### Discussion :: Calendar - General Questions (Q.No.4)

Sumit said: (Oct 18, 2010) | |

11 odd days= 4 odd days. Can't understand. |

Hary said: (Dec 12, 2010) | |

Sumit, 11 days are left and in those 11 days,week days (7) are subtracted so the remaining are 11-7=4 odd days. |

Saurav said: (Jan 25, 2011) | |

How 1600 and 400 gets 0 odd days ? |

Boney said: (Feb 23, 2011) | |

If we divide 1600/400 we will get zero as remainder so it has zero odd days. so in 2000 yrs we have 0 odd days. In 9 yrs there r 2 leap years leap yrs have 2 odd days(366/7 2 as remainder) and non leapyr have 1 odd day totaling we have 2*2+7*1=11 odd days Now cal no: of days in 2010 upto aug 15 then dividing by 7 will give no of odd days 2010 which is 3. |

Rashi said: (Mar 12, 2011) | |

Thanks boney. |

John said: (Mar 16, 2011) | |

Why we are taking the 2000 years as 1600 and 400 years ? |

Sundar said: (Mar 16, 2011) | |

@John You can find the answer for your question in this page. http://www.indiabix.com/aptitude/calendar/discussion-638 Hope this help you. Have a nice day! |

Rakesh said: (Jul 6, 2011) | |

How can I know 9years has (2leap + 7 nonleap)? |

Rupali said: (Aug 27, 2011) | |

Please can you tell me how you write the. Total number of odd days = (0 + 0 + 4 + 3) = 7 odd days. I can't understand this step. |

Trilochan said: (Sep 6, 2011) | |

@Rupali: First you go through the Important formluas which is given in the 1st page . |

Vaibhav said: (Sep 15, 2011) | |

Hey, this type of questions can be easily done by following formula --------------------------------------------------------------------------------- f= K+[(13m-1)/5]+ d+[d/4]+[c/4]-2c Here k= date of which you have to calculate(15 in above que.) m= month (for march, m=1 and so on..for jan m=11) d= last two digit of given year(10 in above que.) c= first two digit of given year(20 in above que.) --------------------------------------------------------------------------------- note: take only integer value after division for ex. 13/5=2 --------------------------------------------------------------------------------- For given question 15th August, 2010? f=15+[(13*6-1)/5]+10+[10/4]+[20/4]-2*20 =15+15+ 10 + 2 + 5 -40 =7 Now 7%7==0 So its SUNDAY. |

Nithya said: (Sep 20, 2011) | |

How for march the value of month is 6? I don't understand the month value. |

Vishnu said: (Sep 27, 2011) | |

9 years = (2 leap years + 7 ordinary years) = (2 x 2 + 7 x 1) = 11 odd days 4 odd days. How com tat 4 odd days ? |

Sudhakar said: (Oct 4, 2011) | |

@vishnu : those 4 days are the result of 11days(divide these 11days into weeks and days, so u get 7+4,here 7days makes a week and do count those 4 extra days.) so finally we get 4ODD DAYS.. |

Malathy said: (Oct 22, 2011) | |

Dear friends, Pls. explain which day will come on 25th dec 1992 with step by step procedure |

Nagaraju said: (Nov 12, 2011) | |

Friends can you explain 7%7=0 taken as sunday how? |

Rupali said: (Apr 9, 2012) | |

Why you are taken one time 1600, 400 or 1600, 300 in sum it's not possible to me to understand please help me? |

Arshed Ahamed said: (Apr 22, 2012) | |

In 2nd and 3rd step, from where does that 1600 and 400 years came? |

Arshad said: (Nov 16, 2012) | |

How to find odd years in 100 years, 80 years or so.... Is their any formula to find it rather to go year by year calculation. |

Nagu said: (Feb 21, 2013) | |

How can I know 9 years = (7 ordinary years +2 leap years). |

Sarcadnya said: (Jun 3, 2013) | |

What the day on 16 may 1992? 1992 = (1991)+(days between 1/1/1992 to 16/5/1992). 1600 = 0. 300 = 1. 91 = (22 leap)+(69 ordinary day). = (22*2)+(69*1). = 113 days. = 16 weeks 1 day = 1 odd day. Total 1991 yrs = 0+1+1 = 2 odd days. jan feb mar april may 31 28 31 30 16 = 136 days. = 19 weeks 3 days. = 3 odd days. Total till 16 may 1992 = 2+3 = 5. 5 = FRIDAY. But in actual calendar it's SATURDAY. CAN I get the correct method to find it? |

