Mickey,

Would you be so kind and check your suggested formula for the years:
1908, 2008, 2108 and 2200(!)

Of course


1908 = leap year
2008 = leap year
2108 - leap year
2200 - Not a leap year


--
Mike

When competing hypotheses are otherwise equal, adopt the hypothesis that
introduces the fewest assumptions while still sufficiently answering the
question.


"מיכאל (מיקי) אבידן" wrote:

Mike,
Would you be so kind and check your suggested formula for the years:
1908, 2008, 2108 and 2200(!)
Micky


"Mike H" wrote:

Micky,

I like the second formula, very neat, but I'm afraid the first gives errors,
you would need to do this to get the correct result using MOD

=IF(OR(MOD(A1,400)=0,AND(MOD(A1,4)=0,MOD(A1,100)< 0)),366, 365)
--
Mike

When competing hypotheses are otherwise equal, adopt the hypothesis that
introduces the fewest assumptions while still sufficiently answering the
question.


"מיכאל (מיקי) אבידן" wrote:

1) =IF(MOD(A1,4)=0,366,365)
2) =IF(ISNUMBER(--(A1&"/2/29")),366,365)
Micky


"igbert" wrote:

Is there a fuction to return the days in a given year?

Entry Return

2008 366
2010 365

Igbert