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 1 of 22.
Mrugesh said:
1 decade ago
Important Points:
An ordinary year has 365 days = 52 weeks and 1 odd day.
A leap year has 366 days = 52 weeks and 2 odd days.
Century = 76 Ordinary years + 24 Leap years.
Century contain 5 odd days.
200 years contain 3 odd days.
300 years contain 1 odd day.
400 years contain 0 odd days.
Last day of a century cannot be Tuesday, Thursday or Saturday.
First day of a century must be Monday, Tuesday, Thursday or Saturday.
Explanation:
100 years = 76 ordinary years + 24 leap years
= 76 odd days + 24 x 2 odd days
= 124 odd days = 17 weeks + 5 days
100 years contain 5 odd days.
No. of odd days in first century = 5
Last day of first century is Friday.
No. of odd days in two centuries = 3
Wednesday is the last day.
No. of odd days in three centuries = 1
Monday is the last day.
No. of odd days in four centuries = 0
Sunday is the last day.
Since the order is continually kept in successive cycles, the last day of a century cannot be Tuesday, Thursday or Saturday.
So, the last day of a century should be Sunday, Monday, Wednesday or Friday.
Therefore, the first day of a century must be Monday, Tuesday, Thursday or Saturday.
Working Rules:
Working rule to find the day of the week on a particular date when reference day is given:
Step 1: Find the net number of odd days for the period between the reference date and the given date (exclude the reference day but count the given date for counting the number of net odd days).
Step 2: The day of the week on the particular date is equal to the number of net odd days ahead of the reference day (if the reference day was before this date) but behind the reference day (if this date was behind the reference day).
Working rule to find the day of the week on a particular date when no reference day is given
Step 1: Count the net number of odd days on the given date
Step 2: Write:
For 0 odd days " Sunday
For 1 odd day " Monday
For 2 odd days " Tuesday
. . . .
. . . .
. . . .
For 6 odd days " Saturday
Examples:
1. If 11th January 1997 was a Sunday then what day of the week was on 10th January 2000?
Sol: Total number of days between 11th January 1997 and 10th January 2000
= (365 " 11) in 1997 + 365 in 1998 + 365 in 1999 + 10 days in 2000
= (50 weeks + 4 odd days) + (52 weeks + 1 odd day) +
(52 weeks + 1 odd day) + (1 week + 3 odd days)
Total number of odd days = 4 + 1 + 1 + 3 = 9 days = 1 week + 2 days
Hence, 10th January, 2000 would be 2 days ahead of Sunday i.e. it was on Tuesday.
2. What day of the week was on 10th June 2008?
Sol: 10th June 2008 = 2007 years + First 5 months up to May 2008 + 10 days of June
2000 years have 0 odd days.
Remaining 7 years has 1 leap year and 6 ordinary years2 + 6 = 8 odd days
So, 2007 years have 8 odd days.
No. of odd days from 1st January 2008 to 31st May 2008 = 3+1+3+2+3 = 12
10 days of June has 3 odd days.
Total number of odd days = 8+12+3 = 23
23 odd days = 3 weeks + 2 odd days.
Hence, 10th June, 2008 was Tuesday.
An ordinary year has 365 days = 52 weeks and 1 odd day.
A leap year has 366 days = 52 weeks and 2 odd days.
Century = 76 Ordinary years + 24 Leap years.
Century contain 5 odd days.
200 years contain 3 odd days.
300 years contain 1 odd day.
400 years contain 0 odd days.
Last day of a century cannot be Tuesday, Thursday or Saturday.
First day of a century must be Monday, Tuesday, Thursday or Saturday.
Explanation:
100 years = 76 ordinary years + 24 leap years
= 76 odd days + 24 x 2 odd days
= 124 odd days = 17 weeks + 5 days
100 years contain 5 odd days.
No. of odd days in first century = 5
Last day of first century is Friday.
No. of odd days in two centuries = 3
Wednesday is the last day.
No. of odd days in three centuries = 1
Monday is the last day.
No. of odd days in four centuries = 0
Sunday is the last day.
