Trick 35 - Find the Day of Any Date II Calendar Problems II Complete Video II Concept & Examples

[@travelogues8895]

Why my maths teachers can't be this patient while explaining..?!

[@jaysong3256]

Cause not every student understands this. They will be more confused :))

[@suryamishra6838]

How lucky I have been to find this vedio🙏 . Simple explaination no short tricks. Thank you sir.

[@LovelySoul510]

Thanks sir for u valuable lessons It's made easy to solve this type of questions I spent 20- 30 minutes to solve the 1 st one but after practicing 2 ,3 questions now I am able to solve this in just 2-3 minutes 🙏❤️🙏❤️🙏

[@prashantsaindane4193]

100% Working. Taking very less time after Practice

[@chilli130]

Such a wonderful video I was searching from many years

[@entertainmentttttttt-o9b]

I found exactly what I needed. Those videos with month codes were frustrating, but this solution is simple and doesn't require memorization. Thanks sirr!❤

[@kAY-yl5en]

Very helpful. Concise explanations and straight to the point with no background music. Just plain mathematics. Love it.

[@diptigidwani6255]

The best explanation. I was so confused regarding this topic bt you explained it so well. Thank you so much for clearing my doubt.

[@srivastavamajjari]

This is a normal method but explanation is simple. It is good for beginners..

[@dylanalexander6172]

Suresh, you are superb. Maths is so much fun with your tricks.My son loves them

[@snowwhite5008]

I had no idea of odd days and even days before ... Now I got the clear idea.... Good explanation sir

[@nopesorry570]

Understood the concept extremely fast. Thank you so much, sir. Very well explained 😊😊

[@Xoxcmasf]

absolutely 💯💯 well explained concept

[@SapnaKumari-mj1uw]

Thanks sir. You cleared the Concept. It's better to understand the Concept rather than going for shortcuts.

[@tayyabaliamer4824]

Very good explanation, this the only video I find without a table to be remembered, your work is appreciated sir, keep on doing the same

[@answhiganswhig377]

Exactly ! because he explains the logic of it. So you don't need to memorize figures you can get them by yourself 🤗

[@drishtibissa]

very helpful sir...thank u so much for this amazing trick and session

[@redditdark_stories]

watching this from Boston right now. Jai hind 🇮🇳

[@Fragmented_2023]

Thankyou very much for helping out the basics with clarity

[@user-lp4um5gv2l]

Now I can find the day of the week of a given date easily....thank you...

[@Shreya-g9j]

Bestest teacher I found on YT

[@prachialdangadi9663]

Not only for children appearing for competitive exams, these tricks can help adults too. Great videos Sir !

[@rojnibibi2975]

L

[@NganthoibaKhangenbam]

Thank your Guruji, I can understand only in this time.

[@AmanRaturi1]

All this in one take, with such great explanation skills, is truly commendable. Keep up the great work sir.

[@bindu277]

😄

[@bindu277]

@Fashion Hub kaha pr?

[@sureshaggarwal]

Thanks a lot

[@rajuchoudhary5506]

Kuch log kya kya tricks yaad karwa dete he, but this video is fully conceptual and remember for a long time, thank u

[@lookingfornirvana9744]

Easly understood sir ..nice way to explain the concepts...

[@nalandabasu8641]

Today morning I learnt this and now you taught.. Ur was better.. The morning session went tangentially

[@varadaradj]

Just fantastic. Very clear explanation. Thank you very much

[@harshchouhan7454]

you made the concept so clear, thank you sir i understood the calendar problems :)

[@ashitagupta3806]

Way of explaining the concepts is awesome!

[@RohitParthasarathy-wt3gt]

Thanks a lot sir. this helped me a lot

[@thecalendarninja]

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 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.

[@surajmahale4187]

Great man !! can you share pre memorized values. i'm very excited

[@surajmahale4187]

Can you please teach me your technique I'll be grateful forever.

