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 🙏❤️🙏❤️🙏
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.
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!❤
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.
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.
@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
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
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.
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...
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 ??
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. 😄
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.
Odd days are the number of days extra which can't form a week. For example, In 10 days there is one week and 3 extra days These extra days are called odd days In 13 days 13/7 remainder= 6. So 6 odd days If there are 7 odd days,it can be counted as one complete week. So odd days can't be greater than 6 Hope it's helpful Thankyou
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 ,, 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..
@@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
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.
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
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
Agarwal sir can you please explain the 14 - 7 - 2010 When I solve the above problem I get the answer as per your theory Thursday but actually answer is Wednesday. Please clarify my doubt
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.
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.
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.
Well, I can't do that in 15 seconds, but here's my approach. After using it many times, I can "think" it a lot faster than I can explain it. 1956 Aug 15 = what day-of week (DOW)? Every 28 years there's an exact repeat of the calendar (so long as no Gregorian century non-leap year is crossed; last one was 1900; next is 2100, so we're safe here). 1956 + 2·28 = 2012; so 1956 Aug 15 = 2012 Aug 15 And every forward jump of one year adds one DOW, so long as no leap day (LD) is crossed. Each LD adds one additional DOW. That means that jumping to the same date, either 5 or 6 years forward, preserves DOW, depending only on how many LD's are crossed (5 y + 2 LDs or 6 y + 1 LD, makes 7 days either way = same DOW). August 2012 is just past a LD. So 2012 Aug 15 + 6 years = 2018 Aug 15 has the same DOW. Now as I write this, it is Jan. 12, 2019, Saturday. Add 3 days to make the day-of-month match: 2019 Jan 12 Sat. + 3 days = 2019 Jan 15 Tue. Now go back to August 2018, for which we must add the days of Dec. backward through August: 31 + 30 + 31 + 30 + 31 = 153 = 7·22 - 1 so subtracting those 153 days, adds one to the DOW. Therfore, 1956 Aug 15 was Tue + 1 = Wednesday. This doesn't show how to do a general DOW problem, however. For that, I have a number of other tricks; such as which months within a calendar year, match: Non-Leap: Jan = Oct Feb = Mar = Nov Apr = July Sep = Dec May, June, and Aug, don't match any others. Leap: Jan = Apr = July Feb = Aug Mar = Nov Sep = Dec May, June, and Oct, don't match any others. Always: May = the following Jan June = the following Feb When the latter year is non-leap: Aug = the following May Sep = the following June Oct = the following July Nov = the following Aug For longer-range problems, you must take account of missing leap days in those Gregorian exception years - century years that aren't 400-divisible. And that the Gregorian calendar repeats every 400 years. (Because 400 Gregorian years = 146097 days = 20871 weeks.) Anyway, this is just what works for me; and although it may not work for you, if it does, go ahead and use it. A few additional notes: 1) A lot of what Suresh is doing here is based on mod 7 arithmetic; if you know how to do that, a lot of his explanations can be greatly simplified. Of course, he's assuming an audience many of whom aren't familiar with modular arithmetic, so he couldn't make those simplifications. 2) If you have access to a JDN (Julian Day Number) calculator, you can just convert the date in question to JDN, then take that mod 7. That remainder corresponds to: 0=Mon, 1=Tue, ..., 6=Sun. Note carefully that JDN is *not* what is often mistakenly called, "Julian Day," which is actually just day-of-year, 1 to 365/366. JDN was invented by historian Joseph Scaliger in 1583 (the year after the introduction of the Gregorian calendar) in order to convert all the multifarious calendars that were used by various nations and empires throughout history, to a single day-counting system. Today it is used in astronomy to time certain types of recurrent astronomical events; and by the various space programs to keep track of what spacecraft are doing. In these arenas, it includes fractional days, the "zero" of which is noon UT. E.g., JDN 1 = Jan.1, 4713 BC (Julian Calendar). JDN of Jan. 12, 2019 (Gregorian Calendar) is 2,458,496. 3) This video really amounts to a way to find a similar sort of count, starting on each Jan.1, 400n, all of which are Saturdays. And I think it's more amenable to general use than my own method. Fred
@@AshokYadav-zq2kd In Hindi? I think you must have meant to ask Suresh that. In any case, I know about as much Hindi as I know Martian. And Suresh's method is more generally suitable than mine, which is more attuned to me personally. Fred
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
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.
