When we move down in yearwise a day decreases. but when a leap year comes 2 days gets decreased..for e.g.12 jan 2008 --- saturday12 jan 2007 --- friday12 jan 2006 --- thursday12 jan 2005 --- wednesday12 jan 2004 --- monday..so in 4 years five days gets reduced... so in this manner we can find out in which year wat day was on 12 jan..ANSWER ----- it was friday on 12 jan 1979......

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Friday

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12th Jan 19791979 = 1900yrs + 78yrs + (period from 1st Jan to 12th Jan 1979) 1900 yrs have 1 odd day(calculated as 1600yrs have 0 odd days, n 300yrs have 15 odd days i.e.15/7= 2 weeks+1 odd day.so,for 300yrs, it's 1 odd day.totally it's 0+1=1odd day for 1900yrs).78yrs = 19leap yrs +59 ordinary yrs = 19 *2 +59 *1 = 97days = 13weeks +6 odd days = 6 odd daysperiod from 1st Jan to 12th Jan 1979 =12 days = 1week + 5odd days = 5 odd days Therefore,12th Jan 1979 = 1+6+5 = 12days =1 week +5 odd days =5 odd daysTherefore, 12th Jan 1979 was a Friday.(Since,Sunday->0,Monday->1,...,Fri->5,sat->6)

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Today is 16/07/2009 - thursday.So, 12.07.2009 -> Sunday.From jan - june, calculate the total no of odd days in each monthJan = 3(31 days... 31%7=3)Feb = 0Mar = 3Apr = 2May = 3Jun = 2Total comes to 13. so, 13%7 is 6 or -1. Since we are back calculating, take -1 as 1, so proceed one day forward from 12.07.2009 ie Sunady.So, 12.01.2009 is Monday.Calculate the total no of odd days in each year, from 1980 to 2009.1 for normal yrs, and 2 for leap yrs.There are 30 years from 1980 to 2009, out of which there are 8 leap yrs.so, 1x22 + 2x8 = 38. now, 38%7 = +3.Take -3 since we are back calculating.go 3 days back from Monday(12.01.2009). its FRIDAY !!!

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12.01.19791600 - 01900 - 1519 leap year = 3859 ordinary yr = 5912total 0+15+38+59+12 = 124/7 = bal 5 so that day is friday

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Its formula to find the day of any date.K+[(13M-1)/5)] + [D/4] + D + (C/4) - 2CK- DATEM-MONTHD-LAST TWO DIGITS IN YEARC-FIRST TWO DIGITS IN YEARFor months:March is considered as first month.April as second month....January as 11 th month.(so decrement year by 1)Febraury as 12th month.(so decrement year by 1)Substitute all those details in formuladivide the answer by 7.for remainder 0-sunday 1-mondayand soon.

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Let me say how to use formula: K+[(13M-1)/5)] + [D/4] + D + (C/4) - 2C K- DATE M-MONTH D-LAST TWO DIGITS IN YEAR C-FIRST TWO DIGITS IN YEAR for months: March is considered as first month. April as second month. . . . January as 11th month (so decrement year by 1) February as 12th month (so decrement year by 1) substitute all those details in formula divide the answer by 7. (no need to consider the numbers after decimal point) for remainder 0-sunday 1-monday and soon. Solution: Given k=12 m=11 (since month is January) c=19 d=78 (since month is January, year is decremented by 1) Substituting in formula: 12 +[{(13*11) -1}/5] + [78/4] + 78 + [19/4] - 2(19) =12+ 28 + 19 +78 +4 - 38 = 103 (discard numbers after the decimal point) Divide answer by 7 Remainder is 5 So solution is Friday

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Jan 12,1979- lets determine the no. of odd days upto year 1978-(1600+300+78)1600 is divisible by 400,therefore the no.of odd days is 0...300 yrs has 1 odd day78yrs consist of (19 leap + 59 ordinary years)=(19x2+59x1)=97 odd days where 2 and 1 are no.of odd days for leap and odd yrs respectively.(97%7=13 ,6 odd days)therefore,jan 12,1979 falls on saturday!!!

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Jan 12,1979- lets determine the no. of odd days upto year 1978-(1600+300+78)1600 is divisible by 400,therefore the no.of odd days is 0...300 yrs has 1 odd day78yrs consist of (19 leap + 59 ordinary years)=(19x2+59x1)=96 odd days where 2 and 1 are no.of odd days for leap and odd yrs respectively.(96%7=13 ,6 odd days)6+12+1=19(18/7 5 odd days)therefore it is friday!!!

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12-01-19791900 - 1 odd day78 - 19 leap years = 19*2 =38 59 ordinary years = 59*1 = 59 -------- 97 = 6 odd days (97/7 remainder =6)date =12/7 = 5 odd days total odd days = 1 + 6 + 5 = 12 odd daysi.e 12/7 = 5 odd days5 means Friday (0- sun, 1- mon, 2- tue, 3- wed,.............)ANS : Friday

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