Formula for the First Day of a YearDate: 03/18/98 at 16:13:36 From: Josh Leonard Subject: Finding first day of a given year. Is there an equation to find the first day of a year if you are given the year? Thanks, Josh Date: 03/19/98 at 11:51:11 From: Doctor Bill Subject: Re: Finding first day of a given year. Josh, Here is a formula for finding the day of the week for ANY date. N = d + 2m + [3(m+1)/5] + y + [y/4] - [y/100] + [y/400] + 2 where d is the number or the day of the month, m is the number of the month, and y is the year. The brackets around the divisions mean to drop the remainder and just use the integer part that you get. Also, a VERY IMPORTANT RULE is the number to use for the months for January and February. The numbers of these months are 13 and 14 of the PREVIOUS YEAR. This means that to find the day of the week of New Year's Day this year, 1/1/98, you must use the date 13/1/97. (It sounds complicated, but I will do a couple of examples for you.) After you find the number N, divide it by 7, and the REMAINDER of that division tells you the day of the week; 1 = Sunday, 2 = Monday, 3 = Tuesday, etc; BUT, if the remainder is 0, then the day is Saturday, that is: 0 = Saturday. As an example, let's check it out on today's date, 3/18/98. Plugging the numbers into the formula, we get; N = 18 + 2(3) + [3(3+1)/5] + 1998 + [1998/4] - [1998/100] + [1998/400] + 2 So doing the calculations, (remember to drop the remainder for the divisions that are in the brackets) we get; N = 18 + 6 + 2 + 1998 + 499 - 19 + 4 + 2 = 2510 Now divide 1510 by 7 and you will get 358 with a remainder of 4. Since 4 corresponds to Wednesday, then today must be Wednesday. You asked about New Year's Day, so let's look at this year, 1/1/98. Because of the "Very Important Rule," we must use the "date" 13/1/97 to find New Year's Day this year. Plugging into the formula, we get; N = 1 + 2(13) + [3(13+1)/5] + 1997 + [1997/4] - [1997/100] + [1997/400] + 2 N = 1+ 26 + 8 + 1997 + 499 - 19 + 4 + 2 = 2518 Now divide 2518 by 7 and look at the remainder: 2518/7 = 359 with a remainder of 5. Since 5 corresponds to Thursday, New Year's Day this year was on a Thursday. This is a very enjoyable formula, I hope you have fun with it. -Doctor Bill, The Math Forum Check out our web site! http://mathforum.org/dr.math/ Date: 08/26/2002 at 16:15:25 From: Jim Colwell Subject: Zeller's congruence Doctor Peterson, In your FAQ, you give the following formula for finding the day of the week for a given date: f = k + [(13*m-1)/5] + D + [D/4] + [C/4] - 2*C But in your archive, you give a different formula: N = d + 2m + [3(m+1)/5] + y + [y/4] - [y/100] + [y/400] + 2 Obviously these two formulas are equivalent, but I can't see why. Why the 2m? Why the +2? Could you be so kind as to explain this equivalence? I guess I'm trying to understand why the formula works; what is the underlying idea. The 2m and the +2 throw me. Thanks. Jim Date: 08/26/2002 at 22:33:56 From: Doctor Peterson Subject: Re: Zeller's congruence Hi, Jim. The first thing to do is to note the differences: In the FAQ, f = k + [(13*m-1)/5] + D + [D/4] + [C/4] - 2*C "m" is the month number starting with March as 1; "k" is the day of the month; the year is given by C, the century, and D, the last two digits; in the result, 0 means Sunday. In the alternative formula, N = d + 2m + [3(m+1)/5] + y + [y/4] - [y/100] + [y/400] + 2 "m" is the month number starting with January as 1; "y" is the year; in the result 0 means Saturday. So if we try to write the latter using the variables of the former, we will have to replace d with k m with m+2 (so March is 3) y with 100C+D and subtract 1 from N to get f (so Sunday is 0). We get f = N-1 = k + 2(m+2) + [3((m+2)+1)/5] + (100C+D) + [(100C+D)/4] - [(100C+D)/100] + [(100C+D)/400] + 2 - 1 = k + 2m + 4 + [3(m+3)/5] + 100C + D + [25C + D/4] - [C + D/100] + [C/4 + D/400] + 1 = k + 2m + 4 + [3m/5 + 9/5] + 100C + D + 25C + [D/4] - C - 0 + [C/4] + 0 + 1 = k + [10m/5 + 4 + 3m/5 + 9/5 + 1] + 124C + D + [D/4] + [C/4] = k + [(13m + 4)/5 + 6] + D + [D/4] + [C/4] + 124C = k + [(13m-1)/5 + 7] + D + [D/4] + [C/4] + 124C This isn't quite right, is it? Unless 7 + 124C = -2C, it can't be the same. Ah! I realized what we're missing when I tried checking this by doing each calculation for a specific date: the answer is not this whole expression, but the remainder when we divide by 7. The two forms will be the same if 7 + 124C = -2C (mod 7). And since 124 = 7*17 + 5 = 7*18 - 2, it is in fact true that the remainder of the left side, 7*(1+18C) - 2C, will be the same as the remainder of -2C. The two formulas are indeed equivalent. Let me know if you need anything more. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ |
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