Path: utzoo!utgpu!water!watmath!clyde!rutgers!husc6!hao!oddjob!gargoyle!ihnp4!occrsh!occrsh.ATT.COM!fubar
From: fubar@squid.UUCP
Newsgroups: comp.sys.ibm.pc
Subject: day of week
Message-ID: <141900013@occrsh.ATT.COM>
Date: 28 Jan 88 11:50:00 GMT
Lines: 48
Nf-ID: #N:occrsh.ATT.COM:141900013:000:1538
Nf-From: squid.UUCP!fubar    Jan 28 05:50:00 1988


From squid!fubar Thu Jan 28 04:22 CDT 1988 remote from occrsh
Subject: day of week

***
*** PLEASE send replies to ihnp4!occrsh!squid!david
***

>   Here's a formula given in the book An Introduction to the Theory of Numbers,
>by Niven and Zuckerman.
>
>   Let M be the number of the month, defined so that Jan = 11, Feb = 12,
>Mar = 1, Apr = 2, ..., Dec = 10.  Let N be the number of the day within the
>month.  Let C be the hundreds in the year, and let Y be the rest of the year.
>Let L be 1 for a leap year and 0 for a non-leap year.
>
>The formula
>
>(N + [2.6M - 0.2] + Y + [Y / 4] + [C / 4] - 2C - (1 + L)[M / 11]) mod 7
>yields 0 for Sunday, 1 for Monday, ..., 6 for Saturday.
>
>(Square brackets denote the greatest integer function.)

So:

Thursday, Jan 28, 1988, should yield 4:

 N = 28   (number of day within month)
 M = 11   (Jan = 11)
 C = 19   (Hundreds w/in year)
 Y = 88   (rest of year??)
 L = 1    (1988 is leap year)

 = (N  +  [2.6M - 0.2]  +  Y +  [Y/4] +  [C/4] -   2C   -  (1+L)[M/11])  mod 7
 = (28 + [(2.6*11)-0.2] + 88 + [88/4] + [19/4] - (2*19) - (1+1)*[11/11]) mod 7
 = (28 +     [28.4]     + 88 +  [22]  + [4.75] -   38   -      2)        mod 7
 = (28 +       29       + 88 +   22   +   5    -   38   -      2)        mod 7
 = 132 mod 7
 = 6 ??

What is the correct formula for this?

David Drexler
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