Aptitude - Calendar - Discussion
Discussion Forum : Calendar - General Questions (Q.No. 2)
2.
What was the day of the week on 28th May, 2006?
Answer: Option
Explanation:
28 May, 2006 = (2005 years + Period from 1.1.2006 to 28.5.2006)
Odd days in 1600 years = 0
Odd days in 400 years = 0
5 years = (4 ordinary years + 1 leap year) = (4 x 1 + 1 x 2) 
6 odd days
Jan. Feb. March April May (31 + 28 + 31 + 30 + 28 ) = 148 days
148 days = (21 weeks + 1 day) 1 odd day.
Total number of odd days = (0 + 0 + 6 + 1) = 7 0 odd day.
Given day is Sunday.
Discussion:
214 comments Page 5 of 22.
Naveen said:
1 decade ago
Hello friends,
I think I have a better sollution. (I haven't read earlier sollutions, so if it is already here, then sorry and credit to the earlier post)
Remember this code for months
622503514624 (Jan, Feb, March,....Dec, for common years)
512503514624 (for leap years, jan code will be 5 and Feb code will be 1, that is the only difference in this code, It should be easy as we remember lots of mobile nos)
Then you'll have to remember years code.
for example 2011 year code will be 6
11+(leap years in 1st 11 years = 2)
so, 11+2 = 13/7, remains 6,the year code (for 2012, year code will be 1 (12+3= 15/7, remains 1)
Days code will be 0 for Sunday, 1 for Monday..like that!
For calculating day the formula is
Year code + month code + date
for example 18th October 2011
year code (6)+month code (6) + date (18)= 6+6+18 = 30/7..remains 2 means Tuesday
Do remember please, you'll be adding extra 0 for centuries divisible by 400 (say 2000), 5 for next century (2100), 3 for another next (2200) and 1 for just after that century (2300), and then 0 agian for 2400!
PS: I'm sorry but I'm not very good in explaining things. But this formula is doing wonders for me! Any help in this case..please contact me naveen.dhanerwal(at)gmail.com
Have a nice day!
I think I have a better sollution. (I haven't read earlier sollutions, so if it is already here, then sorry and credit to the earlier post)
Remember this code for months
622503514624 (Jan, Feb, March,....Dec, for common years)
512503514624 (for leap years, jan code will be 5 and Feb code will be 1, that is the only difference in this code, It should be easy as we remember lots of mobile nos)
Then you'll have to remember years code.
for example 2011 year code will be 6
11+(leap years in 1st 11 years = 2)
so, 11+2 = 13/7, remains 6,the year code (for 2012, year code will be 1 (12+3= 15/7, remains 1)
Days code will be 0 for Sunday, 1 for Monday..like that!
For calculating day the formula is
Year code + month code + date
for example 18th October 2011
year code (6)+month code (6) + date (18)= 6+6+18 = 30/7..remains 2 means Tuesday
Do remember please, you'll be adding extra 0 for centuries divisible by 400 (say 2000), 5 for next century (2100), 3 for another next (2200) and 1 for just after that century (2300), and then 0 agian for 2400!
PS: I'm sorry but I'm not very good in explaining things. But this formula is doing wonders for me! Any help in this case..please contact me naveen.dhanerwal(at)gmail.com
Have a nice day!
(2)
Janvi said:
1 decade ago
2011 year code will be 6. How can you explain this ?
Chinna said:
1 decade ago
Please explain me clearly. I am totally confused.
Jugal said:
1 decade ago
I am totally confused in calander question. Then please expain in a easy lang if any one can.
Satish mohan said:
1 decade ago
Hi Naveen thank you for your information.
Your method is good I like this method.
Please tell me how to know this method and give me some ideas.
Your method is good I like this method.
Please tell me how to know this method and give me some ideas.
Meenu said:
1 decade ago
Hi All,
Deena s solution is correct. Its 106 not 16... But try to substitute 106 instead of 16.. We are getting the same answer. I tried with some other questions also,its working.. Good Solution Deena.... Thanks :-)
Deena s solution is correct. Its 106 not 16... But try to substitute 106 instead of 16.. We are getting the same answer. I tried with some other questions also,its working.. Good Solution Deena.... Thanks :-)
Ajay said:
1 decade ago
Deena's solution is correct just take 106 instead of 16 106/4 you will get quotient 26 and remainder 2 leave the remainder and take 26 you will get the right answer.
Arpit said:
1 decade ago
@NAVEEN. Thanks. !
Vijay singh said:
1 decade ago
Dont know how 1600 and 400 have been accounted.
Number of leap yrs in 2000 yrs= 2000/4-15=485 yrs
(4 is included because leap year comes generally once in 4 yrs.
15 is subtracted because some of yrs namely 100,200,300,500,600 700,900,1000,1100,1300,1400,1500,1700,1800,1900 are not leap yrs in 2000 yrs)
Number of odd days in leap yrs= Remainder of 485*2/7 = 4 days
Number of non leap yrs in 2000 yrs= 2000-485 = 1515 yrs
Number of odd days in non leap yrs= Remainder of 1515/7 = 3 Days
Therefore number of odd days in 2000 yrs= Remainder of(4+3)/7= 0
And for next five yrs from 2001 to 2005 and onwards it is been described in solution.
Number of leap yrs in 2000 yrs= 2000/4-15=485 yrs
(4 is included because leap year comes generally once in 4 yrs.
15 is subtracted because some of yrs namely 100,200,300,500,600 700,900,1000,1100,1300,1400,1500,1700,1800,1900 are not leap yrs in 2000 yrs)
Number of odd days in leap yrs= Remainder of 485*2/7 = 4 days
Number of non leap yrs in 2000 yrs= 2000-485 = 1515 yrs
Number of odd days in non leap yrs= Remainder of 1515/7 = 3 Days
Therefore number of odd days in 2000 yrs= Remainder of(4+3)/7= 0
And for next five yrs from 2001 to 2005 and onwards it is been described in solution.
Subin said:
1 decade ago
Why chosen 2001-2005 please explain as soon as possible ?
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers