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 4 of 22.
Arun Anoop M said:
1 decade ago
What was the day of the week on 28th May, 2006?
Leave 1600,400
it's odd day is zero
----
5=(5/4)leap yr+[5-1]ordinary day
=1 leap year+4 ordinary day
=1*2+4
=6 odd day
---------
Jan Feb Mar Apr May
31 28 31 30 28 ->total->148days->21weeks+1odd day
->1 odd day
Total=0+0+6+1
=7 odd day
Sun Mon Tue Wed Thu Fri Sat
0 1 2 3 4 5 6
7
Day is Sunday.
By Arun Anoop M,MTech IT,Kalasalingam University,TamilNadu.
Leave 1600,400
it's odd day is zero
----
5=(5/4)leap yr+[5-1]ordinary day
=1 leap year+4 ordinary day
=1*2+4
=6 odd day
---------
Jan Feb Mar Apr May
31 28 31 30 28 ->total->148days->21weeks+1odd day
->1 odd day
Total=0+0+6+1
=7 odd day
Sun Mon Tue Wed Thu Fri Sat
0 1 2 3 4 5 6
7
Day is Sunday.
By Arun Anoop M,MTech IT,Kalasalingam University,TamilNadu.
Lateef Ansari said:
1 decade ago
I solved this problem this way.
step 1 : 28 may 2006: take date and year's last two digit number.
28+06 = 34.
step 2: year is divided by 4 so 06/4=1(leave the remainder).
step 3: Codes.
Jan : 0
Feb : 3
Mar : 3
aprl: 6
May : 1
Jun : 2
July; 6
aug : 2
Sep : 5
oct : 0
Nov : 3
Dec : 5
Now add step 1,step 2,step 3 ie,. 34+1+1=36.
divide it by seven 7, so remainder is 0.
Then,
0 = sunday.
1 = monday.
2 = tuesday.
3 = wednesday.
4 = thursday.
5 = friday.
6 = saturday.
step 1 : 28 may 2006: take date and year's last two digit number.
28+06 = 34.
step 2: year is divided by 4 so 06/4=1(leave the remainder).
step 3: Codes.
Jan : 0
Feb : 3
Mar : 3
aprl: 6
May : 1
Jun : 2
July; 6
aug : 2
Sep : 5
oct : 0
Nov : 3
Dec : 5
Now add step 1,step 2,step 3 ie,. 34+1+1=36.
divide it by seven 7, so remainder is 0.
Then,
0 = sunday.
1 = monday.
2 = tuesday.
3 = wednesday.
4 = thursday.
5 = friday.
6 = saturday.
Chirag said:
1 decade ago
Rule 1: Find the remind from complete years by dividing 28
i.e(2005/28 gives 17)
So consider 17 years as left.. it has 4 leap year
as general procedure code for that is 0
Rule 2: Same for month and date calculation :
(31+28+31+30+28)/7 it gives remainder 1
So code is 1
Total code is 1+0=1
Now,
MAGIC rule: add 6 in total cade
i.e (1+6=7)
So code is 7 means 0 means Sunday ! !
It can apply for any type of question...
i.e(2005/28 gives 17)
So consider 17 years as left.. it has 4 leap year
as general procedure code for that is 0
Rule 2: Same for month and date calculation :
(31+28+31+30+28)/7 it gives remainder 1
So code is 1
Total code is 1+0=1
Now,
MAGIC rule: add 6 in total cade
i.e (1+6=7)
So code is 7 means 0 means Sunday ! !
It can apply for any type of question...
Kannaiah said:
9 years ago
MONTH CODE.
J F M A M J J A S O N D
1 4 4 0 2 5 0 3 6 1 4 6
YEAR CODE = 2006 - 6.
Sun = 1. Mon = 2. T = 3. W = 4. Thurs = 5.Fri = 6.Sat = 7 or 0.
28/5/2006.
D + M + Y + Leap year + Cc.
28 + 2 + 6 + 1 + 6 = 43.
43/7 = Q = 6, Re = 1.
SUNDAY = 1.
J F M A M J J A S O N D
1 4 4 0 2 5 0 3 6 1 4 6
YEAR CODE = 2006 - 6.
Sun = 1. Mon = 2. T = 3. W = 4. Thurs = 5.Fri = 6.Sat = 7 or 0.
28/5/2006.
D + M + Y + Leap year + Cc.
28 + 2 + 6 + 1 + 6 = 43.
43/7 = Q = 6, Re = 1.
SUNDAY = 1.
Poonam kowtikwar said:
4 years ago
HERE AS IT IS GIVEN 28 MAY 2006.
28+ 1(CODE OF MAY) +06(LAST TWO DIGITS OF 2006)+06/4=(1),
SO 28+1+6+1 =36 then divide it by 7 for odd days,
36/7 = 1 is reminder.
So according to days
0 - SUNDAY
1 - MONDAY
2 - TUESDAY
3 - WEDNESDAY
4 - THURSDAY
5 - FRIDAY
6 - SATURDAY
7 - SUNDAY.
Here 1 is answer, So MONDAY (But) as 2006 is a leap year - 1 DAY SO (MONDAY - 1 = SUNDAY).