Since the order is continually kept in successive cycles, the last day of a century cannot be Tuesday, Thursday or Saturday.
So, the last day of a century should be Sunday, Monday, Wednesday or Friday.
Therefore, the first day of a century must be Monday, Tuesday, Thursday or Saturday.
Working Rules:
Working rule to find the day of the week on a particular date when reference day is given:
Step 1: Find the net number of odd days for the period between the reference date and the given date (exclude the reference day but count the given date for counting the number of net odd days).
Step 2: The day of the week on the particular date is equal to the number of net odd days ahead of the reference day (if the reference day was before this date) but behind the reference day (if this date was behind the reference day).
Working rule to find the day of the week on a particular date when no reference day is given
Step 1: Count the net number of odd days on the given date
Step 2: Write:
For 0 odd days " Sunday
For 1 odd day " Monday
For 2 odd days " Tuesday
. . . .
. . . .
. . . .
For 6 odd days " Saturday
Examples:
1. If 11th January 1997 was a Sunday then what day of the week was on 10th January 2000?
Sol: Total number of days between 11th January 1997 and 10th January 2000
= (365 " 11) in 1997 + 365 in 1998 + 365 in 1999 + 10 days in 2000
= (50 weeks + 4 odd days) + (52 weeks + 1 odd day) +
(52 weeks + 1 odd day) + (1 week + 3 odd days)
Total number of odd days = 4 + 1 + 1 + 3 = 9 days = 1 week + 2 days
Hence, 10th January, 2000 would be 2 days ahead of Sunday i.e. it was on Tuesday.
2. What day of the week was on 10th June 2008?
Sol: 10th June 2008 = 2007 years + First 5 months up to May 2008 + 10 days of June
2000 years have 0 odd days.
Remaining 7 years has 1 leap year and 6 ordinary years2 + 6 = 8 odd days
So, 2007 years have 8 odd days.
No. of odd days from 1st January 2008 to 31st May 2008 = 3+1+3+2+3 = 12
10 days of June has 3 odd days.
Total number of odd days = 8+12+3 = 23
23 odd days = 3 weeks + 2 odd days.
Hence, 10th June, 2008 was Tuesday.
(1)
VIJAY YETHIRAJAM said:
1 decade ago
Simple solution here is the trick. Just need to follow this step, guys this type of questions is very easy. But you.
Need to be perfect while solving.
First of all.
Month code.
Jan - 1.
Feb - 4.
Mar - 4.
Apr - 0.
May - 2.
June - 5.
July - 0.
August - 3.
September - 6.
October - 1.
November - 4.
December - 6.
Day code:
Sun - 1.
Mon - 2.
Tues - 4.
Wed - 5.
Thur - 7.
Fri - 8.
Sat - no decimal.
Now lets solve one example.
Example :- My DOB is 15th august 1989.
Step 1 - Take the last two digit of your birth year. Here in my case it is ' 89 '.
Step 2 - Now multiply the last two digit of birth year by "5".
89 x 5 = 445.
Step 3 - Now divide the result by "4".
445/4 = 111.25.
Step 4 - in this step remove the part of the result after decimal (RHS).
111.25 - remove 25.
Now we are having 111.
Step 5 - Now add the month code given above to this.
Month code for august is '3'.
So after adding.
3 + 111 = 114.
Step 6 - Now add the date of your birth. To this.
114 + 15 (15th august) = 129.
Step 7 - Don't worry this is the last step. ;-) ;-) ha ha.
Now finally divide this by '7'.
129/7 = 18.428.
Now we have got the final answer 18.428.
Now very important thing to remember. You have to look.
Upon the next digit after the decimal (right hand side).
In 18.428 the digit after the decimal is '4'.
Now go to the day code given above and see there '4' is for Tuesday. So answer is Tuesday.
15th august 1989 was a Tuesday.
Hope you all have understood this trick.
Note :- 1) I f in the " step 7 " you get no decimal than it is definitely " Saturday ". (see the day code column I have Already mentioned there " Saturday - no decimal ").
2) what to do if the date given is from leap year. Do the all step as it is, just.
