Question

What was the day of the week on 28th May, 2006?
Methodology to do this sum?

Answer 1

Wolfram :-P

Answer 2

http://www.wolframalpha.com/input/?i=28th+May+2006

Answer 3

Mathematically how to perform it?

Answer 4

count them. :D

Answer 5

Use a calender of your mobile phone.

Answer 6

no no there is a way lemme show u

Answer 7

4 August ---> Saturday
3 August ---> Friday

Just keep reverse counting.

Answer 8

There's a pretty complicated way.

Answer 9

Gosh! This is an aps ques!

Answer 10

http://en.wikipedia.org/wiki/Determination_of_the_day_of_the_week

Answer 11

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.

Answer 12

well prefer this: http://www.indiabix.com/aptitude/calendar/

Answer 13

lol ^ knew it.
You must use technology for anything.

Answer 14

@ParthKohli if u hav patience to read d wikipedia link. pls go ahead ;) thanku @mathslover i jus saw there n didn understand the odd days concept.

Answer 15

I don't want to learn this method, Shrey, but you do.

Answer 16

see we can write 2005 = 1600+400+5 right?

Answer 17

yes

Answer 18

so first calculate odd days in 1600 that is 0
in 400 = 0
and in 5 years = 4+(1 leap year)=4+2=6

Answer 19

now calculate odd days from 1.1.2006 to 28.05.2006

total days = 148
odd days = 1

hence total odd days :0+0+6+1=7 hence the day on 28.05.2006 will be sunday

Answer 20

Y'know, you still have to use technology in between.

Answer 21

oh alrite! But how come we know that no of odd days in 1600 and 400 is 0??

cuz..Odd days of a year X = Remainder of [No. of days in the year X / 7].

Answer 22

365 mod 7 = 1

Answer 23

> cal 8 2011
August 2011
Su Mo Tu We Th Fr Sa
1 2 3 4 5 6
7 8 9 10 11 12 13
14 15 16 17 18 19 20
21 22 23 24 25 26 27
28 29 30 31

> cal 8 2010
August 2010
Su Mo Tu We Th Fr Sa
1 2 3 4 5 6 7
8 9 10 11 12 13 14
15 16 17 18 19 20 21
22 23 24 25 26 27 28
29 30 31

Answer 24

Oh I see you got there already @mathslover

Answer 25

wait @shrey

Answer 26

@mathslover okies

Answer 27

100 years = 76 ordinary years + 24 leap years

= (76 x 1 + 24 x 2) odd days = 124 odd days.

= (17 weeks + days) 5 odd days.

Number of odd days in 100 years = 5.

Number of odd days in 200 years = (5 x 2) 3 odd days.

Number of odd days in 300 years = (5 x 3) 1 odd day.

Number of odd days in 400 years = (5 x 4 + 1) 0 odd day.

Answer 28

Similarly, each one of 800 years, 1200 years, 1600 years, 2000 years etc. has 0 odd days.

Answer 29

this is the concept behind that:

21/7 = 3 ( 0 => remainder)

hence 0 odd days in 400 and 1600

Answer 30

@shrey got it?

Answer 31

Methodology:

And you need a reference point. You could use

> cal 1 1
January 1
Su Mo Tu We Th Fr Sa
1
2 3 4 5 6 7 8
9 10 11 12 13 14 15
16 17 18 19 20 21 22
23 24 25 26 27 28 29
30 31

Not formatted well, but Jan 1, 1 was a Saturday.

But you still have to worry about which years have an extra day. And you have to worry about the Julian / Gregorian calendar. You have to worry about where you are, because different countries switched in different centuries. And you have to worry about stuff like this:

> cal 9 1752
September 1752
Su Mo Tu We Th Fr Sa
1 2 14 15 16
17 18 19 20 21 22 23
24 25 26 27 28 29 30

Answer 32

Methodology, advance 1 day for same date every year.

Answer 33

gng thru it... tanx a ton :)

Answer 34

ur welcome @shrey