Kavimani said: (Jun 10, 2013) | |

August 15th,2010 = (2000) + (from 2001 to 2009) + (from 1st January 2010 to 15th august 2010). Odd Days in 2000 = 0. Odd Days (from 2001 to 2009) = 2 Leap Year+7 Ordinary Year = (2*2)+(7*1) = 11%7 = 4 Odd Days. Odd Days (from 1st January 2010 to 15th august 2010): Jan = 31. Feb = 28(not a leap year). mar = 31. Apr = 30. May = 31. Jun = 30. Jul = 31. Aug = 15. =========== Tot = 227. =========== Find Odd Days : 227%7 = 3(Odd Days). Answer : Add all Odd Days = 0+4+3 = 7. Notes for Odd Days : Number of days more than a complete week is called as "Odd Days". So, you have to divide 7 by 7. Final Answer = 7%7 = 0 (Sunday) |

Vedasri said: (Jun 13, 2013) | |

friends, Let me introduce a technique which is used to find any day with any date from 1900 to 2999. This is called doomsday technique; doomsday means: Last day of feb. 4th april(04.04) 6th june (06.06) 8th august(08.08) 10th oct(10.10) 12th dec(12.12) To find doomsday of any year we should do, anchor+(y/12)+remainder(y/12)+(remainder(y/12)/4). For any period from 1900 to 1999 anchor = 3(wednesday). For any period from 2000 to 2999 anchor = 2(tuesday) . y means last 2 digits of the given year. Now, our question here is to find the day on 15.08.2010 So,anchor=2, y=10. 2+(10/12)+rem(10/12)+(rem(10/12)/4). 2+0+10+2 =14. 14/7 = 0 so 0 odd days. So dooms day is sunday. For our question aug 8 is nearer. So, 08.08.2010 is surely sunday add another 7 to 0 odd days. because 8+7=15. i.e 8th aug + 7 days = 15th aug. 0+7 odd days=7 odd days=7/7=0. So the day is Sunday. |

Anil Kumar said: (Nov 12, 2013) | |

My dear how take 1600 years and 400 years And how leap year multiply by 2X2? |

Gpvsai said: (Jan 31, 2014) | |

Hello everybody , You must know some codes first. Years, Days, Century code(for every 400 years ). Jan-0, sun-0, 0:99-6. Feb-3, mon-1, 1:199-4. Mar-3, tue-2, 2:299-2. Apr-6, wed-3, 3:399-0. may-1, thr-4, Jun-4, fri-5, Jul-6, sat-6, Aug-2. Sep-5. Oct-0. Nov-3. Dec-5. Now formula is , Day = (Date+month code+no.of years+no.of leap years + century code)/(7). In the given question 15th Aug 2010. Day = (15+2+10+2+6)/7 = 7/7 = 0. Which is 'Sunday'. :). |

Akhil said: (Mar 23, 2014) | |

@Gpvsai : 15+2+10+2+6 = 35. So, dividing 35 by 7 we get 5 which is friday. |

Sunita said: (Jun 20, 2014) | |

I am not able to understand that how you calculate ordinary and leap years in 100 years, 97 years. Is there any formula for that? |

Nishant said: (Jun 22, 2014) | |

@Sunita. 97 year == Divide 97 by 4: = Quotient is 24. = So 24 is leap year and in 1 leap year has 2 odd days. = Than (97-24= 73): 73 is ordinary year and 1 ordinary year has 1 odd day. = 24*2 + 73*1= 121 days. =121/7 = than reminder equal to 2. = So 2 is odd days. |

Arjun said: (Jul 10, 2014) | |

Year Code Month Code Day Code 1600-1699=6 Jan-0 Sun-0 1700-1799=4 Feb-3 Mon-1 1800-1899=2 Mar-3 Tue-2 1900-1999=0 Apr-6 Wed-3 2000-2099=6 May-1 Thu-4 2100-2199=4 Jun-4 Fri-5 July-6 Sat-6 Aug-2 Sept-5 Oct-0 Nov-3 Dec-5 Note: Day+Year+Leap Year+Year Code+Month Code First we need to calculate leap year; it is calculated on 2010 and we need to take last two digits i.e;(10). 2)10(5 10 ---- 0 leave 0 and take only quotient i.e;5 Now come to formula D+Y+L.Y+Y.C+M.C ==> 15+10+5+0+2 = 32. 2)32(16 2 ------ 12 12 ------ 0 The 0 shows day's code. .'.2010 Aug 15th is Sunday. Note: While calculating leap year, take last two digits from the year. And considering year in the problem we need to take last two digits (i.e;10 in the problem). Thank You & Regards, Arjun. |