Consider the earlier day.
Ex:- If at the last step you get " Friday " then Thursday will be your answer.
That's it. Just practice this steps you'll be perfect. Hope you all like this.
Need to be perfect while solving.
First of all.
Month code.
Jan - 1.
Feb - 4.
Mar - 4.
Apr - 0.
May - 2.
June - 5.
July - 0.
August - 3.
September - 6.
October - 1.
November - 4.
December - 6.
Day code:
Sun - 1.
Mon - 2.
Tues - 4.
Wed - 5.
Thur - 7.
Fri - 8.
Sat - no decimal.
Now lets solve one example.
Example :- My DOB is 15th august 1989.
Step 1 - Take the last two digit of your birth year. Here in my case it is ' 89 '.
Step 2 - Now multiply the last two digit of birth year by "5".
89 x 5 = 445.
Step 3 - Now divide the result by "4".
445/4 = 111.25.
Step 4 - in this step remove the part of the result after decimal (RHS).
111.25 - remove 25.
Now we are having 111.
Step 5 - Now add the month code given above to this.
Month code for august is '3'.
So after adding.
3 + 111 = 114.
Step 6 - Now add the date of your birth. To this.
114 + 15 (15th august) = 129.
Step 7 - Don't worry this is the last step. ;-) ;-) ha ha.
Now finally divide this by '7'.
129/7 = 18.428.
Now we have got the final answer 18.428.
Now very important thing to remember. You have to look.
Upon the next digit after the decimal (right hand side).
In 18.428 the digit after the decimal is '4'.
Now go to the day code given above and see there '4' is for Tuesday. So answer is Tuesday.
15th august 1989 was a Tuesday.
Hope you all have understood this trick.
Note :- 1) I f in the " step 7 " you get no decimal than it is definitely " Saturday ". (see the day code column I have Already mentioned there " Saturday - no decimal ").
2) what to do if the date given is from leap year. Do the all step as it is, just.
Consider the earlier day.
Ex:- If at the last step you get " Friday " then Thursday will be your answer.
That's it. Just practice this steps you'll be perfect. Hope you all like this.
(1)
I am Zero said:
1 decade ago
What was the day of the week on 28th May, 2006?
Here is d ? right.
Okay,
Need to calculate 28th May, 2006 rite,
(Please refer important formulas first okay :))
Now 1600 years have 0 odd days because 1600/4= No reminder(we can divide 4 completely in 1600).
Now 400 years have 0 odd days because 400/4= No reminder(we can divide 4 completely in 400).
You confused here right.. okay see 1600+400=2000 so we came till 2000th year we need to calculate for 6 more years to reach 2006. Question is 2006 may right, so calculate till 2005 so 5 more years can calculate now okay then will calculate remaining days in 2006 okay.
So,
2001= its ordinary year so have 54 weeks and 1 odd day as per formula right
2002= its ordinary year so have 54 weeks and 1 odd day as per formula right.
2003= its ordinary year so have 54 weeks and 1 odd day as per formula right.
2004= 54 weeks and 2 odd day.
2005= its ordinary year so have 54 weeks and 1 odd day as per formula right.
1+1+1+2+1=6.
So total=6 odd days.
Now one more year which is 2006, here till 28th may needed so,
jan feb mar apr may
31 28 31 30 28th may ok.
31+28+31+30+28= 148 total days.
So calculate how many weeks and odd days in 148.
For this 148/7= 28 and reminder (1).
So 21 week and (1) odd day.
Add this(1) with (6) odd days that we calculate before of
So total 7 odd days now finish see.
1 2 3 4 5 6 7or(0)
Mo Tu We Th Fr Sat Sun
Total 7 odd days so 7th value OK... for example we got 5 odd days means answer will be Fri OK.
Here Sunday Enjoy :).
Here is d ? right.
Okay,
Need to calculate 28th May, 2006 rite,
(Please refer important formulas first okay :))
Now 1600 years have 0 odd days because 1600/4= No reminder(we can divide 4 completely in 1600).