[@victorynews6184]

Sir i dint understood ua method,, aggarwalz sir tricks also much complicated , plz explain once

[@thecalendarninja]

@@surajmahale4187 Grab any common year (not a leap year) calendar you can write on. Starting on January 1 write -2, then January 2 write -1, then January 3 write 0 The pattern goes -2,-1,0,1,2,3,-3 repeat all the way until December 31 which has a value of -2. Century Codes: 1600's, 2000's, 2400's and all multiple of 4 centuries: code 2 1700's, 2100's, 2500's and so on code 0 1800's, 2200's, 2600's and so on code 5 or -2 1900's, 2300's, 2700's and so on code 3 Year codes: *00 code 0 01 code 1 02 code 2 03 code 3 *04 code 5 or -2 05 code 6 or -1 06 code 0 07 code 1 *08 code 3 09 code 4 or -3 10 code 5 or -2 11 code 6 or -1 *12 code 1 13 code 2 14 code 3 15 code 4 or -3 *16 code 6 or -1 17 code 0 18 code 1 19 code 2 *20 code 4 or -3 21 code 5 or -2 22 code 6 or -1 23 code 0 *24 code 2 25 code 3 26 code 4 or -3 27 code 5 or -2 *28 code 0 29 code 1 30 code 2 31 code 3 *32 code 5 or -2 33 code 6 or -1 34 code 0 35 code 1 *36 code 3 37 code 4 or -3 38 code 5 or -2 39 code 6 or -1 *40 code 1 41 code 2 42 code 3 43 code 4 or -3 *44 code 6 or -1 45 code 0 46 code 1 47 code 2 *48 code 4 or -3 49 code 5 or -2 50 code 6 or -1 51 code 0 *52 code 2 53 code 3 54 code 4 55 code 5 or -2 *56 code 0 57 code 1 58 code 2 59 code 3 *60 code 5 or -2 61 code 6 or -1 62 code 0 63 code 1 *64 code 3 65 code 4 or -3 66 code 5 or -2 67 code 6 or -1 *68 code 1 69 code 2 70 code 3 71 code 4 *72 code 6 or -1 73 code 0 74 code 1 75 code 2 *76 code 4 or -3 77 code 5 or -2 78 code 6 or -1 79 code 0 *80 code 2 81 code 3 82 code 4 or -3 83 code 5 or -2 *84 code 0 85 code 1 86 code 2 87 code 3 *88 code 5 or -2 89 code 6 or -1 90 code 0 91 code 1 *92 code 3 93 code 4 or 03 94 code 5 or -2 95 code 6 or -1 *96 code 1 97 code 2 98 code 3 99 code 4 or -3

[@thecalendarninja]

@@surajmahale4187 Grab any common year (not a leap year) calendar you can write on. Starting on January 1 write -2, then January 2 write -1, then January 3 write 0 The pattern goes -2,-1,0,1,2,3,-3 repeat all the way until December 31 which has a value of -2. Century Codes: 1600's, 2000's, 2400's and all multiple of 4 centuries: code 2 1700's, 2100's, 2500's and so on code 0 1800's, 2200's, 2600's and so on code 5 or -2 1900's, 2300's, 2700's and so on code 3 Year codes: *00 code 0 01 code 1 02 code 2 03 code 3 *04 code 5 or -2 05 code 6 or -1 06 code 0 07 code 1 *08 code 3 09 code 4 or -3 10 code 5 or -2 11 code 6 or -1 *12 code 1 13 code 2 14 code 3 15 code 4 or -3 *16 code 6 or -1 17 code 0 18 code 1 19 code 2 *20 code 4 or -3 21 code 5 or -2 22 code 6 or -1 23 code 0 *24 code 2 25 code 3 26 code 4 or -3 27 code 5 or -2 *28 code 0 29 code 1 30 code 2 31 code 3 *32 code 5 or -2 33 code 6 or -1 34 code 0 35 code 1 *36 code 3 37 code 4 or -3 38 code 5 or -2 39 code 6 or -1 *40 code 1 41 code 2 42 code 3 43 code 4 or -3 *44 code 6 or -1 45 code 0 46 code 1 47 code 2 *48 code 4 or -3 49 code 5 or -2 50 code 6 or -1 51 code 0 *52 code 2 53 code 3 54 code 4 55 code 5 or -2 *56 code 0 57 code 1 58 code 2 59 code 3 *60 code 5 or -2 61 code 6 or -1 62 code 0 63 code 1 *64 code 3 65 code 4 or -3 66 code 5 or -2 67 code 6 or -1 *68 code 1 69 code 2 70 code 3 71 code 4 *72 code 6 or -1 73 code 0 74 code 1 75 code 2 *76 code 4 or -3 77 code 5 or -2 78 code 6 or -1 79 code 0 *80 code 2 81 code 3 82 code 4 or -3 83 code 5 or -2 *84 code 0 85 code 1 86 code 2 87 code 3 *88 code 5 or -2 89 code 6 or -1 90 code 0 91 code 1 *92 code 3 93 code 4 or 03 94 code 5 or -2 95 code 6 or -1 *96 code 1 97 code 2 98 code 3 99 code 4 or -3

