From: b1llt on 26 Mar 2010 17:26 I should have mentioned we're still on Excel 2000. I used these and am getting a #DIV/0! result. Any suggestions thanks, Bill "Ziggy" wrote: > Another 2003 solution is an array formula > > =SUM(($K$15:$K$38=K41)*($M$15:$M$38>=1)) > > Set with CTRLShiftEnter > . >
From: "David Biddulph" groups [at] on 26 Mar 2010 22:18 There are no divide operations in that formula, so if you're seeing a #DIV/0! result it's because you've got a #DIV/0! in the data being used by the formula. Tackle the problem where it's being generated.  David Biddulph "b1llt" <b1llt(a)discussions.microsoft.com> wrote in message news:AF1034DE30024132A3CEBEC4F0199798(a)microsoft.com... > I should have mentioned we're still on Excel 2000. > I used these and am getting a #DIV/0! result. > Any suggestions thanks, > Bill > > "Ziggy" wrote: > >> Another 2003 solution is an array formula >> >> =SUM(($K$15:$K$38=K41)*($M$15:$M$38>=1)) >> >> Set with CTRLShiftEnter >> . >>
From: b1llt on 29 Mar 2010 13:52 My bad! Your correct I did have it pulling a #DIV/O! into the data by mistake. These all work great! Thanks everyone for all your help. B1llt "David Biddulph" wrote: > There are no divide operations in that formula, so if you're seeing a > #DIV/0! result it's because you've got a #DIV/0! in the data being used by > the formula. Tackle the problem where it's being generated. >  > David Biddulph > > > "b1llt" <b1llt(a)discussions.microsoft.com> wrote in message > news:AF1034DE30024132A3CEBEC4F0199798(a)microsoft.com... > > I should have mentioned we're still on Excel 2000. > > I used these and am getting a #DIV/0! result. > > Any suggestions thanks, > > Bill > > > > "Ziggy" wrote: > > > >> Another 2003 solution is an array formula > >> > >> =SUM(($K$15:$K$38=K41)*($M$15:$M$38>=1)) > >> > >> Set with CTRLShiftEnter > >> . > >> > > . >
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