Now 400 years have 0 odd days because 400/4= No reminder(we can divide 4 completely in 400).
You confused here right.. okay see 1600+400=2000 so we came till 2000th year we need to calculate for 6 more years to reach 2006. Question is 2006 may right, so calculate till 2005 so 5 more years can calculate now okay then will calculate remaining days in 2006 okay.
So,
2001= its ordinary year so have 54 weeks and 1 odd day as per formula right
2002= its ordinary year so have 54 weeks and 1 odd day as per formula right.
2003= its ordinary year so have 54 weeks and 1 odd day as per formula right.
2004= 54 weeks and 2 odd day.
2005= its ordinary year so have 54 weeks and 1 odd day as per formula right.
1+1+1+2+1=6.
So total=6 odd days.
Now one more year which is 2006, here till 28th may needed so,
jan feb mar apr may
31 28 31 30 28th may ok.
31+28+31+30+28= 148 total days.
So calculate how many weeks and odd days in 148.
For this 148/7= 28 and reminder (1).
So 21 week and (1) odd day.
Add this(1) with (6) odd days that we calculate before of
So total 7 odd days now finish see.
1 2 3 4 5 6 7or(0)
Mo Tu We Th Fr Sat Sun
Total 7 odd days so 7th value OK... for example we got 5 odd days means answer will be Fri OK.
Here Sunday Enjoy :).
(1)
Mohit Kumar said:
4 years ago
According to me, the simplest way of solution is;
Months - 1440, 2305, 6146
Here, 1 = January, 4 = February, 4 = March, 0= April, May = 2, June = 3, July = 0, August = 5, September = 6, October = 1, November = 4, December = 6.
Learn this table along with the nos of their respective months.
Now come to days, For Days Remember:
1 - Sunday.
2- Monday.
3- Tuesday.
4- Wednesday.
5 - Thursday.
6- Friday .
0 or no remainder = Saturday.
Now come to century part
Remember the small table and understand the concept as to how we can calculate the no that needs to be assigned to a Century.
The No for 2000 Century must be 6, If we go before we need to again Start the chains or simply started from 6 till you reach 6 no.
For Calculating no, after 2000 century like 2100, 2200 just Substract by 2 until you reach 0
Let's see the below example, to go back century just set the 0,2,4, 6 in increasing order and to go after 2000 century just reduce the no by 2.
In below example 2000 is our base.
1600 - 6
1700 - 4
1800- 2
1900 - 0
2000 - 6
2001 - 4
2002 - 2
2003 - 0
And so on
Now come to the question, it Will hardly take 5 to 10 seconds to solve this qsn now, Just learn the above 3 tables
The question is 28 May 2006
Here write in the following manner
28 - Date.
2 - Month(See no assigned in table),
06 - Year(write last 2 digits of year),
6 - Century(2000),
1 - (Divide last 2 digits by 4).
Add on these nos - 43.
Now Divide 43 by 7 and the remainder would be = 1.
Now see the day's table, where 1 = Sunday.
Hence the answer is Sunday.
Months - 1440, 2305, 6146
Here, 1 = January, 4 = February, 4 = March, 0= April, May = 2, June = 3, July = 0, August = 5, September = 6, October = 1, November = 4, December = 6.
Learn this table along with the nos of their respective months.
Now come to days, For Days Remember:
1 - Sunday.
2- Monday.
3- Tuesday.
4- Wednesday.
5 - Thursday.
6- Friday .
0 or no remainder = Saturday.
Now come to century part
Remember the small table and understand the concept as to how we can calculate the no that needs to be assigned to a Century.
The No for 2000 Century must be 6, If we go before we need to again Start the chains or simply started from 6 till you reach 6 no.
For Calculating no, after 2000 century like 2100, 2200 just Substract by 2 until you reach 0
Let's see the below example, to go back century just set the 0,2,4, 6 in increasing order and to go after 2000 century just reduce the no by 2.
In below example 2000 is our base.
