?
India
Hello matho genius
I am big fan you and always waiting for your new video!
I have gain a lot of inspiration of maths from you!
I have maths competition next week and i don't know what to practice
All the best !
Noice
It’s been a year and thanks
Your birthday is on Thursday
Can you give me tips
You can look at the videos on the channel there are tonnes of tips . One thing which I can tell you , your basics should be strong.
Ohh thank you so much sir, i will keep that in my mind
I am so glad that you replied me
thursday
I am in class 10
22 July 1999 is Saturday ig
it could also be tuesday
actually it was Thursday 😂😂
My method, which I discovered myself, for the day in the year, if the year is between 1900 and 2100 (which means no weird or annoying dates). I wrote it c++ style, which should be easy to follow, because I don't know c++ either.
x=months
y=years-1900
If x>2:
x=x-3
Else:
x=x+9
y=y-1
The date is:
z+⌊z/4⌋+2y+⌊(6y+4)/10⌋+days (mod 7)
I can't remember the day it should be on, but using my algorithm, today (Wednesday) is
5+16+122+30+2≡0 mod 7, so Wednesday is at 0 modulo 7. This means that you need to add 3 to get to the intuitive date, where Wednesday is 3 modulo 7.
My algorithm is not onky much shorter, it is also a lot faster. Try it. I won't spoil the date :)
I like it, but mine is even faster! Mine works in a second or less. The problem with my method is that it requires a lot of memorization. The payoff is super fast results that are actually very easy after the memorization and practice has been done.
1.) I simply pre-memorized all 366 possible dates as a value from -3 to 3.
2.) Then I just add that to one of the 4 possible century codes of 2,0,-2, or 3 (7 possible century codes for Julian from -3 to 3) and finally,
3.) Add that to a pre-memorized value for each of the 100 years from 00 to 99.
For example, my birthdate of March 19, 1982 goes:
-2+3-3=-2 so Friday! Or shortcutting, I just cancel the 3 and -3, and -2 is Friday! The fastest way to do this is to pre-memorize all 100 year values from 00 to 99. Then I just add that to the century code and finally add that to a pre-memorized value for each of the possible 366 dates (these 3 values can be added together in any order you like). For example, June 15, 2012 would simply be:
2+2+1=5, so Friday!
June 15 always has a value of 2 (pre-memorized)
2000's has a century code of 2
and,
year 12 always has a value of 1 (pre-memorized)
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 believe my method is the shortest possible, however most probably most people won’t bother to memorize all the items that I did in order to keep everything so short and sweet. After a lot of practice I realized that instead of calculating year codes and month codes with dates, that it can just be memorized to the point of almost no calculation. I got to the point that I can calculate the day of the week for any date much faster than I can even figure out a person’s age based upon their birthdate.
@@thecalendarninja Your method is definitely the fastest, it goes straight to the point. My method is just a compromise between learning time and calculating time.
I shift January and February into the previous year, and the amount of days in the months has a recursive representation, implying that the sum of the months can be written as [ax+b], a linear equation rounded down, or to the nearest integer.
This famous sequence of zeros and ones 010110101101101011... can be expressed recursively by f(0)=01, f(1)=011, where the ratio between the zeros and the ones approaches the golden ratio.
How did you get your rare hobby?
Your birthday is on Thursday