IIT JEE , AIEEE Forum - Mathematics , Algebra

leo

How does one find the day of the year when the date is given?

Is there a shortcut method that helps us to arrive at the answer quickly?

10904him

You must first know the first day of the year.
Find the total no of days to the date from 1st Jan .
Find the reminder after dividing by seven.Seek the day at the distance as the reminder and you get the day.

chimanshu_007

hey man its behind my long book.......................................
tell me i will tell u.........

leo

10904him ,
is there no formula to arrive at the answer?

for eg:- if the qn says that 1895 Jan 2 was a sunday and we have to find the day on 1986 april 9 what do we do?

vasanth

i'm givin it a try

devised ma own formula

{365x + [x/4] + 30y + {[y/2]no. of days} + z}/7 = day

where x- number of years completed

y - number of months completed(the days of the final month can b included in z)

z - number of days

[x] - denotes greatest integer function

[y/2] - denotes the least integer function on y/2

after dividing by 7, if u get the day as a sunday , if u get a remainder of 2

then the day is tuesday

if [x/4] is 0, and there is a leap year between the given dates, add one

similarly [y/2] is also prone to change depending on the months between(because here is confusion when july and august r included)

this is purely first time attempt

simpler solutions are more welcome

plzzzzzzz rate if rite

cheeeeeeers

vasanth

let me compute the answe for the Q given by leo

x = 1986 - 1895 = 91

y = 4-1 = 3

number of days to b added for the leap year factor

[x/4] = [91/4] = 22

as we have jan and mar havin 31 days each

[y/2] = [3/2] = 2------------------as it denotes the leasr integer function

z = 9-2 = 7

now apply

{365(91) +22 + 30(3) + 2 +7}/7 = (33125+ 22+ 90 + 9)/7

= 33246/7

quotient = 4749

remainder = 3/7

but here feb has only 28 days and not 30 days so we subtract 2

so quotient = 4749

remainder = 1/7

so day is sunday + 1 = monday

so april 9 1986 was a monday

u can check it with a calendar

if u want another method, u can use number of weeks in ur formula

i hav used number of days and divided it by 7

plzzzzzzzz rate ma hardwork

cheeeeeeeers

vasanth

the number of months with 31 days r prone 2 change

bcoz of the occurence of july nd aug continuously but it affects much only when july and aug are taken as the startin pt and endin pt. resp

cammon guyzzz show me some gratitude

cheeeeeeeeers

vinod

Hey vasanth u r a genious!!!!!!!!!

good work...wonderfulll....:):)

visioniit

vasanth i have not understood ur solution.pls explain.

vasanth

alryt

x denotes the "number of years"

we hav 365 days in 1 normal year

so we have 365 * x number of days in x years

but we need to take in account the leap years in between which recur once every 4 yrs

as it recurs once every 4 yrs.------the number of "4yrs in x yrs" is x/4

but we do not take into account the remainder...........

let A denote the number of leap years

x/4 ; x=4n where n is a +ve integer

A

{(x/4) - remainder}; x

4n as we take into account only the number of leap years in between

for example if there are 90 years; then we have 90/4 = 22 +2/4 but we hav only 22 years that r leap years in these 90 years, so we exclude the remainder 2/4

observe that we r taking only the greatest integer LESS THAN OR EQUAL TO the QUOTIENT

generalising the value of A for all x , whether x = 4n or not

we have A = [x/4] ; x= or not equal to 4n ; [.] denotes greatest integer function

NOTE: we are adding [x/4] the number of leap years directly......we hav already included the 365 days in all these years given...... we hav not added the extra days provided by the leap years......

as there are only an additional [x/4] number of days for [x/4] leap years we add it directly

i'll post the other explanations in the next post

hope u rate ma patience

cheeeeeeers

vasanth

similarly comin to months

y - denotes the number of months

there r 30 days in a normal month

so there r 30*y number of days in 'y' number of normal months

but we need to take into account the number of months woth 31 days that occur twice in 3 months ( with the exception of august and july)

so the number of months( denoted by B) with 31 days that occur in that "y-month space" r

B = y/2; y= 2k where k is an integer

= (y/2 - remainder)+1 ; 2

2k ; where k is an integer

PS- dont take into account the month space that includes both july and august simultaneously

for instance

if we r computing from january 4 to april 7; y= 3 months completed

so y/2 = 1+ 1/2

but actually we hav 2 months with 31 days in between; so we cannot exclude the remainder but have to round it off to the nearest greatest integer

that is y/2 - remainder +1 = 3/2 - 1/2(remainder) +1 = 2

we hav to include the disturbaces caused by february also

feb might have 28 or 29 days.......as we hav added 30 for feb...we hav to subtract 2 or 1 resp as the case may be......(only for the feb's not included in our leap yr calculations)

by observation we find that

B is given by the least integer greater than or equal to the y/2

so generalising B = [y/2]

where [.] denotes the least integer function here

as we hav included all 30 days even for the months having 31 days; only those extra days are left out------------------------- as we hav to add only 1 day for each 31 day month--------------we add [y/2] number of days for [y/2] number of 31-day months

vasanth

now all that is left is computing the number of days

from april 4 to may17

if z denotes the number of days left then

z = 17-4 = 13

as from april 4 to may 4 we hav included the number of days in y and [y/2] wat is left is 17-4 = 13

so we add this also

now our equation gives the total number of days in the given interval

As the days( sunday or mon or tue or.....) recur once every week (7 days)

we divide the eq by 7 to get the number of weeks in between.......

th quotient gives the number of weeks; and the remainder gives the number of extra days..........

the reminder after dividing by seven is found.the day at the distance as the reminder is found and you get the day.

if we get the remainder as 4

the starting pt as a sunday

then we have the 4th day from sunday as thursday

hope u got it....

plzzzzz rate me

cheeeeeeers

P.S : in the previous post; if july and august r included , then they don not give much difference when taken over a long period......... but when taken alone or with one other month ; correction must b included which can b included to the formula manually

plzzzzzzzz rate me

cheeeeeeeeers

vasanth

hey guyzz

it took me around 30 mins and 3 rounds of scoldings from ma parents to type this; so hopefully u'll rate me........

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