That's it. Thanks.
28+ 1(CODE OF MAY) +06(LAST TWO DIGITS OF 2006)+06/4=(1),
SO 28+1+6+1 =36 then divide it by 7 for odd days,
36/7 = 1 is reminder.
So according to days
0 - SUNDAY
1 - MONDAY
2 - TUESDAY
3 - WEDNESDAY
4 - THURSDAY
5 - FRIDAY
6 - SATURDAY
7 - SUNDAY.
Here 1 is answer, So MONDAY (But) as 2006 is a leap year - 1 DAY SO (MONDAY - 1 = SUNDAY).
That's it. Thanks.
(9)
Meena said:
1 decade ago
400 years is called one century so its value is 0.
Next 2000 year also is value 0.
2005 have one leap year and 4 ordinary year.
1 leap year have 2 odd days 4+2=6.
2006.
Jan-31.
Feb-28.
March-31.
April-30.
May-28 (days only giving).
31+28+31+30+28 = 148.
148/7 days it have 21 weeks 1 odd day.
0 century (2000)+6 odd day (2005)+1 odd day (2006) = 7.
Week day is 7.
Then 7/7=0.
So this answer is sunday.
Next 2000 year also is value 0.
2005 have one leap year and 4 ordinary year.
1 leap year have 2 odd days 4+2=6.
2006.
Jan-31.
Feb-28.
March-31.
April-30.
May-28 (days only giving).
31+28+31+30+28 = 148.
148/7 days it have 21 weeks 1 odd day.
0 century (2000)+6 odd day (2005)+1 odd day (2006) = 7.
Week day is 7.
Then 7/7=0.
So this answer is sunday.
Bikramdevgan said:
9 years ago
Please memorise.
0 Jan
3 Feb
3 March
6 April
1 May
4 June
6 July
2 Aug
5 Sept
0 Oct
3 Nov
5 Dec
&
6 1600-1699
4 1700-1799
2 1800-1899
0 1900-2000
6 2000-2009
15 Aug 1947.
15 + 2 (month key) + 0(year key) + 47 + 11(47/4 "11" quotient) = 78.
78/7 remainder 1.
0 Sunday
1 Monday
2 Tuesday
3 Wednesday
4 Thursday
5 Friday
6 Saturday
Please tell me, If I am wrong.
0 Jan
3 Feb
3 March
6 April
1 May
4 June
6 July
2 Aug
5 Sept
0 Oct
3 Nov
5 Dec
&
6 1600-1699
4 1700-1799
2 1800-1899
0 1900-2000
6 2000-2009
15 Aug 1947.
15 + 2 (month key) + 0(year key) + 47 + 11(47/4 "11" quotient) = 78.
78/7 remainder 1.
0 Sunday
1 Monday
2 Tuesday
3 Wednesday
4 Thursday
5 Friday
6 Saturday
Please tell me, If I am wrong.
Effortless said:
8 years ago
If It is like that, Can anybody tell me what was the day on 1 Jan 1601 ?
When I calculate;
The 1600 year will have 0 odd days ( as it says 400 years has 0 odd days )
1 Jan i.e 1 odd day.
Total odd days till 1 Jan 1601 =1 odd day.
I.e. Monday ( As formula says 0 - Sunday, 1 - Monday ).
But the Actual day on 1 Jan 1601 was Thursday.
So, please explain me to get the solution.
When I calculate;
The 1600 year will have 0 odd days ( as it says 400 years has 0 odd days )
1 Jan i.e 1 odd day.
Total odd days till 1 Jan 1601 =1 odd day.
I.e. Monday ( As formula says 0 - Sunday, 1 - Monday ).
But the Actual day on 1 Jan 1601 was Thursday.
So, please explain me to get the solution.
Lalitha said:
1 decade ago
How can we say the odd day is 0 for 1600 and 400 years...i am getting 1 odd day for total 2000 years....hear is my calculation if wrong pls corret it..
2000/4=500
so 500 leap years..no of odd days for 500 years is (500*2)1000
and 1500 ordinary years having 1500 odd days
total=1500+1000=2500
no of odd days in these 2500 days: 2500/7 = 1 odd day
is it correct procedure?
2000/4=500
so 500 leap years..no of odd days for 500 years is (500*2)1000
and 1500 ordinary years having 1500 odd days
total=1500+1000=2500
no of odd days in these 2500 days: 2500/7 = 1 odd day
is it correct procedure?
Harshitha said:
1 decade ago
I had a little confusion in calender problems.
In order to find what was the day of the week on 16th July, 1776?
Ans: Here chosen no. of odd days in 1600 years = 0.
No. of odd days in 100 years = 5 ?
While in the problems like what was the day of the week on 15th August, 1947?
Odd days in 1600 years = 0.
Odd days in 300 years = 1.
How come? Please explain.
In order to find what was the day of the week on 16th July, 1776?
Ans: Here chosen no. of odd days in 1600 years = 0.
No. of odd days in 100 years = 5 ?
While in the problems like what was the day of the week on 15th August, 1947?
Odd days in 1600 years = 0.
Odd days in 300 years = 1.
How come? Please explain.
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