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
Because when God created the world he finished his work and relaxed on the last day of it was a Sunday point of origin of the world, then the full process started to work on the next day the first day n°1 the Monday 😉 No i'm joking... But it works, right?
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..
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
Why my maths teachers can't be this patient while explaining..?!
Cause not every student understands this. They will be more confused :))
All this in one take, with such great explanation skills, is truly commendable.
Keep up the great work sir.
😄
@Fashion Hub kaha pr?
Thanks a lot
How lucky I have been to find this vedio🙏 . Simple explaination no short tricks. Thank you sir.
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 🙏❤️🙏❤️🙏
Very helpful. Concise explanations and straight to the point with no background music. Just plain mathematics. Love it.
Simply amazing. Thanks a lot.
Not only for children appearing for competitive exams, these tricks can help adults too. Great videos Sir !
L
Such a wonderful video I was searching from many years
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
Exactly ! because he explains the logic of it. So you don't need to memorize figures you can get them by yourself 🤗
100% Working. Taking very less time after Practice
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.
Great man !!
can you share pre memorized values.
i'm very excited
Can you please teach me your technique
I'll be grateful forever.
Sir i dint understood ua method,, aggarwalz sir tricks also much complicated , plz explain once
@@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
@@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
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!❤
The best explanation. I was so confused regarding this topic bt you explained it so well. Thank you so much for clearing my doubt.
Suresh, you are superb. Maths is so much fun with your tricks.My son loves them
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.
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.
@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
Bestest teacher I found on YT
absolutely 💯💯 well explained concept
Kuch log kya kya tricks yaad karwa dete he, but this video is fully conceptual and remember for a long time, thank u
This is a normal method but explanation is simple. It is good for beginners..
Understood the concept extremely fast. Thank you so much, sir. Very well explained 😊😊
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
Go on career will
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Thankyou very much for helping out the basics with clarity
I had no idea of odd days and even days before ... Now I got the clear idea.... Good explanation sir
you made the concept so clear, thank you sir i understood the calendar problems :)
very helpful sir...thank u so much for this amazing trick and session
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.
Easly understood sir ..nice way to explain the concepts...
Good clear teaching.pls keep it coming and enlighten
Really helpful, thanks a lot sir
Commendable board organisation. beautifully explained
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...
Thanks for that....will try it.
Too much to memorise and less to understand in doomsday
@@sureshaggarwal sir 2000 aru 2001,2002,2003 year ka keise
👏👏👏👏👏👏👏ur clear explanation is the fastest way for our learning ❤❤
watching this from Boston right now. Jai hind 🇮🇳
Thanks sir. You cleared the Concept. It's better to understand the Concept rather than going for shortcuts.
Way of explaining the concepts is awesome!
Thanks a lot sir..you made it very simple and effective...
Now I can find the day of the week of a given date easily....thank you...
Thank your Guruji, I can understand only in this time.
Just fantastic. Very clear explanation. Thank you very much
Thank you so much sir for the best explanation 🙏🙇♂️
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 ??
Thanks for posting such an informative video, Mr Aggarwal.
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. 😄
nice explanation. It is very helpful to me in revision of calendar chapter
Great....
Amazing, step by step, without any trick or jumps, very cool..
Sir today i feel happy because this video clear my doubts... Thanks for sharing your concepts
Do share the links with your WhatsApp groups and contacts
Shit, if you were my math teacher during my highschool days. I would have secured some good marks. 😂
it's the best explanation ever seen, thank you, sir.
Way of explaination was really outstanding sir
Today morning I learnt this and now you taught.. Ur was better.. The morning session went tangentially
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.
Very clear teaching sir superb
Wow.. excellent elaboration sir!...such an admirable character!!
6:30
What does it mean by odd days cannot be greater than 7?
