XL: Method to Determine Whether a Year Is a Leap YearID: Q118923
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This article contains information that explains how to determine whether the year in a date used in a Microsoft Excel document is a leap year.
The date system used by Microsoft Excel is based on the Gregorian
calendar, first established in 1582 by Pope Gregory XIII. The Gregorian
calendar was designed to correct the errors introduced by the less
accurate Julian calendar.
In the Gregorian calendar, a normal year consists of 365 days. Because the
actual length of a sidereal year (the time required for the Earth to
revolve once about the Sun) is actually 365.25635 days, a "leap year" of
366 days is used once every four years to eliminate the error caused by
three normal (but short) years. Any year that is evenly divisible by 4 is
a leap year: for example, 1988, 1992, and 1996 are leap years.
However, there is still a small error that must be accounted for. To
eliminate this error, the Gregorian calendar stipulates that a year that
is evenly divisible by 100 (for example, 1900) is a leap year only if
it is also evenly divisible by 400.
For this reason, the following years ARE NOT leap years
1700, 1800, 1900, 2100, 2200, 2300, 2500, 2600
1600, 2000, 2400
=IF(OR(MOD(A1,400)=0,AND(MOD(A1,4)=0,MOD(A1,100)<>0)),"Leap Year",
"NOT a Leap Year")
If the value in cell A1 is this The formula returns
-----------------------------------------------------
1992 Leap Year
2000 Leap Year
1900 NOT a Leap Year
For more information about how Microsoft products are affected by year
2000 (Y2K) issues, please see the following Microsoft World Wide Web site:
http://www.microsoft.com/technet/topics/year2k/default.htm
Additional query words: 2.00 2.20 2.21 4.00a 5.00c 7.00a 97 8.0 98 XL98 XL97 XL7 XL5 XL4 y2k year2000 year 2000
Keywords : kb2000 xlformula
Version : MACINTOSH:2.2,3.0,4.0,5.0,98; WINDOWS:2.1,3.0,4.0,4.0a,5.0,5.0c,7.0,7.0a,97; winnt:5.0
Platform : MACINTOSH WINDOWS winnt
Issue type :
Last Reviewed: May 25, 1999