Ayesha said: (Jul 18, 2014) | |

Please explain how to take 9 years = 2 leap years + 7 ordinary years. |

Kiran said: (Aug 20, 2014) | |

Please explain how Total no of odd days = (0+0+4+3) = 7? |

Selvi said: (Aug 21, 2014) | |

How to find the leap years? Anyone guys explain me I Can't understand Please help ME. |

Priyanka said: (Aug 26, 2014) | |

@Selva. If the year is divisible by 4 that year is called as leap year. |

Priyanka said: (Aug 26, 2014) | |

@Ayesha. 9 years should be multiply by 4. 9/4 = 2 i.e 2 leap years. 9-2 = 7 i.e 7 ordinary years. |

Harshad said: (Nov 29, 2014) | |

What if February of year 2010 has 29 days? We have considered it to be 28 every time? |

Anbu said: (Jan 24, 2015) | |

Please explain 24-2-2015; Above date is what day of the week; |

Smita Deshmukh said: (May 6, 2015) | |

Given that 15th august, 2010. 1) Add day and last two digits of the year 15+10 = 25. 2) Divide last two digit of the year by 4, so 10/4 = 2. 3) Add quotient value to step 1. = 25+2+2(month code)+6(year code) = 35. 4) Divide 35/7 = 5 it means remainder is 0(day code) Sunday. |

Naveen said: (May 27, 2015) | |

We have 365 days for normal year, 366 days for leap year. So how no of odd days in 400 years is '0'. |

Ayushi said: (Jul 4, 2015) | |

I can't understand please anyone understand me in easy language. |

Gopakumar said: (Jul 27, 2015) | |

Is anyone have the simplest way to give answer to this question? How can we understand the year said is leap year? |

Meghna said: (Aug 3, 2015) | |

How you take 1600 and 400? I can't understand please tell me. |

Krishna Chaitu said: (Aug 11, 2015) | |

Let me make it clear 0 means Sunday 1 means Monday 2 means Tuesday 3 means Wednesday 4 means Thursday 5 means Friday 6 means Saturday 7 means 0 in terms of odd days. |

Krishnachaitu said: (Aug 11, 2015) | |

1 year means 365 days it has one odd day after divide with 7 so if this year starts with Sunday then next year starts with Monday but if this year is leap year that means a multiple of 4 then next year starts with Tuesday. |

Krishnachaitu said: (Aug 11, 2015) | |

So each year has one odd day and leap year has 2 odd days for every 100 years 100+24 odd days but after every 7 days the same day repeats for example today is Saturday after 7 days will be Saturday basing on this we neglect multiples of 7 and by dividing 124 with 7 we get 5 odd days. So 100 years have 5 odd days but for every 400 years one day is added based on earth's rotation don't ask me about that so 400 years have 20+1 odd days which means 3 weeks and can be neglected as same day appears after any number of complete weeks. |

Maheshwaree said: (Aug 29, 2015) | |

How to find month code, day code and year code? |

Venkat said: (Oct 12, 2015) | |

What is the day on 8th January 2015 please explain briefly? |

Saikumar said: (Dec 22, 2015) | |

1600 = 0 odd days. 400 years = 0 odd days. 15 years will be remained. 15 year will have (3 leap years + 12 ordinary years). = (3*2+1*12) = (6+12) = 18 = 7+7+4 = 4 odd days. 8 = 1 week+1 day. (0+0+4+1) = 5 days = Friday. |

Datta said: (Jan 1, 2016) | |

2000 to 2010 having 10 years. In which 3 leap year (2000-01), (2004-05), (2008-09) & 7 normal years. 3 leap years have 6 odd days (3 years * 2 odd days) = 6 odd day. Remaining 7 year having 7 odd day (individually 1) & for year (2010-11). January February March April May June July Aug. (31 + 28 + 31 + 30 + 31 + 30 + 31 + 15) = 227 days. No of odd day = 227/7 = Remainder is 3. So total odd days are 3+6+7 = 16. Now 16/7, remainder is 2. So Tuesday will come. So how answer will come (Sunday)? |

Vani said: (Jan 23, 2016) | |

Instead of taking 2000 as reference can we take 2008 as it is also a leap year and closer to 2009? |