Odd days are the number of days extra which can't form a week.
For example,
In 10 days there is one week and 3 extra days
These extra days are called odd days
In 13 days
13/7 remainder= 6. So 6 odd days
If there are 7 odd days,it can be counted as one complete week.
So odd days can't be greater than 6
Hope it's helpful
Thankyou
thank you so much sir...you give best tricks all the time... dislikers are losers..
Thanks for the motivation. ...Kindly do help in sharing
Poonam doll
Hi
It is good but not best.
Have you seen scot flansburg how fast he was... And now i am like him
And I am preparing my self to break the Guinness book of world record of Scot flansburg.
Sir I have learn this in my third standard in 1977 but where is fifteen seconds
You made it very easy sir❤
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!!
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 ???
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..
I can't find the day of 19/01/1988
@@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
@@debkumarbose5683 Tuesday. I don't need a perpetual calendar to tell me that when i already have one in my head.
Thank you sir for explaining in such a simple way, now I can compute the day of any date, thanks a lot sir.
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.
Thank you so much sir...
I got each and every concept...!!
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
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
Thank you 😊😊😊
Nice explaination sir 🙏🙏...
Gd afternoon Sir! Though it is complicated, you explain so well,any lay man can understand. All d best. Tc.
WhatsApp 9896369963 for PDFs of short tricks
7:23 - 8:00
Didn't understand how he calulated the no. of odd days for a span of 300 and 400 years?
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 😊❤
Thanks a lot sir. this helped me a lot
Agarwal sir can you please explain the
14 - 7 - 2010
When I solve the above problem I get the answer as per your theory Thursday but actually answer is Wednesday.
Please clarify my doubt
Watch my method
no it comng wednesday only do it again
Even I got Thursday for the mentioned date.
@@niharikagamedhar1612 no do it precisely you will get thursday
It's Wednesday
3 comes as answer
Finally I can solve these types of questions after seeing this video..
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.
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.
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.
Your all the concept is very nice sir and
You explain very well
Well, I can't do that in 15 seconds, but here's my approach. After using it many times, I can "think" it a lot faster than I can explain it.
1956 Aug 15 = what day-of week (DOW)?
Every 28 years there's an exact repeat of the calendar (so long as no Gregorian century non-leap year is crossed; last one was 1900; next is 2100, so we're safe here).
1956 + 2·28 = 2012; so 1956 Aug 15 = 2012 Aug 15
And every forward jump of one year adds one DOW, so long as no leap day (LD) is crossed. Each LD adds one additional DOW.
That means that jumping to the same date, either 5 or 6 years forward, preserves DOW, depending only on how many LD's are crossed
(5 y + 2 LDs or 6 y + 1 LD, makes 7 days either way = same DOW). August 2012 is just past a LD.
So 2012 Aug 15 + 6 years = 2018 Aug 15
has the same DOW. Now as I write this, it is Jan. 12, 2019, Saturday. Add 3 days to make the day-of-month match:
2019 Jan 12 Sat. + 3 days = 2019 Jan 15 Tue.
Now go back to August 2018, for which we must add the days of Dec. backward through August:
31 + 30 + 31 + 30 + 31 = 153 = 7·22 - 1
so subtracting those 153 days, adds one to the DOW.
Therfore,
1956 Aug 15 was Tue + 1 = Wednesday.
This doesn't show how to do a general DOW problem, however. For that, I have a number of other tricks; such as which months within a calendar year, match:
Non-Leap:
Jan = Oct
Feb = Mar = Nov
Apr = July
Sep = Dec
May, June, and Aug, don't match any others.
Leap:
Jan = Apr = July
Feb = Aug
Mar = Nov
Sep = Dec
May, June, and Oct, don't match any others.
Always:
May = the following Jan
June = the following Feb
When the latter year is non-leap:
Aug = the following May
Sep = the following June
Oct = the following July
Nov = the following Aug
For longer-range problems, you must take account of missing leap days in those Gregorian exception years - century years that aren't 400-divisible.
And that the Gregorian calendar repeats every 400 years. (Because 400 Gregorian years = 146097 days = 20871 weeks.)
