Obtain days in a given year
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From: Brad on 14 May 2010 13:19 You're welcome Another approach that would also work =365+(MONTH(DATE(A2,2,29))=2) Success click yes -- Wag more, bark less "igbert" wrote: > Thanks, formula works preferct for any year. > > > "Brad" wrote: > > > assume 2008 is in a2 > > in b2 > > =DATE(A2+1,1,0)-DATE(A2,1,0) > > > > Success - click yes. > > -- > > Wag more, bark less > > > > > > "igbert" wrote: > > > > > Is there a fuction to return the days in a given year? > > > > > > Entry Return > > > > > > 2008 366 > > > 2010 365 > > > > > > Igbert > > >
From: Mike H on 14 May 2010 13:58 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 > >
From: Bernd P on 14 May 2010 17:18 On 14 Mai, 18:19, Brad <B...(a)discussions.microsoft.com> wrote: > You're welcome > > Another approach that would also work > > =365+(MONTH(DATE(A2,2,29))=2) > > Success click yes > -- > Wag more, bark less > > "igbert" wrote: > > Thanks, formula works preferct for any year. > > > "Brad" wrote: > > > > assume 2008 is in a2 > > > in b2 > > > =DATE(A2+1,1,0)-DATE(A2,1,0) > > > > Success - click yes. > > > -- > > > Wag more, bark less > > > > "igbert" wrote: > > > > > Is there a fuction to return the days in a given year? > > > > > Entry Return > > > > > 2008 366 > > > > 2010 365 > > > > > Igbert =337+DAY(DATE(A1,3,0)) Regards, Bernd
From: Mike H on 14 May 2010 20:16 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 > > > >
From: Mike H on 14 May 2010 20:23 Mickey, For clarification:- A leap year is every 4 years, but not every 100 years, then again every 400 years -- Mike When competing hypotheses are otherwise equal, adopt the hypothesis that introduces the fewest assumptions while still sufficiently answering the question. "Mike H" wrote: > 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 > > > > >
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