[@jagannathbehera4519]

it's the best explanation ever seen, thank you, sir.

[@nargizaxusanova7488]

Thank you 😊😊😊

[@swathiworldkitchen7530]

nice explanation. It is very helpful to me in revision of calendar chapter

[@f3arxtit4n88]

Very good explanation sir ! Saw a similar question come in my bitsat mock test and when i saw it i got worried , though i am not worried anymore because of you! Thank you sir. 😄

[@sapnakaur1200]

Commendable board organisation. beautifully explained

[@CuratedPile]

Sure this works, but here's a simpler method that works for years 1900-2099. You only track one number throughout this method which makes it easy to do quickly in your head. The key is doing the mod 7 (remainder after dividing by 7), you can apply it any time to any number after step 2 through the calculation. 1) Subtract 1900 from the year. 2) multiply that number by 1.25 (add a quarter), round down, if its a leap year subtract 1. Reduce to mod 7. 3) walk through the completed months of the target year and add the mod 7 for that month (31 days=3, 30 days=2...). Reduce to mod 7. 4) add the target day of the month (or the mod 7 of it) and get the mod 7 for the sum. This is your day of the week (0=Sun, 1=Mon,.. 6=Sat). Here's an example for a date later this year, 2019-06-21. 1) 2019-1900 = 119 2) 119 + 29.75 = 148 (rounded down), 148 mod 7 = 1 3) 1 + 3 (Jan) + 3 (Mar) + 2 (Apr) + 3 (May) = 12, 12 mod 7 = 5 4) 5 + 21 = 26, 26 mod 7 = 5, Answer is 5 - Friday (if you prefer, 5 + (21 mod 7) or 5+0) For 1960-03-01 1) 60 2) 60 + 15 - 1 (leap year) = 4 (after mod 7) 3) 4 + 3 (Jan) + 1 (Feb) -> 1 (after mod 7) 4) 1 + 1 = 2 = Tuesday This method works because of 2 factors. Jan 1, 1900 is a Monday (day 1 on date 1), and in that year span there are leap years every 0.25 years so you can calculate your "odd days" with the multiply by 1.25 trick. The other advantage to this method is that it doesn't require any memorization other than the method itself and the number of days per month.

[@CuratedPile]

I had @Sunanda Alapati reply to my comment, but I can't see it anywhere other than my notification. The question was about January 1, 2099 coming up as a Thursday using my method and Sunday using the video method. The correct day of the week is Thursday, I get that with either method, I don't know why you are getting Sunday. For the video method, you have no odd days for the century, 122 odd days for the year (74 normal years, 24 leap years) and 1 more day for the 1st. That leaves a 4 as the remainder which is Thursday.

[@yashrupani1608]

@Sunanda Alapati by the video method : 2099 2000 = 0 odd day 98 = 24x2(98/4*2odd days for leap year) + 74(1odd day for normal year)= 48+74 = 122 @ 122/7 odd day = 3odd days . Jan 1 = 1odd day .. i.e; 3odd day + 1odd day = 4odd day = Thursday ---- SOLUTION BY VIDEO METHOD

[@mohammadahmadrocks]

Thank you so much sir for the best explanation 🙏🙇‍♂️

[@mvjmr7892]

Way of explaination was really outstanding sir

[@kshitijadhongade4179]

Thanks a lot sir..you made it very simple and effective...