Candy said: (Feb 28, 2016) | |

I am not understand please say how will it come? |

Raunak Thakur said: (May 18, 2016) | |

How 100 years has 24 leap years? Please explain. |

Keisha said: (May 27, 2016) | |

How do we know if we have to take for 5 years or for 10 years? |

Keisha said: (May 27, 2016) | |

15 th August 2010. Jan 31 + feb 28 + Mar 31 + Apr 30 + May 31 + June 30 + July 31+ Aug 15. 31 + 28 + 31 + 30 + 31 + 30 + 31 + 15 = 227. 227/7 = 3 odd days. *************************************. For 10 years. 2000 + 2001 + 2002 + 2003 + 2004 + 2005 + 2006 + 2007 + 2008 + 2009. (In 400, 1600, 2000 odd days are zero. ). 0 + 1 + 1 + 1 + 2 + 1 + 1 + 1 + 2 + 1 (odd days). = 11 odd days. Total odd days = 3 + 11 = 14. 14/7= 7. We have 2 leap year + 7 ordinary years so =2 * 2 + 7 = 4 + 7 = 11 = 4 odd days So, 4 + 3 = 7. 7/7 = 0 = Sunday. |

Bidhya Chhetri said: (Jul 27, 2016) | |

Total odd day 0 + 0 + 4 + 3 = 7 = 0 Odd day. |

Bidhya said: (Jul 27, 2016) | |

Please help me to understand. Why a total number of odd days = 0 + 0 + 4 + 3 = 7. |

Salil Chaudhary said: (Jul 31, 2016) | |

@Bidhya. When 7 is divided by 7 it gives remainder as 0. So odd days = 0. |

Amith said: (Aug 1, 2016) | |

Please, can you tell me how you know the century code is 6 for august? |

Murugan.S said: (Aug 30, 2016) | |

We are getting finally 7 (the first day 0 the week is Sunday) so how could we took 7 is Sunday, could you please sort out the same? |

Namzi said: (Sep 9, 2016) | |

Why should we take 1600 and 400? |

Subham said: (Sep 13, 2016) | |

Why not 1600, 1200 = 0 odd days. |

Naresh said: (Oct 27, 2016) | |

If the remainder gets 1 then what is the week? 0 = sun 1 = Monday 2 = Tuesday So on Am I correct? |

Gokila said: (Jan 1, 2017) | |

7/7= 0 remaider. So, 7 ,14, 21, 28, 35 = have been considered as Sunday. |

Pavan said: (Jun 30, 2017) | |

We have an easy method...... Follow this if you feel it simple :) This is a easy code technique. Year cheat code, Month cheat code and date cheat code are the three types. YEAR CHEAT CODE 1600 to 1699 = 6 1700 to 1799 = 4 1800 to 1899 = 2 1900 to 1999 = 0 2000 to 2010 = 6 MONTH CHEAT CODE Jan = 0 Feb = 3 Mar = 3 Apr = 6 May = 1 Jun = 4 Jul = 6 Aug = 2 Sep = 5 Oct = 0 Nov = 3 Dec = 5 DAY CHEAT CODE Sunday = 0 Monday = 1 Tuesday = 2 Wednesday = 3 Thursday = 4 Friday = 5 Saturday = 6 Now lets take an example.... 11/03/1997 which is my birthday Step 1 : Take last two digits of the year = 97 Step 2 : Take the date = 11 Step 3 : Divide 97 by 4 and take the quotient = 24 Step 4 : Take the year cheat code = 0 (For the range 1900 to 1997) Step 5 : Take the Month cheat code = 03 Now add all the results 97+11+24+00+03 = 135. Divide the answer by 7 and consider the remainder = 2 (Subract 1 from this in case of leap year). Which is the Cheat code for Tuesday, therefore the answer is Tuesday. |

Jagadish said: (Jul 14, 2017) | |

How can I find 40 years & how many odd days? |

Narayan said: (Aug 5, 2017) | |

Odd days in 400 years=0? I can't understand it. Please help me. |

Lakshmi said: (Sep 14, 2017) | |

Why we take 1600 and 400 explain in detail? |

Rama Devi said: (Sep 22, 2017) | |

15/8/2010. 15+10+2+2+6/7 =35. 7 )35(5 35 - It's reminder. 0. So, Ans Is Saturday. |

Shagufta said: (Sep 27, 2017) | |

How here are 2 leap year taken from 2000 to 2010? I think there are 3 leap year. 1. 2000 2. 2004 3. 2008 |