1600 - 6
1700 - 4
1800- 2
1900 - 0
2000 - 6
2001 - 4
2002 - 2
2003 - 0
And so on
Now come to the question, it Will hardly take 5 to 10 seconds to solve this qsn now, Just learn the above 3 tables
The question is 28 May 2006
Here write in the following manner
28 - Date.
2 - Month(See no assigned in table),
06 - Year(write last 2 digits of year),
6 - Century(2000),
1 - (Divide last 2 digits by 4).
Add on these nos - 43.
Now Divide 43 by 7 and the remainder would be = 1.
Now see the day's table, where 1 = Sunday.
Hence the answer is Sunday.
(13)
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)
Madhav goutham said:
1 decade ago
Hello Budz.
This is not my technique but I am just explaining the technique, which they explained here. It is only for those who do not understand the technique.
Caution: This can only be used for years above 1900.
So let me start.
If you have to find the day 31st October 1984. 1600 and 300 will give you 1900.
Just remember 1600 has 0 odd day and 300 has 1 odd day. Take the year before 1984 which is 1983.
Then calculate 1983-1900=83; Divide 83 by 4 then you will get quotient as 20, so there will be 20 leap.
Years and 83-20=63 ordinary years will be there. So in leap years there will be 2 odd days and in ordinary years there will be 1 odd day.
So calculate 20*2+63*1 = 103;
Divide 103 by 7 you will get 5 odd days.
Then calculate total no of days from January to October 31. 305 days will be there. Then divide 305 by 7 you will get 4 as result.
Then add 4+5+1+0=10;
Then again divide 10 by 7 will get result as 3;
0-Sunday.
1-Monday.
2-Tuesday.
3-Wednesday.
So answer will be Wednesday.
If you have any doubt regarding this you may please post here I will check out regularly.
This is not my technique but I am just explaining the technique, which they explained here. It is only for those who do not understand the technique.
Caution: This can only be used for years above 1900.
So let me start.
If you have to find the day 31st October 1984. 1600 and 300 will give you 1900.
Just remember 1600 has 0 odd day and 300 has 1 odd day. Take the year before 1984 which is 1983.
Then calculate 1983-1900=83; Divide 83 by 4 then you will get quotient as 20, so there will be 20 leap.
Years and 83-20=63 ordinary years will be there. So in leap years there will be 2 odd days and in ordinary years there will be 1 odd day.
So calculate 20*2+63*1 = 103;
Divide 103 by 7 you will get 5 odd days.
Then calculate total no of days from January to October 31. 305 days will be there. Then divide 305 by 7 you will get 4 as result.
Then add 4+5+1+0=10;
Then again divide 10 by 7 will get result as 3;
0-Sunday.
1-Monday.
2-Tuesday.
3-Wednesday.
So answer will be Wednesday.
If you have any doubt regarding this you may please post here I will check out regularly.
Prasanna Kartik said:
1 decade ago
Hi Guys,
We can solve this by simple formula method and it is called dooms algorithm.
Dooms day = Anchor day+(year/12)+Reminder(year/12)/12+(Reminder(year/12)/12)/4.
For 1900 to 1999 - Anchor will be Wednesday.
2000 to 2099 - Anchor will be Tuesday.
Out come of this doom day will be last day of February, 4th April (4/4), 6 June (6/6), 8th Aug (8/8), 10th Oct (12/12) and 12th Dec (12/12).
By seeing this this formula would be bit confusing but once you have applied this method it will be very handy.
Now the question is 28th May, 2006.
Dooms day = Tuesday+06/12+Remainder(06/12)+Remainder(06/12)/4.
= Tuesday+6+12+3 = Tuesday+21 = Tuesday (There is no ODD days because Rem 21/7 is Zero).
Now from the above method pick a date closest to date which you have to find our problem states May month, so I picked 4th April (4/4).
Now we knew that 4th April is Tuesday (from dooms day algorithm).
We have to find 28th May which is Tuesday+26.
(Remaining days of April)+28(May) = Tuesday+54 = Tuesday+5(ODD days) = Sunday.
We can solve this by simple formula method and it is called dooms algorithm.
Dooms day = Anchor day+(year/12)+Reminder(year/12)/12+(Reminder(year/12)/12)/4.