[@tejasvkumar8059]

you the best teacher compared to the highly paid teachers in the coaching centres like carrer launcher,aakash etc you made my concept crystal clear thank you so much sir i was worried because i was not able to grasp the concept in carrer launcher in 4 long hours and i was searching on the internet for the exact same method for whole 1 day

[@deepakydv209]

Go on career will

[@vijayasreevemula3028]

Y

[@nagajothijothi4772]

Ml

[@RehanKhan-vc1bc]

I0jlqp0q7

[@RehanKhan-vc1bc]

I0jlqp0q7

[@zaaaaays1]

Sir today i feel happy because this video clear my doubts... Thanks for sharing your concepts

[@sureshaggarwal]

Do share the links with your WhatsApp groups and contacts

[@EchoDG04]

Really helpful, thanks a lot sir

[@baskaransubramanian2908]

Good clear teaching.pls keep it coming and enlighten

[@shatakshisingh8508]

Thank you sir you literally saved me bcz after 3 days I had my exams and I was really scared but after watching your video I understood the concept very well

[@assaults614]

👏👏👏👏👏👏👏ur clear explanation is the fastest way for our learning ❤❤

[@ahsanamin3818]

The video is very helpful for me. many many thanks to you sir. but I have a question. when you divide a century by 4, you minus by 1 of the result . you explained that 100 is not a leap year. my question is should we minus by 1 if the year is not divisible by 400 ??

[@rajathowlader127]

Thank you sir for explaining in such a simple way, now I can compute the day of any date, thanks a lot sir.

[@wolferxgo]

Great.... Amazing, step by step, without any trick or jumps, very cool..

[@shubhamgoyal2960]

Best explanation guru ji.

[@priyadey810]

It is really very helpful sir. thank u so much

[@Moonlightcookie123]

Thanks for posting such an informative video, Mr Aggarwal.

[@christophhogl1151]

The Doomsday method (1970 Conway) requiring 3 tables, one sentence and two divisions by 4 and 12 and 3 additions is far easier/faster and much more versatile as it also gives anchoring dates...

[@sureshaggarwal]

Thanks for that....will try it.

[@GoldenElixir8971]

Too much to memorise and less to understand in doomsday

[@jbsstudy1541]

@@sureshaggarwal sir 2000 aru 2001,2002,2003 year ka keise

[@kshitijvishnoi9784]

Ty sir your method of doing this is really easy and there is no need to remember any code

[@soham_chhabra]

Such a great video, I was searching for the same method for so long when I got your video sir ❤ Now the topic is clear 🔮 Thanks a lot sir 😊❤

[@jansirani5663]

Very clear teaching sir superb

[@apongjamir4972]

Shit, if you were my math teacher during my highschool days. I would have secured some good marks. 😂

[@sunayanamohite7243]

Gd afternoon Sir! Though it is complicated, you explain so well,any lay man can understand. All d best. Tc.

[@sureshaggarwal]

WhatsApp [phone] for PDFs of short tricks

[@AJEETKUMAR-dz3kq]

Your all the concept is very nice sir and You explain very well

[@zarinshaikh5574]

Sir you are the best Thank you so much

[@Sanjuhere]

Sir, based on your concept the day for date 22-02-2020 is coming Sunday, (D22+M4+ C6+Y20+Q5= 57/7 remainder is 1 and as per your week num. It is Sunday) where as 22-02-2022 is on Saturday.. please solve..

[@shashikalaraju5769]

I can understand crystal clear. You Sir are a wonderful teacher, I really hope I could be as good as you some day. My heartfelt thanks for setting an example.

[@priyachandran6282]

I searched some videos for this problem... But ur only given the solves from the beginning... Tq so much sir....

[@ssvn.03]

Wow.. excellent elaboration sir!...such an admirable character!!

[@aishwaryaraut4701]

Thank you so much sir... I got each and every concept...!!