Sikandar said: (Nov 12, 2017) | |

Why are we taking 1600 or 400? How do we know what is the year code? What is the basis of codes? How are the codes arrived at? Is there any formula or reasoning for arriving at the different codes for the Months, Days and a particularly century? |

Poonam said: (Dec 15, 2017) | |

Please tell me why always take 1600 and 400? |

Priya said: (Jan 20, 2018) | |

@Poonam. 1600 & 400 are reference years with which we calculate the solution. As they leave 0 odd days. So we can calculate fresh. |

Pramod said: (Feb 22, 2018) | |

2+(10/12)+rem(10/12)+(rem(10/12)/4). 2+0+10+2 = 14. |

Komal said: (Mar 4, 2018) | |

I have a question. How in 100 years, 76 was ordinary years and 24 was leap years? Because generally, every 4 year is a leap year, so 100 is completely divisible by 4 and we get 25, so in general, in 100 years, there were 25 leap years? Please, anyone, explain this to me. |

Himanshu said: (Mar 26, 2018) | |

How 2 Leap Year? Please Explain in detail. |

Biswas said: (Apr 22, 2018) | |

@All. It is simple, zero refers to Sunday in calendar rule, Monday = 1, same as if the total number of odd days is 6 it's your Saturday. |

Saipreethi said: (Jul 5, 2018) | |

How can I find, if they have given the years in 1500 and 1600? Please explain. |

Thinlay Penjor said: (Aug 4, 2018) | |

Hello. Why we are taking the odd number of 1600 years, 400 years and 9 years when it is only asked any the day on 15th August 2010? |

Madhu said: (Feb 6, 2019) | |

How many odd days in 15 years? please tell me the answer. |

Muzige said: (May 17, 2019) | |

@Vaibhav. In the formula f = k + (13m-1)/5 +d + (d/4)+(c/4)-2c, how to determine m? |

Vaishnavi said: (Jun 6, 2019) | |

On what basis is m being determined? |

Thiru said: (Jul 31, 2019) | |

(15+10+2+6+10/4)= 35/7 = 0. |

Nithin Paul said: (Sep 29, 2019) | |

@Madhu. For 15 years we have 3 leap year (12/4=3). And 12 ordinary years, So total num of odd days=(3*2+12*1)=18, (Add 2 point for leap year because 2 days will shift for the next year odd days is the actual concept that if 2016 Jan 1 is sunday then 2017 Jan would be Monday but after a leap year it would be Tuesday so 2 days will shift). And try to divide the number of odd days with 7. Here it is 18 cant divide by 7. So 14/7=2. Then 4 days are remaining then 4 is the num of odd days. *(dividing 7 means a week have seven days). |

Uiyutvbjh said: (Feb 15, 2020) | |

We have an easy method. Follow this if you feel it simple. This is a easy code technique. Year cheat code, Month cheat code, and date cheat code are the three types. YEAR CHEAT CODE: 1600 to 1699 = 6. 1700 to 1799 = 4. 1800 to 1899 = 2. 1900 to 1999 = 0. 2000 to 2010 = 6. MONTH CHEAT CODE: Jan = 0. Feb = 3. Mar = 3. Apr = 6. May = 1. Jun = 4. Jul = 6. Aug = 2. Sep = 5. Oct = 0. Nov = 3. Dec = 5. DAY CHEAT CODE: Sunday = 0. Monday = 1. Tuesday = 2. Wednesday = 3. Thursday = 4. Friday = 5. Saturday = 6. Now, let's take an example, 11/03/1997 which is my birthday. Step 1: Take the last two digits of the year = 97. Step 2: Take the date = 11. Step 3: Divide 97 by 4 and take the quotient = 24. Step 4: Take the year cheat code = 0 (For the range 1900 to 1997). Step 5: Take the Month cheat code = 03. Now add all the results. 97+11+24+00+03 = 135. Divide the answer by 7 and consider the remainder = 2 (Subtract 1 from this in case of a leap year). Which is the Cheat code for Tuesday, therefore the answer is Tuesday. |

Habib said: (Feb 25, 2020) | |

@John. 2000 is splint into a possible and nearest 4th century to the figure and after a remainder is obtained. i.e 2000 = 1600 + 400. |

Aditya said: (Apr 8, 2020) | |

We can get a number of odd days in 9years by dividing 9 by 4 as in every 4 years there is a leap year. The quotient is the number of leap years and rest is ordinary years so, 9years = 2leap years + 7ordinary years. |

Sonam Wangchuk said: (Jul 1, 2020) | |

227 =3 odd days How? Actually 227/7 =4. |

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