Anyway, this is just what works for me; and although it may not work for you, if it does, go ahead and use it.
A few additional notes:
1)
A lot of what Suresh is doing here is based on mod 7 arithmetic; if you know how to do that, a lot of his explanations can be greatly simplified. Of course, he's assuming an audience many of whom aren't familiar with modular arithmetic, so he couldn't make those simplifications.
2) If you have access to a JDN (Julian Day Number) calculator, you can just convert the date in question to JDN, then take that mod 7.
That remainder corresponds to: 0=Mon, 1=Tue, ..., 6=Sun.
Note carefully that JDN is *not* what is often mistakenly called, "Julian Day," which is actually just day-of-year, 1 to 365/366.
JDN was invented by historian Joseph Scaliger in 1583 (the year after the introduction of the Gregorian calendar) in order to convert all the multifarious calendars that were used by various nations and empires throughout history, to a single day-counting system. Today it is used in astronomy to time certain types of recurrent astronomical events; and by the various space programs to keep track of what spacecraft are doing. In these arenas, it includes fractional days, the "zero" of which is noon UT.
E.g., JDN 1 = Jan.1, 4713 BC (Julian Calendar).
JDN of Jan. 12, 2019 (Gregorian Calendar) is 2,458,496.
3) This video really amounts to a way to find a similar sort of count, starting on each Jan.1, 400n, all of which are Saturdays.
And I think it's more amenable to general use than my own method.
Fred
Sir can you explain it in hindi
I couldn't get you
Please sir .I am very eager to know this trick.that will be helpful for my competitive exam
@@AshokYadav-zq2kd In Hindi? I think you must have meant to ask Suresh that.
In any case, I know about as much Hindi as I know Martian.
And Suresh's method is more generally suitable than mine, which is more attuned to me personally.
Fred
@@ffggddss no sir absolutely not I asked you only.i have seen Suresh sir video but I want to know different tricks .so i asked you
I searched some videos for this problem... But ur only given the solves from the beginning... Tq so much sir....
I don't think I can do it in 15 seconds. I lost track in the middle.
Same here
I lost trick at 7;52 sec
Great Video and easy tricks
👍👍👍👍
It is really very helpful sir. thank u so much
the way of your teaching is outstanding....really commendable
Can u please make our CSAT preparation easy by solving the PYQ by using easy tricks ❤️🙏❤️
You r magical, best ever xplanation
Best explanation.. Very much appreciated sir... 🙏🙏
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
Sir you are the best
Thank you so much
Wonderful trick sir
Thank you very much
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.
Thanknyou sir! The video was really really helpful. Keep making such content..
Well explained
Thank you very much
Ty sir your method of doing this is really easy and there is no need to remember any code
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 ?
😉😉
Come and attempt a competitive exam in India....you would require no answer for your question 😄😆
Lol we have this in our exams moreover isn't it a fun skill to have anyway?
@@sureshaggarwal 😀
Best explanation guru ji.
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
7:50 since 400 is a leap year, it should have 2 odd days. 5*4 +2. It should be 22.
it works. it really works. sir, thanks a lot.
Thank you for the maths trick!
This is one of the hard question in competitive exams.This trick is very beneficial for me .
I could help you to fo it in 5-6 seconds.
And For me it takes 1-2 seconds
@@themasterji8252 how ??
Worth share.... V easy..... No cuts
Hello , may I ask?
Why is
0 = Sunday,
1 = Monday,
2 = Tuesday, ect ?
Thank you so much for your Trick .
Kristianto Raharjo because it was Monday on 1-1-1. :)
This is predefined by calendar system we follow ( gregerion calender ) .
The initial Gregorian calendar 1-01-01 Monday..
Because when God created the world he finished his work and relaxed on the last day of it was a Sunday point of origin of the world, then the full process started to work on the next day the first day n°1 the Monday 😉
No i'm joking...
But it works, right?
@@answhiganswhig377 😂😂😂
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..
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
No start with sun=1 mon 2 tue 3 wed 4 thur 5 Fri 6 sat 0
a best explanation to understand these type of competitive problems