For 1900 to 1999 - Anchor will be Wednesday.
2000 to 2099 - Anchor will be Tuesday.
Out come of this doom day will be last day of February, 4th April (4/4), 6 June (6/6), 8th Aug (8/8), 10th Oct (12/12) and 12th Dec (12/12).
By seeing this this formula would be bit confusing but once you have applied this method it will be very handy.
Now the question is 28th May, 2006.
Dooms day = Tuesday+06/12+Remainder(06/12)+Remainder(06/12)/4.
= Tuesday+6+12+3 = Tuesday+21 = Tuesday (There is no ODD days because Rem 21/7 is Zero).
Now from the above method pick a date closest to date which you have to find our problem states May month, so I picked 4th April (4/4).
Now we knew that 4th April is Tuesday (from dooms day algorithm).
We have to find 28th May which is Tuesday+26.
(Remaining days of April)+28(May) = Tuesday+54 = Tuesday+5(ODD days) = Sunday.
Arun Anoop M said:
1 decade ago
I will give you one example:
--------------------------
First of all you have to learn imp things...ie;
Step 1:
Odd days in 100 years is 5
Odd days in 200 years is 3
Odd days in 300 years is 1
Odd days in 400 years is 0
---------------------------
Ok.
Step 2:
1 for Mon , 2 for Tue ..... 7 for sunday.
Step 3:
Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec
31 28 31 30 31 30 31 31 30 31 30 31
Step 4:
Example is
Find a day ?
From "28 May 2006"
2005 years + (01-01-2006 to 28-05-06)
CONVERT 2005 as
2005 is same as 1600+400+5
Odd days in 1600 years is 0
Odd days in 400 years is 0
Calculate 5years?
5years = (5/4)leap year + (5- (5/4)) ordinary day
= 1 leap year + 4 ordinary days
= (1*2) ordinary days + 4 ordinary days
= 6 odd days.
Next Step:
Jan+Feb+Mar+Apr+May=31+28+31+30+28=148 odd days.
Convert 148 as (weeks+days)format.
How?
148/7 = (21 weeks)+(1 odd day)
= 1 odd day.
Final Step:
1600+400+5+months+day = 0+0+6+1 = 7.
So DAY is SUNDAY.
--------------------------
First of all you have to learn imp things...ie;
Step 1:
Odd days in 100 years is 5
Odd days in 200 years is 3
Odd days in 300 years is 1
Odd days in 400 years is 0
---------------------------
Ok.
Step 2:
1 for Mon , 2 for Tue ..... 7 for sunday.
Step 3:
Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec
31 28 31 30 31 30 31 31 30 31 30 31
Step 4:
Example is
Find a day ?
From "28 May 2006"
2005 years + (01-01-2006 to 28-05-06)
CONVERT 2005 as
2005 is same as 1600+400+5
Odd days in 1600 years is 0
Odd days in 400 years is 0
Calculate 5years?
5years = (5/4)leap year + (5- (5/4)) ordinary day
= 1 leap year + 4 ordinary days
= (1*2) ordinary days + 4 ordinary days
= 6 odd days.
Next Step:
Jan+Feb+Mar+Apr+May=31+28+31+30+28=148 odd days.
Convert 148 as (weeks+days)format.
How?
148/7 = (21 weeks)+(1 odd day)
= 1 odd day.
Final Step:
1600+400+5+months+day = 0+0+6+1 = 7.
So DAY is SUNDAY.
Kumar said:
8 years ago
Hi, friends here I'm giving the simple trick to slice these type of questions:
First Remind these codes they are : Monthly Codes
Jan = 0 , Feb = 3 , March = 3
April= 6 , May = 1, June = 4
July = 6 , Aug = 2 , Sept = 5
Octo= 0 , Nov = 3, Dec = 5
Century codes :
Please observe carefully and Apply the century code for the question
1600 = 6 1500 = 0
1700 = 4. 1400 = 4
1800 = 2 1300 = 2
1900 = 0 1200 = 6
Again it repeats
2000 = 6
2100 = 4
2200 = 2
2300 = 0
Weekly codes :
Sun - 0
M-. 1
T- 2
W. 3
Th 4
F-. 5
Sa. 6
Question : 28- may -2016?