[@sidduonline]

The method is good. Thanks for posting. I would like to understand the concept behind the odd days. I mean, the concept behind the circle ( 0-Sun, 1-Mon, 2-Tue ....). Why is 0 always a Sun and 1 always a Mon. It would be great help if you can provide an explanation behind this.

[@themasterji8252]

0 means 7 . When 7 is divided by 7 the remainder is 0 . 0=sunday is a serial number. I use 0 as saturday and use codes January as 0 February as 3 and so on. I can explain to you how January is 0 feb is 3 march 3 April 6 etc.

[@robertowens6654]

Odd days denote the number of days after completing any number of week cycles. We could say that *A week starts on Monday and ends on Sunday by this concept.* When you say year 2000 has 0 odd days it means by the end of the year, an exact number of week cycles has been completed. Which tells us that December 31 is a Sunday. A new week starts on 1 Jan 2001 (1odd day= Monday.) By the next day there would be 2 odd days which means it's the second day of a new week therefore Tuesday and so on.

[@ashishsimon6294]

Wonderful trick sir Thank you very much

[@Swarna_222]

it works. it really works. sir, thanks a lot.

[@VCE_AEshwwaanth]

Simply superb sir..sri guruubyo namaha.

[@jibanpandey9423]

When i watched this vdo immediately i subscribed your channel. Then i wanted to know whether you are still making vdo s or not. So i went through your channel and saw your recent posts. It made me very happy because the world needs the teacher like you. I’m highly impressed by the way you make clear the concept. You are a real and true teacher. Please continue teaching online. The learners from abroad too can benefit from you.Thank you and best of Luck.

[@mallaiahrajubalaraju5533]

Nice explaination sir 🙏🙏...

[@LovelySoul510]

Can u please make our CSAT preparation easy by solving the PYQ by using easy tricks ❤️🙏❤️

[@psbsouri7952]

a best explanation to understand these type of competitive problems

[@The_SSC_LAB]

Well explained Thank you very much

[@dhirajtech4734]

the way of your teaching is outstanding....really commendable

[@akashsri123]

Best explanation.. Very much appreciated sir... 🙏🙏

[@abhishekkarn3111]

You made it very easy sir❤

[@bestfaahimever]

it was truly well explained. but in my opinion i dont really think i can do it without a paper and pen. and this takes a little bit more time then it should. still i learn quite a lot from this and i am thankful for that.

[@williamwheeler4344]

It is easier for students to understand that for century years to be leap years the number of centuries should be divisible by 4.Eg. Look at 1700, 1800, 1900, 2000. At a glance one can see that only 20 is divisible by 4.

[@dahaka47]

Thank you sir i was searching for this method.

[@sureshaggarwal]

Do share the links with your WhatsApp groups and contacts

[@studyforcivilserviceandgov3488]

thank you so much sir...you give best tricks all the time... dislikers are losers..

[@sureshaggarwal]

Thanks for the motivation. ...Kindly do help in sharing

[@mahaboobbasha9298]

Poonam doll Hi

[@themasterji8252]

It is good but not best. Have you seen scot flansburg how fast he was... And now i am like him

[@themasterji8252]

And I am preparing my self to break the Guinness book of world record of Scot flansburg.

[@jaihind5000]

Sir I have learn this in my third standard in 1977 but where is fifteen seconds

[@mrawesomexd6969]

Sir actually I was a fan if the channel dear sir and there I found out the same trick as here but was not able to understand it now by chance I came across this channel and now I was able to understand that trick. Thank you sir from the bottom of my heart

[@jenebraveheart1839]

Sir, you are best and precise in what you teach. Thank you so much for helping us understand this concept so easily and quickly. Thank you so much for your contribution and hard work. I will keep following you and learn from you. Thank you again. Your faithful student

[@mampibiswas6370]

You r magical, best ever xplanation

[@answhiganswhig377]

This explanation is very clear, perfect, thanks for your patience.🤗 But I have a question : what is the use of knowing what day was what date ? 😉😉

[@sureshaggarwal]

Come and attempt a competitive exam in India....you would require no answer for your question 😄😆

[@nihaal7750]

Lol we have this in our exams moreover isn't it a fun skill to have anyway?

