What was the day on 15th August 1947???? Plzz tell me the solution..?
456991
Okay, this is not going to be pretty, but it works, but only for years from A.D. 1582 and later.\n\nFirst, let M equal the number corresponding to the chosen month according to the following list\n\nJan=11\nFeb=12\nMar=1\nApr=2\nMay=3\nJune=4\nJuly=5\nAug=6\nSep=7\nOct=8\nNov=9\nDec=10\n\nNext, let N equal the day of the month for which you are inquiring.  So if you want to look for August 15, M would be 6 and N would be 15 (M=6, N=15).\n\nNow, take the last two digits from the year you are searching for, call that number Y.  For 1947, Y=47.\n\nThe leftover digits of the year make a new number, call it C.  For 1947, after removing 47 we see that C=19.\n\nFinally, if the year is a leap year, let L=1, otherwise, let L=0.\n\nNow we're ready to get down to business.  The following formula makes use of the greatest integer function (ipart) which essentially means that if you get some number with a decimal, chop off the decimal.  3698.2461 becomes 3698.  OK?  Good.\n\nNow we must calculate the following formula:\nN + ipart(2.6M - 0.2) + Y + ipart(Y/4) + ipart(C/4) - 2C - (1 + L)*ipart(M/11)\n\nAfter calculating this number, divide it by seven and look at the remainder.  The remainder corresponds to the day of the week as follows:\n\n0=Sunday\n1=Monday\n2=Tuesday\n3=Wednesday\n4=Thursday\n5=Friday\n6=Saturday\n\nLet's test this with your August 15, 1947.\n\nSo, M=6, N=15, Y=47, C=19, and since 1947 was not a leap year, L=0.\n\nPlugging these values into the formula we get:\n15 + ipart(2.6*6 - 0.2) + 47 + ipart(47/4) + ipart(19/4) - 2*19 - (1 + 0)*ipart(6/11)\n\n=15 + ipart(15.4) + 47 + ipart(11.75) + ipart(4.75) - 38 - 1*ipart(.5454545454)\n\n=15 + 15 + 47 + 11 + 4 - 38 - 0\n\n=54\n\nDividing this by 7, we get a remainder of 5.\n\nThis says that August 15, 1947, was a Friday.
422250
Here are the things that happened on November 15,2000:\n\nhttp://query.nytimes.com/search/query?query=november+15%2C+2000&n=10&dp=0&sort=newest&daterange=full&d=nytdsection%2b&o=e%2b&v=World%2b&c=a%2b
562710
July 1884\n\nI hope this helps ... the link below explains more and is very informative as well.\n\n- Alan
768446
0 degrees fahrenheit ( -18 degrees celsius) on January 15, 1950 and December 29, 1968. After researching this for you, I DEMAND 10 points ! You are welcome.
732347
uhhh.......January 27th
58036
march 2-sst\nmarch 9-maths\nmarch 20-sc\nmarch 24-hindi\nmarch 27-eng
357761
http://en.wikipedia.org/wiki/1991#Events
286427
All Of You Are Idiots... How Do We Know that one Day=24 Hours??? Did They Wear Watches? Was One Year 365 Days?? Did They Have Calendars? For All We Know One Day In The Begining COuld Be Equal To 7 Days In What We Know ....Geesh Some People
171035
Tuesday\nyou can use your computer clock to find out by changing the year
1132704
It was the city of Bhuj in Gujarat and the date was 26  January 2001 when India was celebrating its Republic Day.\n\nCheck it out for yourself at the following site:
94831
Avtomat Kalashnikova 1947
747249
You already know its Tuesday from the other answers.\n\nHere's what I'll add. You can look up AND day with this virtual calendar:
730188
Im from England... HI there. lol\n\nIf someone said next thursday i would think they ment the 11th as the 4th would be this thursday.... but i would always ask to make sure.
399849
Noon. Thanks for the points!!
498462
570AD\n\nHowever, the years don't line up to present since the Islamic Calender only has 354 days (it's lunar, not solar)
514923
i think may 26\nfor delhi region\nat 18:00 hrs\ni saw it on a website..\ni am so sorry i dont remember the name...
48893
August 2, 1776 (not July 4).