Here to get the answer apply a format
First,
28 is the date.
May code - 1.
Century code - 6.
Last 2 digits - 06.
And add to 2 last digits by 4.. that means 06 by 4 take the quotient that is 06/4 = quotient= 1 add to the above format
Format:
28+1+6+6+1 = 42
And lastly, divide by 7
42/7 = 6.
Here take remainder when we are divide by 7 remainder will be 0.
Then 0 is Called Sunday.
First Remind these codes they are : Monthly Codes
Jan = 0 , Feb = 3 , March = 3
April= 6 , May = 1, June = 4
July = 6 , Aug = 2 , Sept = 5
Octo= 0 , Nov = 3, Dec = 5
Century codes :
Please observe carefully and Apply the century code for the question
1600 = 6 1500 = 0
1700 = 4. 1400 = 4
1800 = 2 1300 = 2
1900 = 0 1200 = 6
Again it repeats
2000 = 6
2100 = 4
2200 = 2
2300 = 0
Weekly codes :
Sun - 0
M-. 1
T- 2
W. 3
Th 4
F-. 5
Sa. 6
Question : 28- may -2016?
Here to get the answer apply a format
First,
28 is the date.
May code - 1.
Century code - 6.
Last 2 digits - 06.
And add to 2 last digits by 4.. that means 06 by 4 take the quotient that is 06/4 = quotient= 1 add to the above format
Format:
28+1+6+6+1 = 42
And lastly, divide by 7
42/7 = 6.
Here take remainder when we are divide by 7 remainder will be 0.
Then 0 is Called Sunday.
(1)
Hillihang said:
6 years ago
You can do this in easier way;
2000 Jan 1st is always Sunday so;
2000 Jan 1st Sunday
-2006 May 28
-------------------------------
6+1(leap year i.e. 2004) =7
Jan(31-1)=30 feb=28 march=31 April=30
May=28(as in the question we are asked to find the day of 28th May)
Now add all = 7+30+28+31+30+28 =154.
Now divide this number by 7 to find the odd day 154÷7 =22 as the remainder is zero so the day remains the same Sunday.
If the remainder was to suppose 2 then count two times after the Sunday because 2000 1st Jan is always Sunday which would have been Tuesday.
And for the 12 months for easy odd number counting.
Jan=31÷7=3
Feb=28÷7=0
March=31÷7=3
April=30÷7=2
May=31÷7=3
June=30÷7=2
July=31÷7=3
Aug=31÷7=3
Sept=30÷7=2
Oct=31÷7=3
Nov=30÷7=2
Dec=31÷7=3.
You can remember it as a formula (June Say No 2 April) these four have 2 odd numbers while others have 3 except February(i.e. 0).
Note:that Say is for September (Se) and No for November (No).
2000 Jan 1st is always Sunday so;
2000 Jan 1st Sunday
-2006 May 28
-------------------------------
6+1(leap year i.e. 2004) =7
Jan(31-1)=30 feb=28 march=31 April=30
May=28(as in the question we are asked to find the day of 28th May)
Now add all = 7+30+28+31+30+28 =154.
Now divide this number by 7 to find the odd day 154÷7 =22 as the remainder is zero so the day remains the same Sunday.
If the remainder was to suppose 2 then count two times after the Sunday because 2000 1st Jan is always Sunday which would have been Tuesday.
And for the 12 months for easy odd number counting.
Jan=31÷7=3
Feb=28÷7=0
March=31÷7=3
April=30÷7=2
May=31÷7=3
June=30÷7=2
July=31÷7=3
Aug=31÷7=3
Sept=30÷7=2
Oct=31÷7=3
Nov=30÷7=2
Dec=31÷7=3.
You can remember it as a formula (June Say No 2 April) these four have 2 odd numbers while others have 3 except February(i.e. 0).
Note:that Say is for September (Se) and No for November (No).
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