[@kalia3364]

@@sureshaggarwal 😀

[@dsarkar352]

Sir I liked the video very much. The explanation was amazing , I learnt this trick after watching this video two times. Please upload more interesting tricks . Thank You

[@simranchaturvedi1602]

Such straightforward and to the point explanation, of this entire concept within 20 mins only! Thank you so much Sir, for explaining with such clarity.

[@akshataman226]

such a sweet gentleman. THANK YOU .

[@gaingamgangmei7887]

Calendar which are having the same calendar “Ordinary years” - after every 6 or 11 years. Leap years- after every 28years. Same calendar will be repeated!!

[@dhiraj.govindvira]

Hey, please help 1-11-2020 2020 = 2000 + 20 2000 ->> 0 20 ->> 5 leap + 15 normal so , 10 +15 = 25 25 % 7 = 4 odd days now , 3 + 1 + 3 + 2 + 3 + 2 + 3 + 3 + 2 + 3 + 1 = 26 26 % 7 = 5 so , total = 0 + 4 + 5 = 9 9 % 7 = 2 by 2 its tuesday but it's a sunday ???? why ???

[@gaingamgangmei7887]

Dhiraj ,, let’s us understand that the month is given till Nov which is incomplete year of 2020. For the calculations we have to take complete year ie2019 as per rule 2019 2000 + 19 /4 ( 4 leap year and 15 ordinary years 0+(4x2+15x1)+3+1(lepyears)+3+2+3+2+3+3+2+3+ (***1) 8+15+26= 49/7 = O odd day ie Sunday Follow my method times less consume to be honest Sir method is quit complicated (***) given date may be 15,17,25,35etc whatever may be always divided by 7( applicable when it’s (= or greater than 7) and put the remainder in that month..

[@debkumarbose5683]

I can't find the day of 19/01/1988

[@musaed11]

@@dhiraj.govindvira you shouldn't count the number of years as 2020 but 2019. so 2000>>0 and 19 >> 15 normal+ 4 leap = 15+8= 23----> 23 mod7=2 instead of 4 and then do the rest and it will sunday

[@thebammer5166]

@@debkumarbose5683 Tuesday. I don't need a perpetual calendar to tell me that when i already have one in my head.

[@primechat]

I don't think I can do it in 15 seconds. I lost track in the middle.

[@AmarjeetNagra]

Same here

[@Abhirami1620]

I lost trick at 7;52 sec

[@wolferxgo]

Worth share.... V easy..... No cuts

[@janagarje124]

Sir, you are great

[@epicgamer3174]

Thank you for the maths trick!

[@samarth5776]

This is one of the hard question in competitive exams.This trick is very beneficial for me .

[@themasterji8252]

I could help you to fo it in 5-6 seconds. And For me it takes 1-2 seconds

[@dhiraj.govindvira]

@@themasterji8252 how ??

[@Greenman-qy1cc]

Basically, this is what you do: Take the last two digits of the year. Divide by 4, discarding any fraction. Add the day of the month. Add the month's key value: JFM AMJ JAS OND 144 025 036 146 Subtract 1 for January or February of a leap year. For a Gregorian date, add 0 for 1900's, 6 for 2000's, 4 for 1700's, 2 for 1800's; for other years, add or subtract multiples of 400. For a Julian date, add 1 for 1700's, and 1 for every additional century you go back. Add the last two digits of the year. Divide by 7 and take the remainder

[@chrismann1076]

No start with sun=1 mon 2 tue 3 wed 4 thur 5 Fri 6 sat 0

[@ankitasarkar2627]

Really extremely helpful..... helped me a lot for NTSE preparations

[@Ur_Rahul]

Finally I can solve these types of questions after seeing this video..

[@rockupvideos975]

Ek number..gajab

[@gauravchauhan199]

Great Video and easy tricks 👍👍👍👍

[@bethebestversionofyourself1210]

Thank you sir for all concept cleared in one video....