The number of days more than the complete weeks are called odd days in a given period.

1 ordinary year ≡ 365 days ≡ (52 weeks + 1 day)
Hence the number of odd days in 1 ordinary year= 1.
1 leap year ≡ 366 days ≡ (52 weeks + 2 days)
Hence the number of odd days in 1 leap year= 2.

100 years ≡ (76 ordinary years + 24 leap years )
= (76 x 1 + 24 x 2) odd days = 124 odd days.
=> (17 weeks + 5 days)
≡ 5 odd days.
Hence the number of odd days in 100 years = 5.

Number of odd days—->Day of the week
==>0—->Sunday
==>1—->Monday
==>2—->Tuesday
==>3—->Wednesday
==>4—->Thursday
==>5—->Friday
==>6—->Saturday
last day of a century cannot be Tuesday or Thursday or Saturday.
because 100 century consists 5 odd days, means Friday is the last day.
200 century consists 3 odd days, means Wednesday is the last day.
300 century consists 1 odd day, means Monday is the last day.
400 century doesn’t consist any odd day.
the cycle continues, hence proof.

Solution: First, we count the number of odd days for the leftover days in the given period.

Here,given period is 12.2.1986 to 1.1.1987

Feb Mar Apr May June July  Aug Sept Oct Nov Dec Jan

16   31    30  31   30    31   31   30   31  30   31   1 (left days)

2 + 3 + 2 + 3 + 2 + 3 + 3 + 2 + 3 + 2 + 3 + 1(odd days) = 1 odd day

So,given day Wednesday + 1 = Thursday is the required result.

Repetition of leap year ===> Add +28 to the Given Year.

Repetition of non leap year

Step 1 : Add +11 to the Given Year. If Result is a leap year, Go to step 2.

Step 2:  Add +6 to the Given Year.

Solution :

Given Year is 2024, Which is a leap year.

So, Add +28 to the given year (i.e 2024 + 28) = 2052

Therfore, The calendar of the year 2024 can be used again in the year 2052.

Repetition of leap year ===> Add +28 to the Given Year.

Repetition of non leap year

Step 1 : Add +11 to the Given Year. If Result is a leap year, Go to step 2.

Step 2:  Add +6 to the Given Year.

Solution :

Given Year is 1993, Which is a non leap year.

Step 1 : Add +11 to the given year (i.e 1993 + 11) = 2004, Which is a leap year.

Step 2 : Add +6 to the given year (i.e 1993 + 6) = 1999

Therfore, The calendar for the year 1993 will be same for the year 1999

The number of days more than the complete weeks are called odd days in a given period.

1 ordinary year ≡ 365 days ≡ (52 weeks + 1 day)
Hence the number of odd days in 1 ordinary year= 1.
1 leap year ≡ 366 days ≡ (52 weeks + 2 days)
Hence the number of odd days in 1 leap year= 2.

100 years ≡ (76 ordinary years + 24 leap years )
= (76 x 1 + 24 x 2) odd days = 124 odd days.
=> (17 weeks + 5 days)
≡ 5 odd days.
Hence the number of odd days in 100 years = 5.

Number of odd days—->Day of the week
==>0—->Sunday
==>1—->Monday
==>2—->Tuesday
==>3—->Wednesday
==>4—->Thursday
==>5—->Friday
==>6—->Saturday
last day of a century cannot be Tuesday or Thursday or Saturday.
because 100 century consists 5 odd days, means Friday is the last day.
200 century consists 3 odd days, means Wednesday is the last day.
300 century consists 1 odd day, means Monday is the last day.
400 century doesn’t consist any odd day.
the cycle continues, hence proof.

First,we count the number of odd days for the left over days in the given period.Here,given period is 18.3.1994 to 25.2.1995

Therefore, No. of Odd Days = 6 + 2 + 3 + 2 + 3 + 3 + 2 + 3 + 2 + 3 + 3 + 4 = 36 = 1 odd day

So, given day Friday + 1 = Saturday is the required result.

As day after tomorrow is Monday, today will be Saturday and yesterday will be Friday…

Here, after 8 complete weeks again we will get Monday and 5 more days. So, Monday+5=

Saturday

Since within month ( only date different month and year same), we must count the total day after 2, that is from 3 January to 22 January and then divide by 7. Take remainder as ODD /Extra day. Then add an extra day with the given day. That is, 22-2=20/7, and it will go 2 times and the remainder will be 6. Hence given day +6= Wednesday+6= Tuesday

Total extra days=12 ( Which is greater than 7, so again must divide by 7 to get a total extra day)

First, write down DBY then Y, TD, T, DAT, 1, 2, 3  serially and then count from Thursday to 3.  3 is Friday. Hence the third day after the day after tomorrow is Friday

First of all, we have to find the day on 01.03.2005

01.03.2005 = (2004 years + Period from 01.01.2005 to 01.03.2005)

Number of odd days in 2000 years = 0 —-(1)
Number of odd days in 4 years = 1 leap year + 3 ordinary year
= 1X2 + 3X1 = 5 days —-(2)

From 01.01.2005 to 01.03.2005, we have,
Jan-31days + Feb-28days + Mar-1day = 60days
= 8weeks + 4days = 4 odd days —-(3)

Adding (1),(2)&(3), we get,total no.of odd days = 0+5+4
= 9 days = 1 week + 2days = 2 odd days.
2 odd days corresponding to Tuesday.So 01.03.2005 was Tuesday. Therefore, Friday lies on 04.03.2005.

Hence, the dates of March,2015 on which Fridays fell are 4,11,18&25

16th July,1776 = (1775 years + Period from 01.01.1776 to 16.07.1776)

1775 = 1600 + 100 + 75

No.of odd days in 1600 years = 0 —-(1)
No.of odd days in 100 years = 5 —-(2)
No.of odd days in 75 years = 18 leap years + 57 ordinary years = 18X2+57X1=93days=13weeks+2days=2 odd days —-(3)

From 01.01.1776 to 16.07.1776, we have,

Jan-31days+Feb-29days+Mar-31days+Apr-30days+May-31days+Jun-30days+July-16days = 198days = 28weeks + 2 days = 2 odd days —-(4)

Adding (1),(2),(3) & (4), we get,
Total no.of odd days = 0+5+2+2 = 9days = 1 week + 2days = 2 odd days. 2 odd days corresponding to Tuesday

Hence, the day of the week on 16th July,1776 is Tuesday

For example,if our answer is 2018, we must have the same day on 01.01.07 & 01.01.18 and number of odd days from 01.01.07 to 31.12.17 must be zero. Let us check the same thing for each of the given options.

Option (a)2015: No.of odd days from 01.01.2007 to 31.12.2014 = 2 leap years + 6 ordinary years = 2X2+6X1 = 10days = 1week + 3days = 3 odd days[not correct]

Option (b)2016: No.of odd days from 01.01.2007 to 31.12.2015 = 2 leap years + 7 ordinary years = 2X2+7X1 = 11 days = 1 week + 4 days = 4 odd days [not correct]

Option (c)2017: No.of odd days from 01.01.2007 to 31.12.2016 = 3 leap years + 7 ordinary years = 3X2+7X1 = 13 days = 2 weeks + 1 day = 1 odd day [not correct]

Option (d)2018: No.of odd days from 01.01.2007 to 31.12.2017 = 3 leap years + 8 ordinary years = 3X2+8X1 = 14 days = 2 weeks + 0 day = 0 odd day [correct]

Number of days in “x” weeks = 7x
Number of days in “x” days = x

Number of days in “x”weeks”x”days = 7x+x = 8x

Hence, the number of days in “x”weeks”x”days = 8x days