Aptitude - Calendar - Discussion

Dont know how 1600 and 400 have been accounted.
Number of leap yrs in 2000 yrs= 2000/4-15=485 yrs
(4 is included because leap year comes generally once in 4 yrs.
15 is subtracted because some of yrs namely 100,200,300,500,600 700,900,1000,1100,1300,1400,1500,1700,1800,1900 are not leap yrs in 2000 yrs)
Number of odd days in leap yrs= Remainder of 485*2/7 = 4 days
Number of non leap yrs in 2000 yrs= 2000-485 = 1515 yrs
Number of odd days in non leap yrs= Remainder of 1515/7 = 3 Days
Therefore number of odd days in 2000 yrs= Remainder of(4+3)/7= 0
And for next five yrs from 2001 to 2005 and onwards it is been described in solution.

Because in 100 years there are (96/4) =24 leap years and other72 years are ordinary days and as.

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 in 200 years = (5 x 2) = 10 -> 3 odd days.
Number of odd days in 300 years = (5 x 3) = 15 -> 1 odd days.
Number of odd days in 400 years = (5 x 4 + 1) = 21 -> 0 odd days.
Similarly, the number of odd days in all 4th centuries (400, 800, 1200 etc. ) = 0.

That is why we take 1600 & 400 instead of 1200, 800 etc. Numbers.

Jan=1.
Feb=4.
Mar=4.
Apr=0.
May=2.
Jun=5.
Jul=0.
Aug=3.
Sep=6.
Oct=1.
Nov=4.
Dec=6.

Days:

S=1.
M=2.
T=3.
W=4.
Th=5.
F=6.
Sat=7 or 0.

Day = Given date+Month no.+(x-1900)+(x-1900)/4.

X = The year we have to calculate.

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Important notice for all who follow this formula.

First you should check whether the given year is a leap year or not.

If it is not a leap year our formula is perfect holds for every thing.

But if it is a leap year you should subtract 1 from the obtained answer otherwise you don't get correct answer.

" if you remember this points you can solve the answer in 20 seconds only "

Month Codes:
(J F M A M J J A S O N D)
(0 3 3 6 1 4 6 2 5 0 3 5)
Year codes:

1600-1699 =>6
1700-1799 =>4
1800-1899 =>2
1900-1999 =>0 (repetition)
--------------------------------------------------
2000-2099 =>6
2100-2199 =>4
2200-2299 =>2
2300-2399 =>0

Week day=date+year(last two digits)+leap years+month code+year code.

What was the day of the week on 28th May, 2006?
28+06+01+1+6=46,
week day=46/7=0,
0 means Sunday.

@Jyoti:

Remember 1 Jan 2018 As a Monday.
Now if a year increase or decrease, i.e- 1jan 2019 will Tuesday and 1jan 2017 was Sunday...
*Remember to increase or decrease 2days in leap year.

Now for 1 April 2001:-
As 1 Jan 2018 was Monday.
2018-2001 =17 yr in which 2004,2008,2012,2016 were comes as a leap yr. So totally we have to decrease my 21 ,i.e 21/7 completely divisible. So 1 Jan 2001 is also Monday.

As 1 jan 2001 as a Monday (30days in jan + 28 days in feb + 31days in march + 1Day in April)= 90,

90/7=6 as a remainder so,
1April 2001 is Sunday ..( Mon +6days).

A year in is divisible by 4 and for century years it should be divisible by 400, not 100 then it is said to be a leap year.

For example, 100 is not a leap year even though it is divisible by 4 but as it is a century year it should be divisible by 400.

* Odd days need to be divided the number of days by week -for the normal year 365 days (365/7) reminder — 1 is odd days.
* Similarly for leap years -366 days (366/7) reminder- 2 is odd days.

For 100 years (76 normal years (1 odd day)+ 24 leap years (2 odd days).
76x1 + 24x2= 124 days.
124 days/7 days = 5 odd days.

Year Odd days.
1600 0
1700 5
1900 3
2000 5
2005 6
2006 3

Total 22 = Sunday

Explanation:

1700-1600=100 in 100 yrs 5 odd days.
1900-1700=200 in 200 yrs 3 odd days.
2000-1900=100 in 100 yrs 5 odd days.

Leap year calculation 2000-2005.
(1+1+1+2+1) 2000 non leap yr so add one 2004 leap yr so I have added 2.

2006 odd day calculation.

Jan 3 odd day(31/7= 28 rem 3,so take that three).
Feb 1 odd day(since leap year)(if non leap yr put 0).
March 3 odd day.
Apr 2 odd day.
May 0(since leap yr).

Tot 10(10/7=1 rem 3).

So 3 odd days in 2006.

100 years have 5 odd days,
100 years have 3 odd days,
100 years have 1 odd days,
100 years have 0 odd days,

1 normal year have 1 odd day,

1 leap year have 2 odd days,

On date 1 January 0001 there was a Monday,

One more thing the year containing 00 as a last 2 digits and is not divisible by 400 then it is nota leap year, even though it is divisible by 4.

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I think these points mite help you solving calendar problems.

Lets see that 2000 was a leap year so 1, 2, 3, 5 is non leap year and 2004 is leap year. So counting from 2001, to 5 there will be 6 odd day 1 for each non leap and 2 for 1 leap year next see in 2006 January has 31 day means 3 odd day (31/7), February has 0 (28/7), March has 3, April has 2, and for 28 in May 0 odd days.

So means total 8 days in 2006 means 1 week 1 day hence we total left of 1 odd day. From 2001 to 5 we have 6 odd and 1 in 2006 total 7. So 1 week again so starting day Sunday is answer.

Hi there is a formula for this type of problem.

Plz memorize this

jan=1
feb=4
mar=4
apr=0
may=2
jun=5
jul=0
aug=3
sep=6
oct=1
nov=4
dec=6

days:
s=1
m=2
t=3
w=4
th=5
f=6
sat=7 or 0

formula=day+month no.+(x-1900)+(x-1900)/4
x= the year we have to calculate

Eg:
28-may-2006

for=28+2+(2006-1900)+(2006-1900)/4
=28+2+16+16/4
=28+2+16+4 note:
=50 (here we should not take the no
behind decimal i.e 17/4=4.25)
50/7=1(reminder)

sun =1