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

From the notes given on this web page, we have

Number of odd days in 100 years (1st century) = 5
So, the last day of 1st century is Friday

Number of odd days in 200 years (2nd century) = 3
So, the last day of 2nd century is Wednesday

Number of odd days in 300 years (3rd century) = 1
So, the last day of 3rd century is Monday

Number of odd days in 400 years (4th century) = 0
So, the last day of 4th century is Sunday

This pattern is repeated

Therefore, the last day of a century can not be Tuesday or Thursday or Saturday.

1988 april 4th to 1987 november 3rd we have 27+31+31+29+31+4 = 153 = 153/7 = 21 weeks + 6 odd days.
rnHere is where everyone got confused,As per rules we take
rn[0 -sun 1 -mon 2 -tues 3 -wed 4 – thurs 5 – fri 6 -sat]
rnAs no.of odd days = 6 you may think the answer is saturday…
rnBut here later date is given and former date is asked,so the solution should be something like this..
rnwe have six odd days which means that we are short of one day to get the complete week i.e.., november 2nd would be monday and hence november 3rd is tuesday

This being a leap year none of the next 3 years is a leap year. So, the day of the week will be 3 days beyond Monday ie, it will be Thursday.

→ 4th June, 2002 = (2001 years + Period from 1.1.2002 to 4.6.2002)

→ Odd days in 1600 years = 0

→ Odd days in 400 years = 0

→ Odd days in 1 ordinary year = 1

→ Odd days in 2001 years = (0+0+1) = 1

→ Jan(31) Feb(28) March(31) April(30) May(31) June(4)

→ (31 + 28 + 31 + 30 + 31 + 4) = 155 days

→ 22 weeks + 1day ≡ 1 odd day

→ Total number of odd days = (1+1) = 2

→ Required day is Tuesday

61 days = 8 weeks 5 days = 5 odd days

Hence if today is Monday, After 61 days, it will be = (Monday + 5 odd days)
= Saturday